Question: As shown in the figure, it is known that $AB$ is the diameter of the circle $\odot O$, $\angle ACD$ is the angle subtended by arc $AD$ on the circumference of the circle, and $\angle ACD=30^\circ$. A line is drawn through point $D$ such that $DE\perp AB$ with $E$ being the foot of the perpendicular, and the extension of $DE$ intersects $\odot O$ at point $F$. If $AB=4$, what is the length of $DF$?

Solution:
\textbf{Solution:} Extend line OD, \\
$\because \angle DAB=60^\circ$ and OD=OA=2,\\
$\therefore \triangle AOD$ is an equilateral triangle,\\
$\therefore AD=2$,\\
$\because DE \perp AB$,\\
$\therefore \angle ADE=30^\circ$ and DF=2DE,\\
$\therefore AE=\frac{1}{2}AD=1$,\\
In $\triangle ADE$, $DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{3}$,\\
$\therefore DF=2\sqrt{3}=\boxed{2\sqrt{3}}$.


Question: As shown in the figure, within circle $O$, the diameter $AB \perp$ chord $CD$ at point $M$, and point $E$ is the midpoint of arc $AD$. Connect $AD$ and $CE$, and extend $AE$ and $CD$ to meet at point $F$. It is given that $\angle AEC = 2\angle C$. When $AM=4$ and $CD=6$, what is the length of $AF$?

Solution:
\textbf{Solution:} Given the problem statement,\\
$\because \angle AEC = 2\angle C$ and $\angle AEC = \angle C + \angle F$,\\
$\therefore \angle C = \angle F$,\\
Also, $\because \angle DAE = \angle C$,\\
$\therefore \angle DAE = \angle F$,\\
$\therefore AD=DF$,\\
$\because AM=4$, $CD=6$, and diameter $AB\perp$ chord CD,\\
$\therefore AD=DF=5$,\\
$\therefore AF=\sqrt{4^2+8^2}=\boxed{4\sqrt{5}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $C$ is on $\odot O$, extend $BC$ to point $D$ such that $DC=CB$, extend $DA$ to meet $\odot O$ at another point $E$, and connect $AC$, $CE$. If $\angle E = 32^\circ$, what is the measure of $\angle D$ in degrees?

Solution:
\textbf{Solution:} Given: $\because$ AB is the diameter of $\odot$O,\\
$\therefore$ $\angle$ACB=$90^\circ$,\\
$\because$ CD=CB,\\
$\therefore$ AC is the perpendicular bisector of BD.\\
$\therefore$ AB=AD,\\
$\therefore$ $\angle$B=$\angle$D,\\
$\because$ $\angle$B=$\angle$E,\\
$\therefore$ $\angle$D=$\angle$E,\\
$\because$ $\angle$E=$32^\circ$,\\
$\therefore$ $\angle$D=$\boxed{32^\circ}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, point C is on $\odot O$, extend BC to point D such that DC=CB, extend DA to another intersection point E with $\odot O$, and connect AC, CE. If AB=13 and BC−AC=7, what is the length of CE?

Solution:
\textbf{Solution:} Let $AC=x$,\\
since $BC-AC=7$,\\
therefore $BC=x+7$,\\
in $\triangle ABC$, where $AB=13$,\\
thus $AB^2=AC^2+BC^2=13^2$, that is $x^2+(x+7)^2=13^2$,\\
hence $x_1=5$, $x_2=-12$ (discard),\\
therefore $AC=5$,\\
therefore $BC=CD=12$,\\
since $\angle D=\angle E$,\\
thus $CD=CE$,\\
therefore $CE=\boxed{12}$.


Question: As shown in the figure, in the sector $ABC$ with a radius of 6, $\angle AOB=120^\circ$, C is a moving point on the arc $\overset{\frown}{AB}$ (not coinciding with A or B), $OD\perp AC$, $OE\perp BC$, with the feet of the perpendiculars being points D and E, respectively. What is the length of $DE$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $AB$, and draw $OF\perp AB$ at $F$,\\
$\because \angle AOB=120^\circ, OA=OB$,\\
$\therefore \angle OAB=\angle OBA=\frac{180^\circ-\angle AOB}{2}=30^\circ$,\\
$\therefore OF=\frac{1}{2}OA=3$,\\
In $\triangle AOF$, by the Pythagorean theorem, $AF=\sqrt{OA^{2}-OF^{2}}=3\sqrt{3}$,\\
$\therefore$, by the perpendicular diameter theorem, $AB=2AF=6\sqrt{3}$,\\
$\because OD\perp AC, OE\perp BC$,\\
$\therefore$ D and E are respectively the midpoints of $AC$ and $BC$,\\
$\therefore DE$ is the midline of $\triangle ABC$,\\
$\therefore DE=\frac{1}{2}AB=3\sqrt{3}$;\\
Thus, the length of $DE$ is $\boxed{3\sqrt{3}}$.


Question: As shown in the figure, in the sector $ABC$ with a radius of $6$, where $\angle AOB = 120^\circ$, $C$ is a moving point on the arc $\overset{\frown}{AB}$ (not coinciding with $A$ or $B$), $OD \perp AC$, $OE \perp BC$, with the feet of the perpendiculars being points $D$ and $E$, respectively. Find the measures of each interior angle of the quadrilateral $ODCE$.

Solution:
\textbf{Solution:} As the figure shows, take a point $H$ on the major arc $AB$, and connect $AH$, $BH$, \\
$\therefore \angle AHB=\frac{1}{2}\angle AOB=60^\circ$, \\
$\because$ quadrilateral $ACBH$ is a cyclic quadrilateral, \\
$\therefore \angle DCE=180^\circ-\angle AHB=\boxed{120^\circ}$, \\
$\because OD\perp AC$, $OE\perp BC$, \\
$\therefore \angle ODC=\angle OEC=90^\circ$, \\
$\therefore \angle BOD=360^\circ-\angle ODC-\angle OEC-\angle DCE=\boxed{60^\circ}$;


Question: In the circle $\odot O$, $AB$ is a diameter, and point C is a point on the circle. The minor arc is folded along the chord $AC$ intersecting $AB$ at point D, and $CD$ is connected. As shown in Figure 1, if point D coincides with the center O and $AC = \sqrt{3}$, what is the radius $r$ of $\odot O$?

Solution:
\textbf{Solution:} Let the point $D$ be symmetric to the chord $AC$, and call this symmetric point $F$. Connect $DF$ and let it intersect $AC$ at point $E$,

thus $DE=EF, DF\perp AC, AE=EC$,

since $AC=\sqrt{3}$,

therefore $AE=EC=\frac{\sqrt{3}}{2}$,

let $DE=EF=\frac{r}{2}$,

then $AD=DF=r$,

according to the Pythagorean theorem, we get ${r}^{2}-\left(\frac{r}{2}\right)^{2}=\left(\frac{\sqrt{3}}{2}\right)^{2}$,

solving this yields $r=\boxed{1}$,

therefore, the radius $r$ of the circle is \boxed{1}.


Question: In circle $ \odot O$, $ AB$ is a diameter, and point $C$ is a point on the circle. Fold the minor arc along chord $ AC$ to intersect $ AB$ at point $D$. Connect $ CD$. As shown in Figure 2, if point $D$ and the center $O$ do not coincide, and $ \angle BAC=20^\circ$, find the degree measure of $ \angle DCA$.

Solution:
\textbf{Solution:} Let the symmetrical point of point D with respect to chord $AC$ be F, and connect $AF$ and $CB$, \\
According to the problem, we get $\angle BAC=\angle FAC=20^\circ$, $CD=CF$, \\
Hence, $CB=CF=CD$, \\
Thus, $\angle B=\angle CDB$; \\
Since $AB$ is the diameter, \\
Therefore, $\angle ACB=90^\circ, \angle B=\angle CDB=70^\circ$, \\
Therefore, $\angle DCA=\angle CDB-\angle BAC=70^\circ-20^\circ=\boxed{50^\circ}$.


Question: In circle $ \odot O$, where $AB$ is the diameter and point C is a point on the circle. The minor arc is folded along chord $AC$ to intersect $AB$ at point D. Connect $CD$. As shown in Figure 2, if $AD=6$ and $DB=2$, find the length of $AC$.

Solution:
\textbf{Solution:} As shown in the figure, connect $OC$ and $CB$, and draw $CG\perp AB$ at point G,\\
according to (2) we have $CB=CD$,\\
$\therefore BG=DG$,\\
$\because AD=6$, $DB=2$,\\
$\therefore BG=DG=\frac{1}{2}DB=1$, $AB=AD+DB=6+2=8$,\\
$\therefore r=OC=\frac{1}{2}AB=4$,\\
$\therefore OD=AD-OB=6-4=2$, $OG=OD+DG=1+2=3$,\\
$\therefore CG=\sqrt{OC^{2}-DG^{2}}=\sqrt{4^{2}-3^{2}}=\sqrt{7}$, $AG=AD+DG=6+1=7$,\\
$\therefore AC=\sqrt{AG^{2}+CG^{2}}=\sqrt{7^{2}+(\sqrt{7})^{2}}=\boxed{2\sqrt{14}}$.


Question: Given in the diagram, $AB$ is the diameter of $\odot O$, and points $C$, $D$ are on $\odot O$, with $BC\perp AC$ and $OD\parallel BC$, $AC$ intersects with $BD$, $OD$ at point $E$, $F$ respectively. Point $D$ is the midpoint of $\overset{\frown}{AC}$; if $DF=7$, $AC=24$, what is the diameter of $\odot O$?

Solution:
Solution: From the given information, we have

\[
\because OF\perp AC,
\]
\[
\therefore AF=\frac{1}{2}AC=12,
\]
\[
\because DF=7,
\]
\[
\therefore OF=OD-DF=OA-7,
\]
\[
\because OA^2=AF^2+OF^2,
\]
\[
\therefore OA^2=12^2+(OA-7)^2,
\]
\[
\therefore OA=\frac{193}{14},
\]
\[
\therefore \text{The diameter of the circle } O \text{ is } \boxed{\frac{193}{7}}.
\]


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, point $E$ is on $AC$, the circle $O$ passing through points $A$, $B$, and $E$ intersects $BC$ at point $D$, and point $D$ is the midpoint of arc $BE$, with $AB$ being the diameter of the circle; if $AB=8$ and $\angle C=60^\circ$, what is the area of the shaded region?

Solution:
\textbf{Solution:}

Let $OF \perp BE$,\\
$\therefore \angle OFB=90^\circ$, $BE=2BF$,\\
$\because AB=AC$, $\angle C=60^\circ$,\\
$\therefore \triangle ABC$ is an equilateral triangle,\\
$\therefore \angle BAC=60^\circ$,\\
$\because \overset{\frown}{BE}=\overset{\frown}{BE}$,\\
$\therefore \angle BOE=2\angle BAC=120^\circ$, $\angle AOE=180^\circ-120^\circ=60^\circ$,\\
$\because AB=8$,\\
$\therefore$ the radius of the circle is 4,\\
$\because OB=OE$, $\angle BOE=120^\circ$, $OF\perp BE$,\\
$\therefore \angle BOF= \frac{1}{2}\angle BOE=60^\circ$,\\
$\therefore \angle OBF=90^\circ-60^\circ=30^\circ$,\\
$\therefore OF= \frac{1}{2}OB=2$,\\
$\therefore BF=\sqrt{4^2-2^2}=2\sqrt{3}$,\\
$\therefore BE=4\sqrt{3}$,\\
$\therefore S_{\text{shaded part}}= S_{\text{sector} AOE}+S_{\triangle BOE}=\frac{60^\circ \pi \times 4^2}{360}+\frac{1}{2}\times 2\times 4\sqrt{3}=\boxed{\frac{8}{3}\pi +4\sqrt{3}}$


Question: As shown in the figure, AB is the diameter of $\odot O$, point C is the midpoint of $\overset{\frown}{BD}$, CF is a chord of $\odot O$, and CF $\perp$ AB with the foot of the perpendicular at point E. Connect BD and CF at point G, and connect CD, AD, BF. $\triangle BFG \cong \triangle CDG$; if AD=10 and EF=15, find the length of BE.

Solution:
\textbf{Solution:} Draw line OF,\\
$\because$ arc CD = arc CB = arc BF,\\
$\therefore$ arc CF = arc DB,\\
$\therefore$ BD = CF = 2, EF = 30,\\
$\because$ AD = 10, $\angle ADB = 90^\circ$,\\
$\therefore$ AB = $10\sqrt{10}$,\\
$\therefore$ OF = $5\sqrt{10}$,\\
$\because$ CF $\perp$ AB,\\
$\therefore$ OE = 5, BE = $5\sqrt{10} - 5 = \boxed{5\sqrt{10} - 5}$.


Question: As shown in the figure, within circle $\odot O$, the diameter $AB$ intersects with the chord $CD$ at point $P$, $\angle CAB=40^\circ$, $\angle APD=65^\circ$. What is the measure of $\angle DAB$?

Solution:
\textbf{Solution:} Given $\because \angle APD = \angle C + \angle CAB$,\\
$\therefore \angle C = \angle APD - \angle CAB = 65^\circ - 40^\circ = 25^\circ$,\\
$\therefore \angle B = \angle C = 25^\circ$.\\
Because $AB$ is the diameter of $\odot O$, $\therefore \angle ADB = 90^\circ$,\\
$\therefore \angle DAB = 90^\circ - \angle B = 90^\circ - 25^\circ = \boxed{65^\circ}$.


Question: As shown in the figure, within circle $O$, the diameter $AB$ intersects the chord $CD$ at point $P$, $\angle CAB=40^\circ$, $\angle APD=65^\circ$. If $AD=6$, what is the distance from the center of the circle $O$ to $BD$?

Solution: \textbf{Solution:} It is insufficient to deduce the distance from the center $O$ to $BD$ based on the given information, therefore a specific solution cannot be provided. $\boxed{\text{Insufficient information}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and $CD$ is a chord of $O$ with $CD\perp AB$ at point $E$. $\angle BCO = \angle D$; If $BE=9\text{cm}$ and $CD=6\text{cm}$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Let the radius be $x$, then $OE=9-x$.\\
Since diameter $AB \perp CD$ and $CD= 6$,\\
it follows that $CE=3$.\\
In $\triangle OCE$, we have $OC^{2}=OE^{2}+CE^{2}$,\\
therefore $x^{2}=(9-x)^{2}+3^{2}$,\\
hence $x=\boxed{5}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in a circle $\odot O$, where $\overset{\frown}{AD}=\overset{\frown}{CD}$, $\angle ABD=33^\circ$, and $\angle ACB=44^\circ$. What is the degree measure of $\angle BAC$?

Solution:
\textbf{Solution:} Given $\because \overset{\frown}{AD}=\overset{\frown}{CD}$,\\
$\therefore \angle ABD=\angle DBC=\angle DAC=33^\circ$,\\
$\therefore \angle ABC=\angle ABD+\angle DBC=33^\circ+33^\circ=66^\circ$,\\
$\therefore \angle BAC=180^\circ-\angle ABC-\angle ACB=180^\circ-66^\circ-44^\circ=\boxed{70^\circ}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $O$, with $\overset{\frown}{AD} = \overset{\frown}{CD}$, $\angle ABD = 33^\circ$, and $\angle ACB = 44^\circ$. What is the measure of $\angle BAD$?

Solution:
Solution: Since $\overset{\frown}{CD}=\overset{\frown}{CD}$,\\
$\therefore \angle DAC=\angle DBC=33^\circ$,\\
$\therefore \angle BAD=\angle DAC+\angle CAB=33^\circ+70^\circ=\boxed{103^\circ}$.


Question: As shown in the diagram, $AB$ and $CD$ are diameters of $\odot O$, with chords $DE$ and $BF$ intersecting the radii $AO$ and $CO$ at points $G$ and $H$, respectively, and $\angle FBA=\angle EDC$. Given $DE=BF$, and if $\overset{\frown}{AE}=\overset{\frown}{EF}=\overset{\frown}{FC}$, and $\angle DOB=\angle EGO$, what is the degree measure of $\overset{\frown}{AC}$?

Solution: \textbf{Solution:} As shown in the figure, since $OB = OD$, it follows that $\angle 1 = \angle 2$, therefore $\angle DOB = 180^\circ - 2\angle 1$. Since $\angle EGO = \angle EDB + \angle ABD = \angle 3 + \angle 1 + \angle 2 = \angle 3 + 2\angle 1$, and $\angle DOB = \angle EGO$, it follows that $180^\circ - 2\angle 1 = \angle 3 + 2\angle 1$, therefore $\angle 3 = 180^\circ - 4\angle 1$. Since $\overset{\frown}{AE} = \overset{\frown}{EF} = \overset{\frown}{FC}$, it follows that $\angle 3 = 2\angle ADE$, therefore $\angle ADE = \frac{1}{2}\angle 3$. Since $CD$ is the diameter of $\odot O$, it follows that $\overset{\frown}{AD} + \overset{\frown}{AE} + \overset{\frown}{CE} = 180^\circ$, therefore $\angle 2 + \angle ADE + \angle 3 = 90^\circ$, hence $\angle 1 +  \frac{1}{2} \times (180^\circ - 4\angle 1) + (180^\circ - 4\angle 1) = 90^\circ$, therefore $5\angle 1 = 180^\circ$, thus $\angle 1 = 36^\circ$, hence $\angle DOB = 180^\circ - 36^\circ \times 2 = 108^\circ$, therefore $\angle AOC = 108^\circ$, hence the degree of $\overset{\frown}{AC}$ is $\boxed{108^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $\odot O$, and $C,D$ are two points on the circle, with $OD\parallel AC$ and $OD$ intersecting $BC$ at point $E$. $E$ is the midpoint of $BC$. If $BC=6$ and $DE=2$, what is the length of $AB$?

Solution: \textbf{Solution:} From (1), we have $BE=3$. Let $OB=OD=x$, then $OE=x-2$. In right triangle $OEB$, we have $BE^2+OE^2=OB^2$, that is $3^2+(x-2)^2=x^2$. Solving this, we get: $x=\frac{13}{4}$, so $OB=\frac{13}{4}$, therefore $AB=2OB=\boxed{\frac{13}{2}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, the chord $CD \perp AB$ at point $E$. $G$ is a moving point on arc $AC$ and does not coincide with points $A$ and $C$. The extensions of $AG$ and $DC$ intersect at point $F$. Connect $BC$. Given $CD = 4\sqrt{3}$ and $BE = 2$, find the length of the radius.

Solution: \textbf{Solution:} As shown in the figure, connect OD and let OD = OB = r. Since AB is the diameter and AB $\perp$ CD, with CD = $4\sqrt{3}$, it follows that DE = EC = $2\sqrt{3}$. In $\triangle ODE$, by the Pythagorean theorem, we have $OD^2 = OE^2 + DE^2$. Therefore, $r^2 = (r-2)^2 + 12$. Solving this, we obtain $r = \boxed{4}$. Hence, the radius of circle O is \boxed{4}.


Question: As shown in the diagram, AB is the diameter of the circle $\odot O$. The chord CD is perpendicular to AB at point E. G is a moving point on the arc AC and does not coincide with points A and C. The extensions of AG and DC intersect at point F. Connect BC. The length of CD is $4\sqrt{3}$ and BE is 2. What is the area of the sector DOC?

Solution:
\textbf{Solution:} As shown in the figure, connect OC, \\
$\because$ in $\triangle ODE$, $DO=4$, $OE=2$, \\
$\therefore \angle ODE=30^\circ$, \\
$\because$ $OD=OC$, \\
$\therefore \angle OCD=30^\circ$, \\
$\therefore \angle DOC=120^\circ$, \\
$\therefore$ the area of sector DOC $=\frac{120^\circ \cdot \pi \cdot 16}{360^\circ}=\boxed{\frac{16\pi}{3}}$.


Question: As shown in the figure, $E$ is a point on semicircle $O$, $C$ is the midpoint of arc $\overset{\frown}{BE}$, and diameter $AB \parallel$ chord $DC$, intersecting $AE$ at point $F$. Prove that: $CF=AF$.

Solution:
\textbf{Solution:} Proof: \\
$\because C$ is the midpoint of $\overset{\frown}{BE}$, \\
$\therefore \angle EAC = \angle CAB$, \\
$\because$ diameter AB$\parallel$chord DC, \\
$\therefore \angle DCA = \angle CAB$, \\
$\therefore \angle DCA = \angle EAC$, \\
$\therefore \boxed{CF = AF}$.


Question: As shown in the figure, E is a point on the semicircle with center O, C is the midpoint of the arc $\overset{\frown}{BE}$, and the diameter $AB$ is parallel to chord $DC$, intersecting $AE$ at point $F$. It is given that $CF=AF$. When $AB=4$ and $OE \perp CD$, find the value of $EF$.

Solution:
\textbf{Solution:} As shown in the figure, connect OC, OE.\\
$\because$ OE$\perp$CD, $\therefore$ $\overset{\frown}{AE}=\overset{\frown}{BE}$, $\therefore$ $\angle AOE=\angle BOE=90^\circ$.\\
$\because$ AB=4, $\therefore$ AO=EO=2, $\therefore$ AE= $2\sqrt{2}$, $\angle OAC=\angle OCA$. By (1) we have $\angle DCA=\angle EAC=\angle CAB$, $\therefore$ $\angle FCA=\angle FAC=\angle OAC=\angle OCA$. Also, $\because$ AC=AC, $\therefore$ $\triangle AOC\cong \triangle AFC$, $\therefore$ AF=AO=2, $\therefore$ EF=AE−AF= $2\sqrt{2}-2=\boxed{2\sqrt{2}-2}$.


Question: As shown in the figure, $\triangle ABC$ is inscribed in $\odot O(AC>BC)$, with $AB$ being the diameter of $\odot O$. Points E, C, D lie on $\odot O$, and the arcs $\overset{\frown}{EC}=\overset{\frown}{CB}=\overset{\frown}{BD}$. Lines $ED$ intersect $AC, AB$ at points F and G, respectively. $\triangle EFA\sim \triangle BCA$. If $BC=5, BG=4$, calculate the length of $AE$.

Solution:
\textbf{Solution:} Given that $AB$ is the diameter of $\odot O$,\\
$\therefore \angle ACB=90^\circ$,\\
$\because \triangle EFA\sim \triangle BCA$,\\
$\therefore \angle EFA=\angle C=90^\circ$, $\frac{EF}{AE}=\frac{BC}{AB}$,\\
Also $\because \angle CAE=\angle CAB$, $AF=AF$,\\
$\therefore \triangle AEF\cong \triangle AFG$ (ASA),\\
$\therefore AE=AG$, $EF=FG$,\\
$\because \angle AEG=\angle ABD$, $\angle AGE=\angle BGD$,\\
$\therefore \triangle AEG\sim \triangle DBG$,\\
$\therefore \frac{BD}{BG}=\frac{AE}{EG}=\frac{5}{4}$,\\
Let $EF=FG=2x, AE=AG=5x$,\\
$\therefore AB=5x+4$,\\
$\therefore \frac{BC}{AB}=\frac{EF}{AE}=\frac{5}{5x+4}=\frac{2}{5}$,\\
$\therefore x=\frac{17}{10}$,\\
$\therefore AE=5x=\boxed{\frac{17}{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, CD is a chord, and CD $\perp$ AB at point E. Connect DO and extend it to intersect $\odot O$ at point F. Connect AF, which intersects CD at point G, with CG = AG, and connect AC. AC $\parallel$ DF; if AB = 12, what are the lengths of AC and GD?

Solution:
\textbf{Solution:} As shown in the figure, connect CO,\\
since AB$\perp$CD,\\
thus $\overset{\frown}{AC} $= $\overset{\frown}{AD}$, CE=DE,\\
since $\angle$DCA=$\angle$CAF,\\
thus $\overset{\frown}{AD} $= $\overset{\frown}{CF}$,\\
thus $\overset{\frown}{AC} $= $\overset{\frown}{AD} $= $\overset{\frown}{CF}$,\\
thus $\angle$AOD=$\angle$AOC=$\angle$COF,\\
since DF is the diameter,\\
thus $\angle$AOD=$\angle$AOC=$\angle$COF=$60^\circ$,\\
since OA=OC,\\
thus $\triangle$AOC is an equilateral triangle,\\
thus AC=AO=6, $\angle$CAO=$60^\circ$,\\
since CE$\perp$AO,\\
thus AE=EO=3, $\angle$ACD=$30^\circ$,\\
thus CE=3$\sqrt{3}$=DE,\\
since AG$^{2}$=GE$^{2}$+AE$^{2}$,\\
thus AG$^{2}$=（3$\sqrt{3}$−AG）$^{2}$+9,\\
thus AG=2$\sqrt{3}$,\\
thus GE=$\sqrt{3}$,\\
thus DG=4$\sqrt{3}$.\\
Therefore, the lengths of AC and GD are respectively AC=\boxed{6} and DG=\boxed{4\sqrt{3}}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ passes through points $A(1,0)$ and $B(0,2)$. What is the analytical expression of line $l_{1}$?

Solution:
\textbf{Solution:} Let the equation of line $l_{1}$ be $y=kx+b$. According to the problem, we have
\[
\left\{
\begin{array}{l}
k+b=0, \\
b=2.
\end{array}
\right.
\]
Solving these equations, we find
\[
\left\{
\begin{array}{l}
k=-2, \\
b=2.
\end{array}
\right.
\]
$\therefore$ The equation of line $l_{1}$ is \boxed{y=-2x+2}.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_{1}$ passes through point $A(1,0)$ and point $B(0,2)$. The moving point $P(m,n)$ is on the line $l_{1}$. When $-2<m<4$, what is the range of values for $n$?

Solution: \textbf{Solution:} Since $-2<0$, it follows that $y$ decreases as $x$ increases. When $x=m=-2$, we have $n=y=-2x+2=6$; when $x=m=4$, we find $n=y=-2x+2=-6$. Therefore, for $-2<m<4$, it holds that $\boxed{-6<n<6}$.


Question: As shown in the Cartesian coordinate system, line $l_{1}$ passes through points $A(1,0)$ and $B(0,2)$. By translating line $l_{1}$ downward by 4 units, we obtain line $l_{2}$. Line $l_{2}$ intersects the $x$ axis and the $y$ axis at points $C$ and $D$, respectively. Connect $AD$, $BC$, and $CP$. If segment $CP$ divides the quadrilateral $ABCD$ into two parts with an area ratio of $1:3$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Since the line $l_{1}$ is translated 4 units downward to obtain line $l_{2}$,\\
the equation of line $l_{2}$ is $y=-2x-2$.\\
When $y=-2x-2=0$, we solve to find $x=-1$; when $x=0$, $y=-2x-2=-2$.\\
Thus, points C$(-1,0)$ and D$(0,-2)$,\\
hence OA$=$OC$=1$, OB$=$OD$=2$.\\
Therefore, quadrilateral ABCD is symmetric about the $x$ axis and the $y$ axis.\\
Therefore, when line CP passes through the midpoint of AB or AD, it divides quadrilateral ABCD into two parts with an area ratio of $1:3$.\\
When point P is the midpoint of AB (denoted as $P_{1}$), take the midpoint of OA, denoted as E, and draw line $P_{1}E$ such that $P_{1}E \parallel y$ axis, and $P_{1}E= \frac{1}{2} OB=1$,\\
thus $P_{1}\left( \frac{1}{2}, 1\right)$.\\
Let M be the midpoint of AD, by symmetry, M$\left( \frac{1}{2}, -1\right)$,\\
the equation of line CM can be obtained from the coordinates of C and M as $y=-\frac{2}{3}x-\frac{2}{3}$.\\
By solving $\begin{cases}y=-\frac{2}{3}x-\frac{2}{3}\\ y=-2x+2\end{cases}$, we find $\begin{cases}x=2\\ y=-2\end{cases}$\\
Therefore, $P_{2}(2, -2)$\\
In conclusion, the coordinates of point P are $\boxed{\left( \frac{1}{2}, 1\right)}$ or $\boxed{(2, -2)}$.


Question: As shown in the figure: It is known that the line $y_{1}=kx+b$ passes through points $A(5,0)$ and $B(1,4)$, and intersects with line $y_{2}=2x-4$ at point $C$. Find the equation of line $y_{1}$ as well as the coordinates of point $D$, the intersection point of $y_{2}$ and the $x$-axis.

Solution:
\textbf{Solution:} Based on the given problem, we have:
\[
\left\{
\begin{array}{l}
5k+b=0 \\
k+b=4
\end{array}
\right.
\]
Solving this, we find:
\[
\left\{
\begin{array}{l}
k=-1 \\
b=5
\end{array}
\right.
\]
Thus, the equation for $y_{1}$ is $y_{1}=-x+5$. Setting $y_{2}=0$ in $y_{2}=2x-4$, we get:
$2x-4=0$,
hence $x=2$,
which means the coordinates of point D are $\boxed{(2,0)}$.


Question: As shown in the figure: it is known that the line $y_1=kx+b$ passes through point $A(5,0)$ and $B(1,4)$, and intersects with the line $y_2=2x-4$ at point $C$. What are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given the problem, we have:
\[
\left\{
\begin{array}{l}
y = -x + 5 \\
y = 2x - 4
\end{array}
\right.
\]
Solving these equations, we get:
\[
\left\{
\begin{array}{l}
x = 3 \\
y = 2
\end{array}
\right.
\]
$\therefore C(\boxed{3, 2})$


Question: As shown in the diagram: It is known that the line $y_1=kx + b$ passes through points $A(5,0)$ and $B(1,4)$, and intersects with the line $y_2=2x-4$ at point $C$. Based on the diagram, write down the range of values for $x$ when the inequality $y_1>y_2>0$ holds.

Solution: \textbf{Solution:} As shown in the figure, let us draw perpendicular lines $l$ and $m$ from $D$ and $C$ respectively to the x-axis. The segment between lines $l$ and $m$ satisfies $y_1>y_2>0$. In this case, $2<x<3$. Therefore, the range of $x$ that satisfies the inequality $y_1>y_2>0$ is $x\in\boxed{(2,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, for rectangle $ABCD$, edge $AB$ lies on the $x$-axis, $AB = 3$, and $AD = 2$. The line passing through point $C$ described by $y = x - 2$ intersects the $x$-axis and the $y$-axis at points $E$ and $F$, respectively. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} Let the coordinates of point $C$ be $(4, 2)$,\\
$\because$ point $C$ is on the line $y=x-2$,\\
$\therefore 2=4-2$,\\
which means the coordinates of point $C$ are $(4, 2)$,\\
$\because$ quadrilateral $ABCD$ is a rectangle,\\
$\therefore AB=CD=3$, $AD=BC=2$,\\
$\therefore$ the coordinates of point $D$ are $\boxed{(1, 2)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rectangle $ABCD$ has its side $AB$ on the $x$-axis, with $AB=3$ and $AD=2$. The line passing through point $C$ and described by $y=x-2$ intersects the $x$-axis and $y$-axis at points $E$ and $F$, respectively. The question is whether there exists a point $P$ on the line $y=x-2$ such that $\triangle PDC$ is an isosceles right-angled triangle. Determine the coordinates of point $P$.

Solution:
\textbf{Solution:} There exists.
\\
$\because \triangle EBC$ is an isosceles right-angled triangle,\\
$\therefore \angle CEB=\angle ECB=45^\circ$,\\
Also, $\because DC\parallel AB$,\\
$\therefore \angle DCE=\angle CEB=45^\circ$,\\
$\therefore \triangle PDC$ can only be an isosceles right-angled triangle with $P$ and $D$ as the right-angled vertices.\\
As shown in the figure, (1) when $\angle D=90^\circ$, extend $DA$ to intersect the line $y=x-2$ at point $P_1$,\\
$\because$ the coordinates of point $D$ are $(1, 2)$,\\
$\therefore$ the $x$-coordinate of point $P_1$ is 1,\\
Substituting $x=1$ into $y=x-2$ gives, $y=-1$,\\
$\therefore$ the point $P_1(1, -1)$;\\
(2) when $\angle DPC=90^\circ$, draw the perpendicular bisector of $DC$ to intersect the line $y=x-2$ at point $P_2$,\\
Therefore, the $x$-coordinate of point $P_2$ is $\frac{1+4}{2}=\frac{5}{2}$,\\
Substituting $x=\frac{5}{2}$ into $y=x-2$ gives, $y=\frac{1}{2}$,\\
Therefore, the point $P_2\left(\frac{5}{2}, \frac{1}{2}\right)$,\\
In summary, the coordinates of the suitable point $P$ are \boxed{(1, -1)} or \boxed{\left(\frac{5}{2}, \frac{1}{2}\right)}.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the rectangle $ABCD$ has side $AB$ on the $x$ axis, with $AB=3$ and $AD=2$. The line through point $C$, given by $y=x-2$, intersects the $x$ axis and the $y$ axis at points $E$ and $F$, respectively. Determine the coordinate of point $M$ in the Cartesian coordinate system such that the quadrilateral formed by points $M$, $D$, $C$, and $E$ is a parallelogram. Please directly provide the coordinates of point $M$.

Solution: \textbf{Solution:} When $y=0$, we have $x-2=0$, solving this gives $x=2$, thus $OE=2$. Since the quadrilateral with vertices $M$, $D$, $C$, $E$ is a parallelogram, if $DE$ is a diagonal, then $EM=CD=3$, therefore $OM=EM-OE=3-2=1$, in this case, the coordinates of point $M$ are $(-1, 0)$. If $CE$ is a diagonal, then $EM=CD=3$, and $OM=OE+EM=2+3=5$, in this case, the coordinates of point $M$ are $(5, 0)$. If $CD$ is a diagonal, then the center of the parallelogram has coordinates $\left(\frac{5}{2}, 2\right)$, let the coordinates of point $M$ be $(x, y)$, then we have $\frac{x+2}{2}=\frac{5}{2}$ and $\frac{y+0}{2}=2$, solving these gives $x=3$ and $y=4$, in this case, the coordinates of point $M$ are $(3, 4)$. In summary, the coordinates of point $M$ can be $(-1, 0)$, $(5, 0)$, or \boxed{(3, 4)}.


Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point C, and they intersect the y-axis at points A and B, respectively. What are the coordinates of points A, B, and C?

Solution: \textbf{Solution:} For the line $y=x+7$, when $x=0$, we have $y=0+7=7$, therefore $A(0, 7)$. On the line $y=-2x-2$, when $x=0$, we have $y=0-2=-2$, therefore $B(0, -2)$. Solving the system of equations $\begin{cases}y=x+7\\ y=-2x-2\end{cases}$ yields $\begin{cases}x=-3\\ y=4\end{cases}$, therefore $C(-3, 4)$. Consequently, the coordinates of points $A$, $B$, and $C$ are $\boxed{A(0, 7)}$, $\boxed{B(0, -2)}$, and $\boxed{C(-3, 4)}$, respectively.


Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point $C$. They intersect the y-axis at points $A$ and $B$, respectively. Point $F$ is on the positive x-axis, such that the area of $\triangle ABC$ equals the area of $\triangle AFC$. Find the coordinates of point $F$.

Solution:
\textbf{Solution:} Since $S_{\triangle AFC} = S_{\triangle ABC}$, \\
it follows that $BF \parallel AC$, and thus the equation of line $BF$ is $y=x-2$. When $y=0$, $x-2=0$, which implies $x=2$, \\
hence, the coordinates of $F$ are $\boxed{(2, 0)}$.


Question: As shown in the figure, the line $y=x+7$ intersects with the line $y=-2x-2$ at point C, and they intersect the y-axis at points A and B, respectively. Point P is on the x-axis, such that $\angle PBO=2\angle PAO$. Directly write down the coordinates of point P.

Solution:
\textbf{Solution:} Take point $B'$(0, 2) on the positive semi-axis of $y$, then $OB=OB'=2$, connect $PB'$,\\
$\therefore PB=PB'$, hence $\angle PBO=\angle PB'O=2\angle PAO$,\\
$\therefore \angle B'AP=\angle B'PA$,\\
therefore $PB'=B'A=7-2=5$,\\
Assume point $P(x, 0)$,\\
then $PB'^2=2^2+x^2=5^2$, solving for $x$ yields $x=\pm \sqrt{21}$,\\
Therefore, the coordinates of point P are $\boxed{P_1(\sqrt{21}, 0)}$ or $\boxed{P_2(-\sqrt{21}, 0)}$.


Question: As shown in Figure 1, the analytical expression of line $AB$ is $y=kx+6$. The coordinates of point $D$ are $\left(8, 0\right)$. The symmetric point $C$ of point $O$ with respect to line $AB$ lies on line $AD$. What are the analytical expressions of lines $AD$ and $AB$?

Solution:
\textbf{Solution:} From the problem, we have
\[
\because y=kx+6,
\]
\[
\therefore A\left(0, 6\right),
\]
that is, $OA=6$. Also,
\[
\because D\left(8, 0\right),
\]
\[
\therefore OD=8,
\]
Suppose the equation of line $AD$ is $y=nx+6$. Substituting point $D\left(8, 0\right)$ into it, we get the equation of line $AD$ as $y=-\frac{3}{4}x+6$.

In $\triangle ABD$, $AD=\sqrt{6^2+8^2}=10$,

\[
\because \text{Point $O$ and point $C$ are symmetric about line $AB$},
\]
\[
\therefore \text{Let $OB=BC=a$, $OA=AC=6$, $CD=4$},
\]
\[
\therefore BD=8-a,
\]
In $\triangle BCD$, $a^2+4^2=(8-a)^2$,

\[
\therefore a=3 \text{(area method is also applicable)}
\]
\[
\therefore B\left(3, 0\right),
\]
\[
\therefore \text{The equation of line $AB$ is } y=-2x+6.
\]

Therefore, the equation of line $AD$ is $\boxed{y=-\frac{3}{4}x+6}$, and the equation of line $AB$ is $\boxed{y=-2x+6}$.


Question: As shown in Figure 1, the equation of line $AB$ is $y=kx+6$, and the coordinate of point $D$ is $(8, 0)$. The symmetric point of point $O$ with respect to line $AB$ is point $C$, which lies on line $AD$. As shown in Figure 2, if $OC$ intersects $AB$ at point $E$, is there a point $F$ on segment $AD$ such that the area of $\triangle ABC$ is equal to the area of $\triangle AEF$? Determine the coordinates of point $F$.

Solution: Solution: From (1), we have $BC=OB=3$, $CD=4$, and $BD=5$. Thus, $\frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\times y_c$. Solving this gives: $y_c=\frac{12}{5}$. Substituting $y_c=\frac{12}{5}$ into $y=-\frac{3}{4}x+6$ yields: $x_c=\frac{24}{5}$. Therefore, $C\left(\frac{24}{5}, \frac{12}{5}\right)$. The equation of line $AD$ is: $y=-\frac{3}{4}x+6$. $\therefore$ The equation of line $OC$ is: $y=\frac{1}{2}x$. $\because S_{\triangle ABC}=S_{\triangle AEF}$, $\therefore S_{\triangle BEC}=S_{\triangle ECF}$, $\therefore BF\parallel OC$. Let the equation of line $BF$ be: $y=\frac{1}{2}x+b$. Point $B(3, 0)$ lies on line $BF$, therefore $\frac{1}{2}\times 3+b=0$, hence $b=-\frac{3}{2}$. Thus, the equation of line $BF$ is: $y=\frac{1}{2}x-\frac{3}{2}$. Solving the system of equations $\begin{cases}y=\frac{1}{2}x-\frac{3}{2}\\ y=-\frac{3}{4}x+6\end{cases}$, we get $\begin{cases}x=6\\ y=\frac{3}{2}\end{cases}$. Therefore, point $\boxed{F\left(6, \frac{3}{2}\right)}$ exists.


Question: As shown in the figure, the equation of line $AB$ is $y=kx+6$, the coordinates of point $D$ are $(8, 0)$, and the symmetric point $C$ of point $O$ with respect to line $AB$ lies on line $AD$. Consider a line $l: y=mx+b$ passing through point $D$. When the angle between line $l$ and line $AB$ is equal to $45^\circ$, find the corresponding value of $m$.

Solution:
\textbf{Solution:} Set if the angle between line $DE$, $DF$ and line $AB$ is $45^\circ$, i.e., $\triangle DEF$ is an isosceles right triangle. Construct $EM\perp DM$ at $M$, $FN\perp DN$ at $N$, then $\triangle DEM\cong \triangle FDN$,

$\therefore EM=DN$, $DM=FN$. Line $l$ passes through $D(8, 0)$, hence $0=8m+b$, solving this gives: $b=-8m$,

$\therefore$ the equation of line $l$ is: $y=mx-8m$. Let the coordinates of $E$ be $(t, -2t+6)$, then $EM=DN=8-t$, $DM=FN=-2t+6$,

$\therefore$ the coordinates of $F$ are $(2+2t, t-8)$. Point $F$ lies on line $AB$,

$\therefore t-8=-2(2+2t)+6$, solving this gives: $t=2$,

$\therefore E(2, 2)$, $F(6, -6)$. When line $l$ passes through point $E$, $2m-8m=2$, solving this gives: $m=\boxed{-\frac{1}{3}}$, when line $l$ passes through point $F$, $6m-8m=-6$, solving this gives: $m=\boxed{3}$.

Therefore, $m=\boxed{3}$ or $\boxed{-\frac{1}{3}}$.


Question: As shown in the Cartesian coordinate system $xOy$, given $A\left(0, 4\right)$ and $B\left(4, 0\right)$, the graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. Find the coordinates of point $P$.

Solution: \textbf{Solution:} According to the problem, let the equation of line $AB$ be: $y=kx+m$. Also, since $A\left(0, 4\right)$ and $B\left(4, 0\right)$ are on line $AB$, it follows that $m=4$, and $4k+4=0$. Solving this gives $k=-1$. Therefore, the equation of line $AB$ is $y=-x+4$. By solving the system of equations, we have $\begin{cases}y=-x+4\\ y=-2x\end{cases}$, which yields $\begin{cases}x=-4\\ y=8\end{cases}$. Therefore, the coordinates of point $P$ are $\boxed{(-4, 8)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, it is known that $A(0, 4)$ and $B(4, 0)$. The graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. If point $M$ is located on the $y$ axis, and the area of $\triangle PMA$ is equal to $10$, what are the coordinates of point $M$?

Solution:
\textbf{Solution:} From the problem, we have
\[
\because S_{\triangle PMA} = \frac{1}{2}AM \times |x_{P}|,
\]
\[
\therefore \frac{1}{2} \times AM \times 4 = 10,
\]
\[
\therefore AM = 5,
\]
\[
\because A(0, 4),
\]
\[
\therefore M(0, 9) \text{ or } M(0, -1).
\]

Thus, the coordinates of point $M$ are $\boxed{M(0, 9)}$ or $\boxed{M(0, -1)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, given $A\left(0, 4\right)$ and $B\left(4, 0\right)$, the graph of the linear function $y=-2x$ intersects with line $AB$ at point $P$. If the line $y=-2x+b$ has exactly two intersection points with the three sides of $\triangle AOB$, what is the range of values for $b$?

Solution:
\textbf{Solution:} Let the range of $b$ be $0<b<80$, \\
$\therefore$ the range of $b$ is $\boxed{0<b<80}$.


Question: As shown in Figure 1, the line $y=-\frac{1}{2}x+2$ intersects with the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. Point $E$ is a point on the negative half of the $y$ axis, and the area of $\triangle ABE$ is 12. What is the equation of the line AE?

Solution: \textbf{Solution:} Since for $y=-\frac{1}{2}x+2$, we set $y=-\frac{1}{2}x+2=0$ and find $x=4$. Setting $x=0$ gives $y=2$. Therefore, the coordinates of points $A$ and $B$ are $(4, 0)$ and $(0, 2)$, respectively. Hence, $OB=2$. The area of triangle $\triangle ABE$ is given by $12=\frac{1}{2}\times EB\times AO=\frac{1}{2}\times (2+OE)\times 4$. Solving this yields $OE=4$. Thus, point $E$ is $(0, -4)$. Let the equation of line $AE$ be $y=kx-4$. Substituting the coordinates of point $A$ into the equation yields $0=4k-4$. Solving this, we find $k=1$. Therefore, the equation of line $AE$ is given by $\boxed{y=x-4}$.


Question: As shown in Figure 1, the line $y=-\frac{1}{2}x+2$ intersects the x-axis and the y-axis at points $A$ and $B$ respectively, and point $E$ is a point on the negative half-axis of the y-axis, with the area of $\triangle ABE$ being $12$. As shown in Figure 2, the line $y=mx$ intersects line $AB$ at point $M$ and intersects line $AE$ at point $N$. When the area of $\triangle OEN$ is twice the area of $\triangle OBM$, what is the value of $m$?

Solution:
\textbf{Solution:} From equation (1), we know that $OE=4$,\\
since $S_{\triangle OEN}=2S_{\triangle OBM}$, i.e., $\frac{1}{2}\times OB\times |x_{M}|=\frac{1}{2}\times \frac{1}{2}\times OE\times x_{N}$,\\
which implies $\frac{1}{2}\times 2\times |x_{M}|=\frac{1}{4}\times 4\times x_{N}$, i.e., $x_{M}=-x_{N}$, then $y_{M}=-y_{N}$,\\
let the coordinates of point $N$ be $(n, n-4)$, then the coordinates of point $M$ would be $(-n, -n+4)$,\\
Substituting the coordinates of point $M$ into $y=-\frac{1}{2}x+2$ yields:\\
$-n+4=-\frac{1}{2}(-n)+2$,\\
solving this gives $n=\frac{4}{3}$,\\
therefore, the coordinates of point $N$ are $\left(\frac{4}{3}, -\frac{8}{3}\right)$,\\
Substituting the coordinates of point $N$ into $y=mx$ yields:\\
$-\frac{8}{3}=\frac{4}{3}m$,\\
solving this gives $m=\boxed{-2}$.


Question: In the figure, the line $y=-\frac{1}{2}x+2$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. Point $E$ is on the negative half-axis of the $y$ axis, and the area of $\triangle ABE=12$. In the figure, point $P$ is on the line $y=-x-1$. If $\angle ABP=45^\circ$, please directly write down the coordinates of point $P$.

Solution: \textbf{Solution:} The coordinates of point $P$ are $\left( \frac{3}{2}, -\frac{5}{2} \right)=\boxed{\left( \frac{3}{2}, -\frac{5}{2} \right)}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. It intersects the graph of the function $y=kx$ at point $M(1, 2)$. What are the values of $k$ and $b$?

Solution:
\textbf{Solution:} Given point $M(1, 2)$ lies on the line $y=-\frac{1}{2}x+b$, \\
thus $2=-\frac{1}{2}\cdot 1+b$, \\
solving for $b$ yields $b=\boxed{\frac{5}{2}}$, \\
since the graph of the function $y=kx$ passes through point $M(1, 2)$, \\
it follows that $2=k$ or $k=\boxed{2}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+b$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and intersects the graph of the function $y=kx$ at point $M(1, 2)$. Point $C$ is on the line segment $MA$, and a perpendicular line to the $x$-axis is drawn through point $C$, intersecting the graph of the function $y=kx$ at point $D$. If $2CD=OB$, find the coordinates of point $C$.

Solution:
\textbf{Solution:} According to the problem,\\
$\because b=\frac{5}{2}$,\\
$\therefore OB= b=\frac{5}{2}$,\\
$\because 2CD=OB$,\\
$\therefore CD=\frac{1}{2}OB=\frac{1}{2}\times \frac{5}{2}=\frac{5}{4}$,\\
$\because$ point C is on the line $y=-\frac{1}{2}x+\frac{5}{2}$,\\
$-\frac{1}{2}x+\frac{5}{2}=\frac{5}{4}$,\\
Solving this, we get $x=\frac{5}{2}$,\\
The coordinates of point C are $\boxed{\left(\frac{5}{2}, \frac{5}{4}\right)}$.


Question: As shown in the Cartesian coordinate system $xOy$, the graph of the linear function $y=-x+6$, denoted as ${l}_{1}$, intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct proportionality function, ${l}_{2}$, intersects ${l}_{1}$ at point C$(m,5)$. Find the value of $m$ and the equation of ${l}_{2}$.

Solution:
\[ \text{{Solution:}} \] \[ \because \] the graph of the linear function \(y = -x + 6\), denoted as \(l_1\), passes through the point \(C(m, 5)\), \\
\[ \therefore \] \(-m + 6 = 5\), \\
\[ \therefore \] \(m = \boxed{1}\), \\
Let the equation of \(l_2\) be \(y = kx\), \\
\[ \therefore \] \(k = 5\), \\
\[ \therefore \] the equation of \(l_2\) is \(y = \boxed{5x}\).


Question: In the Cartesian coordinate system xOy, the graph of the linear function $y=-x+6$, denoted as ${l}_{1}$, intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct variation function, denoted as ${l}_{2}$, intersects ${l}_{1}$ at point C ($m$, 5). What is the area of $\triangle AOC$?

Solution:
\textbf{Solution:} The solution for the linear function $y=-x+6$ and its intersection point A with the x-axis is found by setting $y=0$, and solving $-x+6=0$ to get $x=6$,\\
$\therefore$OA=6,\\
The area of $\triangle AOC$ is $S_{\triangle AOC}=\frac{1}{2}OA\cdot y_c=\frac{1}{2}\times 6\times 5=\boxed{15}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -x + 6$ is denoted as ${l}_{1}$, which intersects the $x$ and $y$ axes at points A and B, respectively. The graph of the direct proportion function, denoted ${l}_{2}$, intersects ${l}_{1}$ at point C$(m,5)$. A line perpendicular to the $x$ axis, $x = a$, intersects lines ${l}_{1}$ and ${l}_{2}$ at points P and Q, respectively. If the length of segment PQ is 2, find the value of $a$.

Solution:
\textbf{Solution:} When $x=a$, it intersects with line $l_1$ at point $P(a, 6-a)$ and with line $l_2$ at point $Q(a, 5a)$. The distance PQ= $\left|5a-(6-a)\right|=6\left|a-1\right|=2$, \\
$\therefore \left|a-1\right|=\frac{1}{3}$, \\
$\therefore a-1=\pm \frac{1}{3}$, \\
$a=\boxed{\frac{4}{3}}$ or $a=\frac{2}{3}$


Question: Given that in the Cartesian coordinate system, point $A(1, 2)$ lies on the graph of the inverse proportion function $y=\frac{k_1}{x}$, and the line passing through point $A$ intersects the other branch of the hyperbola at point $B(-2, m)$. What is the analytical equation of line $AB$?

Solution:
\textbf{Solution:} To solve, substitute point A$(1,2)$ into $y=\frac{2}{x}$, solving for $k_1=2$,\\
$\therefore$ the inverse proportion function is $y=\frac{2}{x}$.\\
Substituting point B$(-2,m)$ into $y=\frac{2}{x}$ yields $m=-1$,\\
$\therefore$ the coordinates of point B are $(-2,-1)$.\\
Let the equation for line AB be $y=kx+b$,\\
Substituting A$(1,2)$ and B$(-2,-1)$ into the equation results in
\[
\left\{
\begin{array}{l}
k+b=2 \\
-2k+b=-1
\end{array}
\right.
\]
Solving this system gives
\[
\left\{
\begin{array}{l}
k=1 \\
b=1
\end{array}
\right.
\]
$\therefore$ the functional expression for line AB is $\boxed{y=x+1}$.


Question: Given that in the Cartesian coordinate system, point $A(1,2)$ lies on the graph of the inverse proportion function $y=\frac{k_1}{x}$, the line passing through point $A$ intersects the other branch of the hyperbola at point $B(-2,m)$. If point $C$ is a moving point on the $x$-axis, what are the coordinates of point $C$ when the area of $\triangle ABC$ is 6?

Solution: \textbf{Solution:} Let the line AB intersect the x-axis at point D. Therefore, the coordinates of point D are (-1,0). Therefore, the area $S_{\triangle ABC}=S_{\triangle ADC}+S_{\triangle BDC}=6$, that is, $\frac{1}{2} \times DC \times 2 + \frac{1}{2} \times DC \times 1 = 6$ $\therefore CD = 4$. Therefore, the coordinates of point C are $\boxed{(3,0)}$ or $\boxed{(-5,0)}$.


Question: As shown in the figure, the graph of a linear function intersects the x-axis at point A and the y-axis at point B. What is the analytical expression of the linear function?

Solution:
Solution: From the graph, we can find $A(-6, 0)$ and $B(0, 2)$. Assume the equation of the linear function is $y=kx+b$. Substituting the coordinates of $A$ and $B$ into the equation, we obtain
\[
\begin{cases}
-6k+b=0 \\
b=2
\end{cases}
\]
From which we can solve
\[
\begin{cases}
k=\frac{1}{3} \\
b=2
\end{cases}
\]
Therefore, the equation of the linear function is $\boxed{y=\frac{1}{3}x+2}$


Question: As shown in the figure, the graph of a linear function intersects the x-axis at point A and the y-axis at point B. What is the length of $AB$?

Solution:
\textbf{Solution:} Given $A(-6, 0)$ and $B(0, 2)$, \\
$\therefore OA=6$ and $OB=2$, \\
In the right-angled triangle $AOB$, $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{6^{2}+2^{2}}=\boxed{2\sqrt{10}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=3x-3$ intersects the $x$ axis at point $A$, and intersects the graph of the direct proportion function $y=kx$ at point $B(a, 3)$. What is the analytic expression of the direct proportion function?

Solution:
\textbf{Solution:} Since point $B\left(2, 3\right)$ lies on the graph of the linear function $y=3x-3$,\\
we have $3=3\cdot2-3$. Solving this gives: $a=2$,\\
Hence, for point $B\left(2, 3\right)$, substituting the coordinates of point $B$ into $y=kx$ gives $2k=3$,\\
Solving this yields: $k=\frac{3}{2}$,\\
Therefore, the equation of the direct proportion function is $\boxed{y=\frac{3}{2}x}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=3x-3$ intersects the $x$-axis at point $A$, and intersects the graph of the direct proportion function $y=kx$ at point $B(a, 3)$. Taking point $A$ as the center and the length of $AB$ as the radius, an arc is drawn that intersects the $x$-axis at point $C$. Please directly write down the coordinates of point $C$.

Solution: \textbf{Solution:} From the problem statement, we have $y=3x-3$. When $y=0$, it follows that $3x-3=0$, which solves to: $x=1$. Therefore, $A(1, 0)$ and hence $\left|AB\right|=\sqrt{(3-0)^2+(2-1)^2}=\sqrt{10}$. Draw an arc with point $A$ as the center and the length of $AB$ as the radius, intersecting the x-axis at point $C$. Therefore, $C\left(\sqrt{10}+1, 0\right)$ or $\left(-\sqrt{10}+1, 0\right)$. Hence, the coordinates of point $C$ are $\boxed{C\left(\sqrt{10}+1, 0\right)}$ or $\boxed{C\left(-\sqrt{10}+1, 0\right)}$.


Question: As shown in the figure, the line $l_{1}: y=2x+2$ intersects the $x$-axis at point $A$ and the $y$-axis at point $C$; the line $l_{2}: y=kx+b$ intersects the $x$-axis at point $B(3, 0)$ and intersects line $l_{1}$ at point $D$, with the ordinate of point $D$ being 4. What is the solution set of the inequality $kx+b>2x+2$?

Solution:
\textbf{Solution:} Given the inequality $kx+b>2x+2$.

First, convert the inequality into standard form, that is, move all terms to one side to obtain $kx - 2x + b - 2 > 0$.

To solve the inequality, we need to determine the values of $k$ and $b$. However, these values are not provided in the question, hence we cannot directly find the solution set for this inequality.

However, the given information is $x<1$, which does not directly assist us in resolving the original problem, as there is insufficient information to determine the solution set of the inequality $kx+b>2x+2$.

\boxed{\text{Indeterminate}}, due to the lack of specific information about $k$ and $b$.


Question: As shown in the figure, the line $l_{1}: y=2x+2$ intersects the $x$-axis at point $A$ and the $y$-axis at point $C$; the line $l_{2}: y=kx+b$ intersects the $x$-axis at point $B(3, 0)$ and intersects with the line $l_{1}$ at point $D$, and the $y$-coordinate of point $D$ is 4. Find the equation of line $l_{2}$ and the area of $\triangle CDE$.

Solution:
\textbf{Solution:} Substitute $B(3, 0)$ and $D(1, 4)$ into $y=kx+b$ to obtain: $\begin{cases}3k+b=0\\ k+b=4\end{cases}$\\
$\therefore \begin{cases}k=-2\\ b=6\end{cases}$\\
$\therefore$ the equation of line $l_{2}$ is: $y=-2x+6$\\
Let $x=0$\\
$\therefore y=6$\\
$\therefore E(0, 6)$\\
From $y=2x+2$, we know: $C(0, 2)$\\
$\therefore Area_{\triangle CDE}=\frac{1}{2}\times (6-2)\times 1=\boxed{2}$\\
$\therefore$ The area of $\triangle CDE$ is $\boxed{2}$


Question: As shown in the figure, in the Cartesian coordinate system, the line $AC$ intersects the x-axis at point $C$ and the y-axis at point $A$, the line $AB$ intersects the x-axis at point $B$ and the y-axis at point $A$, with $A(0,4)$, $B(2,0)$ known. What is the analytical expression of line $AB$?

Solution:
\textbf{Solution:} Let the equation of line AB be $y=kx+b$.\\
Since line AB passes through A$(0, 4)$ and B$(2, 0)$,\\
therefore,
\[
\left\{
\begin{array}{l}
b=4 \\
2k+b=0
\end{array}
\right.
\]
Solving this, we get $k=-2$ and $b=4$,\\
therefore, the equation of line AB is $y=-2x+4=\boxed{y=-2x+4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line AC intersects the x-axis at point C, and intersects the y-axis at point A. The line AB intersects the x-axis at point B, and intersects the y-axis at point A. Given that A(0, 4) and B(2, 0). If $S_{\triangle ABC} = 12$, find the coordinates of point C.

Solution: \textbf{Solution:} Let C$(x,0)$, since A$(0,4)$ and B$(2,0)$, therefore OA$=4$, OB$=2$. Given that the area $S_{\triangle ABC}=12$, thus $\frac{1}{2}$BC$\cdot$OA$=12$, therefore BC$=6$, hence $|x-2|=6$. Solving this gives $x=8$ or $x=-4$. Therefore, the coordinates of C are $\boxed{(-4,0)}$ or $\boxed{(8,0)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, line $l_1: y = -2x + 1$ intersects the y-axis at point A, and line $l_2$ intersects the y-axis at point B $(0, -2)$ and intersects line $l_1$ at point C. The ordinate of point C is -1. Point D is an arbitrary point on line $l_2$. A perpendicular line to the x-axis is drawn from point D and intersects line $l_1$ at point E. What is the equation of line $l_2$?

Solution:
\textbf{Solution:} Since the ordinate of point $C$ is $-1$ and point $C$ lies on line $l_1$,\\
we have $-1=-2x+1$, solving this gives $x=1$,\\
thus the coordinates of point $C$ are $(1, -1)$,\\
Let the equation of line $l_2$ be $y=kx+b$,\\
since line $l_2$ intersects the $y$-axis at point $B(0, -2)$, and intersects line $l_1$ at point $C$,\\
we obtain $\begin{cases}k+b=-1\\ b=-2\end{cases}$ solving this gives $\begin{cases}k=1\\ b=-2\end{cases}$\\
Therefore, the equation of line $l_2$ is $y=x-2$, thus the equation of line $l_2$ is $\boxed{y=x-2}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_1\colon y=-2x+1$ intersects with the y-axis at point A, and the line $l_2$ intersects with the y-axis at point B(0, -2) and intersects with line $l_1$ at point C, where the y-coordinate of point C is -1. Point D is any point on line $l_2$. A perpendicular line is drawn from point D to the x-axis, intersecting line $l_1$ at point E. When $DE = 2AB$, find the coordinates of point D?

Solution:
\textbf{Solution:} Let $x=0$, we get $y=-2\times 0+1=1$, \\
$\therefore$ the coordinates of point $A$ are $\left(0, 1\right)$, \\
$\therefore$ $AB=3$, let the abscissa of point $D$ be $m$, then the coordinates of point $D$ are $\left(m, m-2\right)$, \\
$\because$ $DE$ is parallel to the y-axis, \\
$\therefore$ the coordinates of point $E$ are $\left(m, -2m+1\right)$, \\
$\therefore$ $DE=\left|\left(m-2\right)-\left(-2m+1\right)\right|=\left|3m-3\right|$, \\
$\because$ $DE=2AB=6$, \\
$\therefore$ $\left|3m-3\right|=6$, solving gives $m=3$ or $m=-1$, \\
When $m=3$, the coordinates of point $D$ are $\left(3, 1\right)$, \\
When $m=-1$, the coordinates of point $D$ are $\left(-1, -3\right)$. \\
To summarize: The coordinates of point $D$ are $\boxed{\left(3, 1\right)}$ or $\boxed{\left(-1, -3\right)}$.


Question: As shown in the figure, the line $l_1$ passing through point $B(2, 0)$ intersects with the line $l_2: y=2x+8$ at point $P(-1, n)$. What is the value of $n$?

Solution: \textbf{Solution:} Substitute $P(-1, n)$ into $y=2x+8$ to get $n=2\times (-1)+8$. Solving this, we find $n=\boxed{6}$.


Question: The line $l_1$ passing through point $B(2, 0)$ intersects with the line $l_2: y=2x+8$ at point $P(-1, n)$. Try to find the length of $PB$.

Solution:
\textbf{Solution:} Construct $PI\perp x$-axis at point $I$,

since the coordinates of point $P$ are $(-1, 6)$,

and point $B(2, 0)$,

therefore $BP=\sqrt{PI^{2}+BI^{2}}=\sqrt{6^{2}+3^{2}}=\boxed{3\sqrt{5}}$.


Question: As shown in the Cartesian coordinate system, the line $y=\frac{1}{2}x+m$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, where the coordinates of point $A$ are $\left(-2, 0\right)$, and the coordinates of point $C$ are $\left(3, 0\right)$. What are the coordinates of point $B$?

Solution: \textbf{Solution:} Substitute $A(-2, 0)$ into $y=\frac{1}{2}x+m$, yielding $\frac{1}{2}\times (-2)+m=0$. Solving for $m$ gives $m=1$. Therefore, the equation of line $AB$ is $y=\frac{1}{2}x+1$. Letting $x=0$, we find $y=1$, thus $B(0, 1)$. Hence, the coordinates of point $B$ are $\boxed{B(0, 1)}$.


Question: As shown in the Cartesian coordinate system, the line $y=\frac{1}{2}x+m$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively. The coordinates of point $A$ are $\left(-2, 0\right)$, and the coordinates of point $C$ are $\left(3, 0\right)$. If points $B$ and $C$ are symmetrical about line $l$, draw line $l$ in the auxiliary figure and then find the analytical expression of line $l$.

Solution: \textbf{Solution:} As shown in Figure 1, let the intersection of line $l$ with the x-axis be $E$, and it intersects the y-axis at $F$. Let $OE=x$, and connect $BE$, then $BE=CE=3-x$. In $\triangle BOE$, we have $BE^2=OB^2+OE^2$, $\therefore (3-x)^2=1+x^2$. Solving this, we find $x=\frac{4}{3}$, $\therefore E\left(\frac{4}{3}, 0\right)$. Let $H$ be the intersection of $BC$ and $l$, then $H\left(\frac{3}{2}, \frac{1}{2}\right)$. Let the equation of line $l$ be $y=kx+b$, then $\begin{cases}\frac{3}{2}k+b=\frac{1}{2}\\ \frac{4}{3}k+b=0\end{cases}$, solving this gives $\begin{cases}k=3\\ b=-4\end{cases}$, $\therefore$ the equation of line $l$ is \boxed{y=3x-4}.


Question: As shown in the Cartesian coordinate system, the line $y=\frac{1}{2}x+m$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$ respectively, with point $A$ having the coordinates $\left(-2, 0\right)$, and point $C$ having the coordinates $\left(3, 0\right)$. Point $M$ is the intersection of the line $y=x+m$ and line $l$, and point $N$ is in the first quadrant. When $\triangle BMN$ forms an isosceles right triangle, directly write out the coordinates of point $N$.

Solution:
\textbf{Solution:} From (1) we know $m=1$. Solving the system of equations $\begin{cases}y=x+1\\ y=3x-4\end{cases}$ gives $x=\frac{5}{2}$ and $y=\frac{7}{2}$, that is $M\left(\frac{5}{2}, \frac{7}{2}\right)$,\\
$\therefore BM=\sqrt{\left(\frac{5}{2}-0\right)^{2}+\left(\frac{7}{2}-1\right)^{2}}=\frac{5\sqrt{2}}{2}$. Point $N$ is in the first quadrant. When $\triangle BMN$ is an isosceles right triangle, the following cases are considered;\\
(1) When $BM$ is the hypotenuse, it is evident from the diagram that when $\triangle BMN$ is an isosceles right triangle, $N\left(\frac{5}{2}, 1\right)$, fulfilling: $BM^{2}=BN^{2}+MN^{2}$ and $BN=MN=\frac{5}{2}$, thus $\triangle BMN$ is an isosceles right triangle;\\
(2) When $BM$ is a leg, based on symmetry, the point $N$ at this instance, with $M$ being symmetrical to the point $N$ mentioned in (1), $\therefore N\left(\frac{5}{2}, -\frac{3}{2}\right)$, which contradicts with point $N$ being in the first quadrant and is thus discarded,\\
(3) When $BM$ is a leg, based on symmetry, the point $N$ at this instance, with $B$ being symmetrical to the point $N$ mentioned in (1), $\therefore N(5, 1)$, fulfilling $BN^{2}=MB^{2}+MN^{2}$ and $MN=MB$. When $N$ is on the upper left side of $BM$, there exists a point $N$ that meets the conditions,\\
In summary: $N\left(\frac{5}{2}, 1\right)$ and $N(5, 1)$ are the coordinates of point $N$, directly presented as \boxed{N\left(\frac{5}{2}, 1\right)} and \boxed{N(5, 1)}.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ passes through point $A(-2, 4)$, and intersects with the graph of the direct proportion function $y=-\frac{2}{3}x$ at point $B(a, 2)$. What is the value of $a$ and what is the equation of the linear function $y=kx+b$?

Solution:
\textbf{Solution:} Given the graph of the direct proportion function $y=-\frac{2}{3}x$ passes through point $B(a, 2)$,\\
$\therefore 2=-\frac{2}{3}a$, solving for $a$, we find $a=\boxed{-3}$,\\
$\therefore B(-3, 2)$.\\
Given that the graph of the linear function $y=kx+b$ passes through points $A(-2, 4)$, $B(-3, 2)$,\\
$\therefore \left\{\begin{array}{l}
-2k+b=4 \\
-3k+b=2
\end{array}\right.$, solving for, $\left\{\begin{array}{l}
k=\boxed{2} \\
b=\boxed{8}
\end{array}\right.$,\\
$\therefore$ the equation of the linear function $y=kx+b$ is $\boxed{y=2x+8}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ passes through point $A(-2, 4)$, and intersects with the graph of the direct proportion function $y=-\frac{2}{3}x$ at point $B(a, 2)$. Write down directly the solution set of the inequality $-\frac{2}{3}x>kx+b$ with respect to $x$.

Solution: \textbf{Solution:} From the given problem, it is known based on the graph that the solution set for $-\frac{2}{3}x > kx + b$ is: $\boxed{x < -3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the coordinates of point $A$ are $\left(5, 0\right)$, and point $B$ is on the positive semi-axis of the $y$-axis ($OB<OA$). The segment $AB$ is rotated clockwise by $90^\circ$ around point $A$ to obtain the segment $AC$. Perpendiculars are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$, respectively. What is the area of the quadrilateral $ABEC$?

Solution: \textbf{Solution:} Given that $A\left(5, 0\right)$, it follows that $OA=5$. Connect $AE$ and construct $\angle EAG=90^\circ$, with $AG$ intersecting the extension of $EC$ at point $G$, as shown in the diagram. Consequently, $\angle BAC=\angle EAG=90^\circ$, which implies $\angle CAG=\angle BAE$. Since $\angle CEO+\angle BAC=180^\circ$, it means that $\angle AEB+\angle AEC+\angle BAE+\angle CAE=180^\circ$. In $\triangle ACE$, we have $\angle AEC+\angle CAE+\angle ACE=180^\circ$. Given that $\angle AEB+\angle BAE=180^\circ-\angle ABE$, it follows that $\angle ABE+\angle ACE=180^\circ$. Additionally, since $\angle ACG+\angle ACE=180^\circ$, it implies $\angle ABE=\angle ACG$. Given $AB=AC$, it follows that $\triangle ABE\cong \triangle ACG$. Since $AE=AG$, we conclude $\angle AEG=\angle AGE=45^\circ$ and therefore $\angle AEO=\angle EAO=45^\circ$. Consequently, $OE=OA=5$, thus $AE=5\sqrt{2}$ and the area of quadrilateral $ABEC$, $S_{\text{quadrilateral }ABEC}=S_{\triangle AEG}=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=\boxed{25}$.


Question: As shown in the figure, in the Cartesian coordinate system, point $A$ has coordinates $\left(5, 0\right)$, and point $B$ lies on the positive half of the $y$-axis (with $OB<OA$). The line segment $AB$ is rotated $90^\circ$ clockwise about point $A$ to obtain the line segment $AC$. Perpendicular lines are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$, respectively. If $CE=4BE$, what is the equation of line $AC$?

Solution: \textbf{Solution:} Let $OB=a$. According to equation (1), we know that $OA=OE$. Since $OE=CD$, it follows that $OA=CD$. Since $\triangle AOB$ and $\triangle CDA$ are both right-angled triangles, and $AB=AC$, it therefore follows that $\triangle AOB \cong \triangle CDA$. Thus, $AD=OB=a$, and hence $CE=OD=a+5$, $BE=5-a$. Since $CE=4BE$, we have $a+5=4(5-a)$. Solving this, we find $a=3$. Therefore, $C(8, 5)$. Also, since $A(5, 0)$, let the equation of line AC be $y=kx+b$. Then $\begin{cases}5k+b=0\\ 8k+b=5\end{cases}$, from which we find $\begin{cases}k=\frac{5}{3}\\ b=-\frac{25}{3}\end{cases}$. Therefore, the equation of line AC is $y=\frac{5}{3}x-\frac{25}{3}$. Thus, the expression for line AC is $\boxed{y=\frac{5}{3}x-\frac{25}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the coordinate of point $A$ is $(5, 0)$. Point $B$ is located on the positive half of the $y$-axis ($OB<OA$). Line segment $AB$ is rotated clockwise around point $A$ by $90^\circ$ to obtain line segment $AC$. Perpendiculars are drawn from point $C$ to the $x$-axis and the $y$-axis, with the feet of the perpendiculars being $D$ and $E$ respectively. Under the conditions of (2), point $P$ is a point on the extension of $OE$. Line $PC$ is connected, and the bisector of $\angle PCD$ intersects the $x$-axis at point $F$. If $\triangle PCF$ is an isosceles triangle, what are the coordinates of point $F$?

Solution:
\textbf{Solution:} Let the coordinates of point $F$ be $\left(m, 0\right)$,

$\because$ $CF$ bisects $\angle PCD$,

$\therefore$ $\angle PCF=\angle DCF$,

(1) When $PF=PC$, then $\angle PFC=\angle PCF=\angle DCF$,

$\therefore$ $PF\parallel CD$,

$\therefore$ $F$ coincides with $O$,

$\therefore$ $F\left(0, 0\right)$;

(2) When $PF=CF$, draw $FM\perp CP$ at point $F$ with the foot $M$,

then $PM=CM$, $\angle CMF=\angle CDF=90^\circ$,

Also $\because$ $\angle PCF=\angle DCF$, $CF=CF$,

$\therefore$ $\triangle CMF\cong \triangle CDF$,

$\therefore$ $CM=CD=5$,

$\therefore$ $PC=10$, in $Rt\triangle PCE$, by the Pythagorean theorem, $PE=6$,

$\therefore$ $OP=11$, in $Rt\triangle POF$, $PF^{2}=OF^{2}+OP^{2}$,

in $Rt\triangle CDF$, $CF^{2}=CD^{2}+DF^{2}$,

$\therefore$ $OF^{2}+OP^{2}=CD^{2}+DF^{2}$,

$\therefore$ $\left(-m\right)^{2}+11^{2}=5^{2}+\left(8-m\right)^{2}$,

Solving gives $m=-2$,

$\therefore$ $F\left(-2, 0\right)$;

(3) When $CP=CF$, extend $CF$ to intersect the $y$-axis at point $N$,

$\because$ $CD\parallel PN$, and $\angle PCF=\angle DCF$,

$\therefore$ $\angle CNP=\angle PCF=\angle DCF$,

$\therefore$ $PN=PC$, draw $PQ\perp CF$ at point $P$ with the foot $Q$,

then $CQ=QN$, $\angle PQC=\angle CDF=90^\circ$,

$\therefore$ $\triangle PCQ\cong \triangle FCD$,

$\therefore$ $CQ=CD=5$,

$\therefore$ $CN=10$, in $Rt\triangle CEN$, by the Pythagorean theorem, $EN=6$,

$\therefore$ $ON=1$,

$\therefore$ $N\left(0, -1\right)$,

$\because$ $C\left(8, 5\right)$, let the equation of line $CN$ be $y=kx+b$,

then $\begin{cases}8k+b=5\\ b=-1\end{cases}$, solving gives $\begin{cases}k=\frac{3}{4}\\ b=-1\end{cases}$,

$\therefore$ the equation of line $CN$ is $y=\frac{3}{4}x-1$,

When $y=0$, solving gives $x=\frac{4}{3}$,

$\therefore$ $F\left(\frac{4}{3}, 0\right)$.

In summary, when $\triangle PCF$ is an isosceles triangle, the coordinates of point $F$ are $\boxed{\left(0, 0\right)}$, or $\boxed{\left(-2, 0\right)}$, or $\boxed{\left(\frac{4}{3}, 0\right)}$.


Question: As shown in the figure, the line $l_{1}\colon y=-x-3$ intersects with the line $l_{2}$ passing through the point $A(0,3)$ at point $C(m,1)$, and intersects with the y-axis at point $B$. Find the analytical expression of the line $l_{2}$.

Solution:
\textbf{Solution:} Substitute C($m$,1) into $y=-x-3$ to get $-m-3=1$,\\
$\therefore m=-4$,\\
$\therefore$ C($-4$,1). Let the equation of line $l_{2}$ be $y=kx+b$. Substituting A(0,3) and C($-4$,1) into it, we get
\[
\begin{cases}
b=3, \\
-4k+b=1,
\end{cases}
\]
which, upon solving, gives
\[
\begin{cases}
k=\frac{1}{2}, \\
b=3,
\end{cases}
\]
$\therefore$ The equation of line $l_{2}$ is $\boxed{y=\frac{1}{2}x+3}$;


Question: As shown in the figure, the line $l_{1}\colon y=-x-3$ intersects with the line $l_{2}$ passing through point $A(0,3)$ at point $C(m,1)$, and intersects with the y-axis at point $B$. Point $P$ is on the line $l_{1}$, $PQ\parallel x$-axis, and intersects line $l_{2}$ at point $Q$. If $PQ=AB$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Solve: In the line $l_{1}\colon y=-x-3$, let $x=0$, we get $y=-3$,\\
$\therefore B(0,-3)$,\\
$\therefore AB=3-(-3)=6$,\\
Let $P(-b-3,b)$, since $PQ\parallel x$ axis, we get $Q(2b-6,b)$,\\
$PQ=\left|-b-3-\left(2b-6\right)\right|=AB=6$,\\
Solving, we get $b=3$ or $b=-1$,\\
$\therefore P=\boxed{(-6,3)}$ or \boxed{(-2,-1)}.


Question: As shown in the diagram, in the Cartesian coordinate system, it is known that point $A(1,0)$. The $x$-axis is rotated clockwise around point $A$ by $60^\circ$ to intersect the $y$-axis at point $B$, and then point $B$ is rotated clockwise around point $A$ by $90^\circ$ to obtain point $C$. What is the equation of line $BC$?

Solution:
\textbf{Solution}: Given the problem statement, since the $x$-axis rotates $60^\circ$ clockwise about point $A$ and intersects the $y$-axis at point $B$,

therefore $\angle OAB = 60^\circ$,

since point $A(1, 0)$,

therefore $OA=1$,

therefore $AB=2$, $OB=\sqrt{3}$,

therefore $B(0, \sqrt{3})$,

since point $B$ rotates $90^\circ$ clockwise about point $A$ to get to point $C$,

therefore $AC=AB=2$, $\angle BAC=90^\circ$,

draw $CG \perp x$-axis at point $G$,

therefore $\angle CAG=30^\circ$,

therefore $CG=1$, $AG=\sqrt{3}$,

therefore $C(1+\sqrt{3}, 1)$,

let the equation of line $BC$ be $y=kx+b$,

then $\left\{\begin{array}{l}b=\sqrt{3}\\ (1+\sqrt{3})k+b=1\end{array}\right.$,

solving gets $\left\{\begin{array}{l}k=-2+\sqrt{3}\\ b=\sqrt{3}\end{array}\right.$,

therefore $y=(-2+\sqrt{3})x+\sqrt{3}$;

thus, the equation of line $BC$ is $\boxed{y=(-2+\sqrt{3})x+\sqrt{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, given point $A(1,0)$, the $x$-axis is rotated clockwise by $60^\circ$ about point $A$ to intersect the $y$-axis at point $B$, and then point $B$ is rotated clockwise by $90^\circ$ about point $A$ to obtain point $C$. If point $Q$ is a point in the Cartesian coordinate system, and it satisfies that the quadrilateral $ABCQ$ is a parallelogram, what are the coordinates of point $Q$?

Solution:
\textbf{Solution:} Let $Q(x, y)$, \\
since quadrilateral $ABCQ$ is a parallelogram, \\
therefore $AC$ and $BQ$ are the diagonals of the parallelogram, \\
the midpoint of $AC$ is $\left(\frac{2+\sqrt{3}}{2}, \frac{1}{2}\right)$, the midpoint of $BQ$ is $\left(\frac{x}{2}, \frac{y+\sqrt{3}}{2}\right)$, \\
therefore $\frac{2+\sqrt{3}}{2}=\frac{x}{2}$, $\frac{1}{2}=\frac{y+\sqrt{3}}{2}$, \\
therefore $x=2+\sqrt{3}$, $y=1-\sqrt{3}$, \\
therefore $Q(x, y)=\boxed{(2+\sqrt{3}, 1-\sqrt{3})}$;


Question: As shown in the Cartesian coordinate system, it is known that point $A(1,0)$. The $x$-axis is rotated clockwise about point $A$ by $60^\circ$ to intersect with the $y$-axis at point $B$. Then, point $B$ is rotated clockwise about point $A$ by $90^\circ$ to obtain point $C$. Are there points $M$ on line $BC$ and point $N$ on the $y$-axis, such that the quadrilateral formed by points $M$, $N$, $A$, and $C$ is a parallelogram? Write down the coordinate of point $M$.

Solution:
\textbf{Solution:} \\
Given that point $M$ is on line $BC$ and point $N$ is on the $y$-axis, let $M(t, (-2+\sqrt{3})t+\sqrt{3})$ and $N(0, n)$, \\
(1) When diagonals $AC$ and $MN$ form a parallelogram, the $x$-coordinate of the midpoint of $AC$ is $\frac{2+\sqrt{3}}{2}$, and the $x$-coordinate of the midpoint of $MN$ is $\frac{t}{2}$, \\
Thus, $\frac{2+\sqrt{3}}{2}=\frac{t}{2}$, \\
Therefore, $t=2+\sqrt{3}$, \\
Hence, $M(2+\sqrt{3}, \sqrt{3}-1)$; \\
(2) When diagonals $AM$ and $CN$ form a parallelogram, the $x$-coordinate of the midpoint of $AM$ is $\frac{t+1}{2}$, and the $x$-coordinate of the midpoint of $CN$ is $\frac{1+\sqrt{3}}{2}$, \\
Thus, $\frac{t+1}{2}=\frac{1+\sqrt{3}}{2}$, \\
Therefore, $t=\sqrt{3}$, \\
Hence, $M(\sqrt{3}, 3-\sqrt{3})$; \\
(3) When diagonals $AN$ and $CM$ form a parallelogram, the $x$-coordinate of the midpoint of $AN$ is $\frac{1}{2}$, and the $x$-coordinate of the midpoint of $CM$ is $\frac{1+\sqrt{3}+t}{2}$, \\
Thus, $\frac{1}{2}=\frac{1+\sqrt{3}+t}{2}$, \\
Therefore, $t=-\sqrt{3}$, \\
Hence, $M(-\sqrt{3}, -3+3\sqrt{3})$; \\
In conclusion, the coordinates of point $M$ are $\boxed{(2+\sqrt{3}, \sqrt{3}-1)}$, $\boxed{(\sqrt{3}, 3-\sqrt{3})}$, or $\boxed{(-\sqrt{3}, -3+3\sqrt{3})}$.


Question: As shown in the figure, the line $y = \frac{2}{3}x + 4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. What are the coordinates of point $A$ and point $B$?

Solution:
\textbf{Solution:} Given $y= \frac{2}{3}x+4$, let $x=0$, then $y=4$,\\
$\therefore A(0,4)$;\\
Let $y=0$, then $\frac{2}{3}x+4=0$, solving for $x$ gives $x=-6$,\\
$\therefore B(-6,0)$;\\
Hence, the coordinates of point $A$ are $A=\boxed{(0,4)}$, and the coordinates of point $B$ are $B=\boxed{(-6,0)}$.


Question: As illustrated, the line $y = \frac{2}{3}x + 4$ intersects the $x$-axis at point $B$ and the $y$-axis at point $A$. Translate line $AB$ to intersect the positive half of the $x$-axis at point $E$ and the $y$-axis at point $F$. Let $P$ be a point in the third quadrant on line $EF$. Draw $PG \perp x$-axis at point $G$. Given that the area of $\triangle PAB = 20$ and $GE = 12$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} As shown in the diagram, since $PE \parallel AB$,\\
it follows that $S_{\triangle PAB} = S_{\triangle EAB} = 20$,\\
hence $\frac{1}{2} AO \times BE = 20$, that is $4 \times (6 + OE) = 40$,\\
therefore $OE = 4$.\\
Thus, $E(4, 0)$.\\
Since $GE = 12$,\\
it follows that $GO = 8$.\\
Thus, $G(-8, 0)$.\\
Let the equation of EF be $y = \frac{2}{3}x + b$. Substituting the coordinates of E yields $0 = \frac{2}{3} \times 4 + b$,\\
solving for $b$ gives $b = -\frac{8}{3}$.\\
Therefore, the equation of line EF is $y = \frac{2}{3}x - \frac{8}{3}$.\\
Substituting $x = -8$ results in $y = \frac{2}{3} \times (-8) - \frac{8}{3} = -8$,\\
thus, $P(-8, -8)$.\\
Hence, the coordinates of point $P$ are $P = \boxed{(-8, -8)}$.


Question: As shown in the graph, in the Cartesian coordinate system, the coordinates of vertices A and C of parallelogram OABC are (2,0) and (1,2), respectively. Find the equation of line AC.

Solution:
\textbf{Solution:} In the parallelogram $OABC$, we have $BC=OA$ and $BC\parallel OA$. Given that the coordinates of vertices $A$ and $C$ are $(2, 0)$ and $(1, 2)$, respectively, it follows that the coordinates of point $B$ are $(3, 2)$. Let the equation of line $AC$ be $y=kx+b$. Substituting $(2, 0)$ and $(1, 2)$ into this equation, we have:
\[
\begin{cases}
2k+b=0\\
k+b=2
\end{cases}
\]
Solving the system, we find:
\[
\begin{cases}
k=-2\\
b=4
\end{cases}
\]
Therefore, the equation of line $AC$ is: $y=-2x+4=\boxed{-2x+4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the parallelogram OABC have coordinates (2, 0) and (1, 2), respectively. If $D$ is a moving point on the line segment $AC$ (with $D$ not coinciding with $A$ or $C$), and the area of $\triangle DBC$ is 3, determine the coordinates of point $D$.

Solution:
\textbf{Solution:} Given that $BC=OA=2$, and the area of $\triangle DBC$ is 3,\\
$\therefore$ Let the height from $D$ to $BC$ in $\triangle DBC$ be $h$,\\
then $\frac{1}{2}BC\cdot h=3$,\\
Solving gives: $h=3$,\\
$\therefore$ The y-coordinate of point $D$ is $2+3=5$ or $2-3=-1$,\\
Also, since $D$ is a point on line $AC$,\\
$\therefore$ In the equation $y=-2x+4$,\\
when $y=5$, $-2x+4=5$ gives $x=-\frac{1}{2}$,\\
when $y=-1$, $-2x+4=-1$ gives $x=\frac{5}{2}$,\\
$\therefore$ The coordinates of point $D$ are $\boxed{\left(-\frac{1}{2}, 5\right)}$ or $\boxed{\left(\frac{5}{2}, -1\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices A and C of the parallelogram OABC are located at (2,0) and (1,2), respectively. Is there a point P on the y-axis such that $\triangle PAC$ forms a right triangle with P as the right angle vertex? What are the coordinates of P?

Solution:
\textbf{Solution:} Let the coordinates of point $P$ be $(0, m)$. Based on the problem statement, we have: $PC^{2}=1^{2}+(2-m)^{2}$, $AC^{2}=1^{2}+2^{2}=5$, $PA^{2}=2^{2}+m^{2}$,

(1) When point $C$ is the right angle vertex, $PC^{2}+AC^{2}=PA^{2}$,

$\therefore 1^{2}+(2-m)^{2}+1^{2}+2^{2}=2^{2}+m^{2}$,

Solving this, we get: $m=\frac{3}{2}$,

At this point, the coordinates of $P$ are $(0, \frac{3}{2})$;

(2) When point $A$ is the right angle vertex, $PA^{2}+AC^{2}=PC^{2}$,

$\therefore 2^{2}+m^{2}=1^{2}+(2-m)^{2}+1^{2}+2^{2}$,

Solving this, we get: $m=-1$,

At this point, the coordinates of $P$ are $(0, -1)$;

In summary, the coordinates of point $P$ are $\boxed{(0, \frac{3}{2})}$ or $\boxed{(0, -1)}$.


Question: As shown in the figure, the line $y=kx+b$ ($k\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. Find the values of $k$ and $b$, and the coordinates of point $C$?

Solution:
\textbf{Solution:} Given the line $y=kx+b$ ($k\neq 0$) intersects the x-axis at point A(6, 0) and the y-axis at point B(0, 3),\\
$\therefore \left\{\begin{array}{l}
6k+b=0\\
b=3
\end{array}\right.$\\
Solving this system, we get: $\left\{\begin{array}{l}
k=-\frac{1}{2}\\
b=3
\end{array}\right.$,\\
Hence, the equation of line $AB$ is: $y=-\frac{1}{2}x+3$,\\
By solving the system: $\left\{\begin{array}{l}
y=-\frac{1}{2}x+3\\
y=x
\end{array}\right.$,\\
We find: $\left\{\begin{array}{l}
x=2\\
y=2
\end{array}\right.$,\\
Therefore, $C(2, 2).$\\
$\therefore k=\boxed{-\frac{1}{2}}, b=\boxed{3}, \boxed{C\left(2, 2\right)}.$


Question: As shown in the figure, the line $y=kx+b$ ($k\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. What is the area of $\triangle BOC$?

Solution:
\textbf{Solution:} Given the conditions, we have $B(0, 3)$, $O(0, 0)$, $C(2, 2)$,\\
$\therefore S_{\triangle BOC}=\frac{1}{2}OB\cdot|x_{C}|=\frac{1}{2}\times 3\times 2=\boxed{3}$.


Question: As shown in the figure, the line $y=kx+b$ ($k\neq 0$) intersects the x-axis at point $A(6,0)$ and the y-axis at point $B(0,3)$, and intersects the line $y=x$ at point $C$. Directly write down the solution to the inequality $kx+b\geq x$.

Solution:
\textbf{Solution:} Upon observing the graph, in conjunction with $C(2, 2)$,

$\therefore$ the solution set for the inequality $kx+b\geq x$ is: $x\boxed{x \le 2}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y_1=kx+b$ ($k\ne 0$, $k$, $b$ are constants), denoted as $l_1$, intersects with the $x$ axis and $y$ axis at points $A$ and $B$ respectively, where the coordinates of point $B$ are (0,5). The graph of the direct proportion function $y_2=2x$, denoted as $l_2$, intersects with $l_1$ at point $C(m, 4)$. Find the expression of the linear function?

Solution:
\textbf{Solution:} Substitute point C$(2, 4)$ into the direct proportion function $y=2x$, and we find $m=2$.\\
$\therefore$C$(2, 4)$.\\
Substitute points C$(2, 4)$ and B$(0, 5)$ into the linear function $y=kx+b$, yielding the system of equations
\[
\left\{
\begin{array}{l}
2k+b=4 \\
b=5
\end{array}
\right.
\]
Solving this, we find $k=-\frac{1}{2}$ and $b=5$.\\
$\therefore$The equation of the linear function is $y=-\frac{1}{2}x+5$, thus the equation is $\boxed{y=-\frac{1}{2}x+5}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y_1=kx+b$ (where $k\ne 0$, and $k$, $b$ are constants) $l_1$ intersects the $x$ axis and $y$ axis at points $A$ and $B$ respectively. The coordinates of point $B$ are $(0, 5)$. The graph of the directly proportional function $y_2=2x$, $l_2$, intersects $l_1$ at point $C(m, 4)$. For what value of $x$ is $y_1 > y_2$?

Solution:
\textbf{Solution:} Given the problem, the line $l_{1}: y=-\frac{1}{2}x+5$ and the line $l_{2}: y_{2}=2x$ intersect at point $C(2, 4)$,\\
$\therefore$ when $x<2$, $y_{1}>y_{2}$.\\
Thus, the answer is $\boxed{x<2}$.


Question: Synthesis and Exploration: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y_{1}=-\frac{1}{2}x+6$ intersects with the x-axis and y-axis at point B and C, respectively, and intersects with the line $l_{2}: y_{2}=\frac{1}{2}x$ at point A. Determine the coordinates of point A.

Solution: \textbf{Solution:} Solve the system of equations for the two lines to get $\begin{cases}y=-\frac{1}{2}x+6\\ y=\frac{1}{2}x\end{cases}$, and we find: $\begin{cases}x=6\\ y=3\end{cases}$, $\therefore$ the coordinates of point A are $\boxed{(6,3)}$;


Question: Synthesis and Inquiry: As shown in the figure, in the Cartesian coordinate system, the line $l_1\colon y_1 = -\frac{1}{2}x + 6$ intersects the x-axis and y-axis at points B and C, respectively, and intersects with the line $l_2\colon y_2 = \frac{1}{2}x$ at point A. When $y_1 < y_2$, write down directly the range of values for x.

Solution: \textbf{Solution:} Given point A$(6,3)$ and the graph, when $y_1<y_2$, then $\boxed{x>6}$.


Question: In the Cartesian coordinate system, the line $l_{1}\colon y_{1}=-\frac{1}{2}x+6$ intersects the x-axis and y-axis at points B and C, respectively, and intersects with the line $l_{2}\colon y_{2}=\frac{1}{2}x$ at point A. Is there a point Q in the plane such that the quadrilateral with vertices O, C, A, and Q is a parallelogram? Find the coordinates of point Q.

Solution:
\textbf{Solution:} There exists a point Q such that the quadrilateral with vertices O, C, A, Q is a parallelogram; as shown in the diagram, consider three cases:
- $\because C(0, 6)$, A(6,3)
  - (1) When the quadrilateral OAQ$_{1}$C is a parallelogram, $OC=AQ_{1}$, \\
    $\therefore$ the x-coordinate of point Q$_{1}$ is 6, and the y-coordinate is $3+6=9$,\\
    $\therefore$ the coordinates of Q$_{1}$ are (6,9),
  - (2) When the quadrilateral OQ$_{2}$AC is a parallelogram, $OC=AQ_{2}$,\\
    the x-coordinate of point Q$_{2}$ is 6, and the y-coordinate is $3-6=-3$,\\
    $\therefore$ the coordinates of Q$_{2}$ are (6, −3),
  - (3) When the quadrilateral OACQ$_{3}$ is a parallelogram,\\
    $\because AC=\sqrt{6^{2}+(6-3)^{2}}=3\sqrt{5}$, $AO=\sqrt{3^{2}+6^{2}}=3\sqrt{5}$,\\
    $\therefore AC=AO$,\\
    $\therefore$ the quadrilateral OACQ$_{3}$ is a rhombus,\\
    Given that a rhombus is an axis-symmetric figure, with $OC$ as a diagonal,\\
    point Q$_{3}$ is symmetric to point A with respect to OC,\\
    $\therefore$ the coordinates of Q$_{3}$ are (−6, 3),\\
Summarizing, the coordinates of point Q are: $\boxed{(6,9)}$ or $\boxed{(6, −3)}$ or $\boxed{(−6, 3)}$.


Question: As shown in the diagram, it is known that in parallelogram $OABC$, point $O$ is the coordinate vertex, point $A(3, 0)$, $C(1, 2)$, and the graph of the function $y=\frac{k}{x} \ (k\ne 0)$ passes through point $C$. Determine the value of $k$ and the equation of line $OB$.

Solution:
\textbf{Solution:} According to the problem, point $C(1, 2)$ is on the graph of the inverse proportion function $y=\frac{k}{x}(k\ne 0)$, \\
hence $k=xy=2$, \\
since $A(3, 0)$, \\
hence $CB=OA=3$, \\
also, $CB$ is parallel to the x-axis, \\
hence $B(4, 2)$, \\
let the equation of line $OB$ be $y=ax$, \\
hence $2=4a$, \\
hence $a=\frac{1}{2}$, \\
therefore, the equation of line $OB$ is $\boxed{y=\frac{1}{2}x}$;


Question: As shown in the figure, it is known that in the parallelogram $OABC$, point $O$ is the origin of the coordinates, point $A(3, 0)$, and point $C(1, 2)$. The graph of the function $y=\frac{k}{x} (k \neq 0)$ passes through point $C$. What is the perimeter of the parallelogram $OABC$?

Solution:
\textbf{Solution:} Construct $CD\perp OA$ at point $D$, \\
$\because$ $C(1, 2)$, \\
$\therefore$ $OC=\sqrt{1^2+2^2}=\sqrt{5}$, \\
In parallelogram $OABC$, $CB=OA=3$, $AB=OC=\sqrt{5}$, \\
$\therefore$ the perimeter of parallelogram $OABC$ is: $3+3+\sqrt{5}+\sqrt{5}=6+2\sqrt{5}$, \\
hence, the perimeter of parallelogram $OABC$ is $\boxed{6+2\sqrt{5}}$.


Question: Given, as shown in the figure, the graph of a linear function passes through points $P(3, 2)$ and $B(0, -2)$, and intersects the x-axis at point $A$. What is the analytical expression of the linear function?

Solution:
\textbf{Solution:} Let the linear function be $y=kx+b$. Substituting points $P(3,2)$ and $B(0, -2)$ into $y=kx+b$, we get
\[
\begin{cases}
3k+b=2\\
b=-2
\end{cases}
\]
Solving these equations, we obtain
\[
\begin{cases}
k=\frac{4}{3}\\
b=-2
\end{cases}
\]
$\therefore$ The equation of the linear function is $\boxed{y=\frac{4}{3}x-2}$.


Question: As shown in the figure, the graph of a linear function passes through points $P(3, 2)$ and $B(0, -2)$, and intersects with the x-axis at point $A$. Point $M$ is on the y-axis, and the area of $\triangle ABM$ is $\frac{15}{4}$. Find the coordinates of point $M$.

Solution:
\textbf{Solution:} When $y=0$, $\frac{4}{3}x-2=0$, solving for $x$ gives $x=\frac{3}{2}$, so $A\left(\frac{3}{2},0\right)$,\\
$\therefore OA=\frac{3}{2}$.\\
$\because$ point M is on the y-axis, and the area of $\triangle ABM$ is $\frac{15}{4}$,\\
$\therefore S_{\triangle ABM}=\frac{1}{2}BM\cdot OA=\frac{15}{4}$,\\
$\therefore BM=5$,\\
$\because B(0,-2)$,\\
$\therefore \boxed{M(0,3)} or \boxed{M(0,-7)}$.


Question: In the Cartesian coordinate system $xOy$, line $y=-2x+a$ intersects the $y$-axis at point $A$ and intersects the line $y=x+1$ at point $P(3, b)$. $B$ is a point on the line $y=x+1$. Find the values of $a$ and $b$?

Solution:
\textbf{Solution:} Substituting point P$(3, b)$ into the line equation $y=x+1$, we get: $b=4$. Substituting P$(3, 4)$ into $y=-2x+a$, we derive: $a=10$.\\
Hence, $a=\boxed{10}$ and $b=\boxed{4}$.


Question: In the Cartesian coordinate system $xOy$, the line $y=-2x+a$ intersects with the $y$-axis at point $A$ and intersects with the line $y=x+1$ at point $P(3,b)$. $B$ is a point on the line $y=x+1$. What are the coordinates of point $B$ when the segment $AB$ is at its shortest length?

Solution:
\textbf{Solution:} When $AB\perp$ line $y=x+1$, the segment $AB$ is shortest. Label the intersection point of line $y=x+1$ with the $y$-axis $(0,1)$ as $E$. Given the conditions, we have $A(0,10)$, and $\angle AEB=45^\circ$, $\triangle AEB$ is an isosceles right triangle,\\
$\therefore AE=9$, $AB=BE=\frac{9\sqrt{2}}{2}$\\
$\therefore$ The x-coordinate of B is $\frac{9\sqrt{2}}{2}\times \frac{\sqrt{2}}{2}=\frac{9}{2}$, and the y-coordinate is $\frac{9}{2}+1=\frac{11}{2}$,\\
$\therefore \boxed{B\left(\frac{9}{2}, \frac{11}{2}\right)}$.


Question: In the Cartesian coordinate system $xOy$, the line $y=-2x+a$ intersects the $y$ axis at point $A$ and intersects the line $y=x+1$ at point $P(3,b)$. Point $B$ lies on the line $y=x+1$. Find a point $C$ on the $x$ axis such that the value of $AC-PC$ is maximized. Please write the coordinates of point $C$ and find the maximum value.

Solution:
\textbf{Solution:} When points $A$, $P$, and $C$ are collinear, the maximum value of $AC-PC$ is equal to $AP$. Solving $0 = -2x + 10$ gives $x = 5$, thus $C(5,0)$.\\
Since $P(3,4)$,\\
$AP= \sqrt{AH^{2}+PH^{2}}=\sqrt{(10-4)^{2}+3^{2}}=\boxed{3\sqrt{5}}$.


Question: As shown in the figure, a rectangular paper ABCO is placed in the Cartesian coordinate system, with BC = 10. After folding the paper, point B exactly falls on the x-axis, denoted as $B'$, and the fold line is CE. Given that the ratio $OC:OB'=4:3$, what are the coordinates of point $B'$?

Solution: \textbf{Solution:} Since quadrilateral OABC is a rectangle, $\angle$AOC$=90^\circ$. Given OC$\colon$ OB'$=4\colon 3$, it follows that B$^{\prime }$C$\colon$ OB$^{\prime }$=$5\colon 3$. Given B$^{\prime }$C=BC=10, it implies OB$^{\prime }$=6. Therefore, the coordinates of point B$^{\prime}$ are: $\boxed{(6,0)}$.


Question: As shown in the diagram, a rectangular paper ABCO is placed in a Cartesian coordinate system, where BC = 10. The paper is folded so that point B precisely lands on the x-axis, denoted as $B'$, with CE being the crease. Given that the ratio $OC:OB'=4:3$, what is the equation of the line on which the crease CE lies?

Solution:
\textbf{Solution:} After folding the paper, point B lands precisely on point B' on the x-axis, with CE being the fold line. Therefore, $\triangle CBE \cong \triangle CB^{\prime}E$, which implies that BE=B$^{\prime}E$, and CB$^{\prime}=CB=OA$. Given OB$^{\prime}=6$ and the ratio OC:OB'=4:3, it follows that OC=8. Let AE=a, then EB$^{\prime}=EB=8-a$, and AB$^{\prime}=AO-OB^{\prime}=10-6=4$. By the Pythagorean theorem, we obtain $a^{2}+4^{2}=(8-a)^{2}$. Solving for a yields $a=3$. Therefore, the coordinates of point E are $(10,3)$, and the coordinates of point C are $(0,8)$. Let the equation of line CE be $y=kx+b$. According to the given conditions, we have
\[
\begin{cases}
3=10k+b\\
8=b
\end{cases}
\]
Solving this, we find:
\[
\begin{cases}
k=-\frac{1}{2}\\
b=8
\end{cases}
\]
Therefore, the equation of line CE is $y= -\frac{1}{2}x+8$, so the answer is $\boxed{y= -\frac{1}{2}x+8}$.


Question: As shown in the figure, a rectangular paper ABCO is placed in the Cartesian coordinate system, with $BC=10$. After folding the paper, point B falls exactly on the x-axis, denoted as $B'$, and the crease is CE. Given that the ratio $OC:OB'=4:3$. If point P is a moving point on the y-axis, when $\triangle CPE$ is an isosceles triangle, find the coordinates of point P.

Solution:
\textbf{Solution:} Let the coordinates of point E be (10, 3), and those of C be (0, 8), $\therefore$ the coordinates of B are (10, 8), $\therefore$ CE= $\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt{5}$,\\
(1) When PC=CE, the coordinates of P are ($8+5\sqrt{5}$, 0) and ($8-5\sqrt{5}$, 0);\\
(2) When PE=CE, the coordinates of P are ($-2$, 0);\\
(3) When PC=PE, let the coordinates of P be (0, y), then PC=$8-y$, PE= $\sqrt{10^2+(3-y)^2}$, hence $(8-y)^2=100+(3-y)^2$, solving this gives: $64-16y+y^2=100+9-6y+y^2$, $-10y=45$, $y=-\frac{9}{2}$, the coordinates of P are $\left(-\frac{9}{2}, 0\right)$;\\
Overall, the coordinates of P are ($8+5\sqrt{5}$, 0), ($8-5\sqrt{5}$, 0), ($-2$, 0), $\boxed{\left(-\frac{9}{2}, 0\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system xOy, OA = OB, $\angle AOB = 90^{\circ}$. A line parallel to the x-axis is drawn through point B, which intersects with the line passing through point A, $y = -x + b$, at point C. Given that the coordinates of point A are (2, 1), what are the coordinates of points C and D?

Solution:
\textbf{Solution:} Let the line $y=-x+b$ pass through point A, with the coordinates of point A being $(2,1)$.\\
Therefore, we have $1=-2+b$,\\
which implies $b=3$,\\
thus, the equation of the line is: $y=-x+3$,\\
Given that $\angle AOB=90^\circ$,\\
we find $B(-1,2)$,\\
Constructing a line parallel to the x-axis through point B,\\
implies that the y-coordinates of points B and C are the same, both are 2,\\
Given that C lies on the line,\\
the coordinates of point C are $\boxed{(1,2)}$,\\
The line $y=-x+3$ intersects the x-axis at point D,\\
thus, the coordinates of point D are $\boxed{(3,0)}$;


Question: As shown in the graph, in the Cartesian coordinate system $xOy$, we have $OA=OB$ and $\angle AOB=90^\circ$. A line parallel to the $x$-axis is drawn through point B, and it intersects with the line passing through point A, which is $y=-x+b$, at point C. Given that the coordinates of point A are (2, 1), what is the degree measure of $\angle ACB$?

Solution:
\textbf{Solution:} Let the line intersect the y-axis at point E, as shown in the figure:\\
$\therefore OD=OE$,\\
$\therefore \angle OEC=\angle ODC=45^\circ$,\\
Also, since $BC\parallel OD$,\\
$\therefore \angle BCA+\angle CDO=180^\circ$,\\
$\therefore \angle BCA=135^\circ$,\\
Therefore, the measure of $\angle ACB$ is $\boxed{135^\circ}$.


Question: The graph of the linear function $y=kx+b$ passes through points $A(0, 2)$ and $B(-2, 0)$. Find the analytic expression of this linear function.

Solution:
Based on the information given in the question, we have
\[
\left\{
\begin{array}{l}
b=2 \\
-2k+b=0
\end{array}
\right.
\]
Solving this, we obtain
\[
\left\{
\begin{array}{l}
k=1 \\
b=2
\end{array}
\right.
\]
Hence, the equation of the linear function is \boxed{y=x+2}.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through point $A(0, 2)$ and point $B(-2, 0)$. What is the area of the triangle formed by the linear function and the coordinate axes?

Solution:
\textbf{Solution:} Given that $A(0, 2)$, $B(-2, 0)$,\\
it follows that $OB=OA=2$,\\
thus, the area of $\triangle AOB$ is given by ${S}_{\triangle AOB}=\frac{1}{2}OA\cdot OB=\frac{1}{2}\times 2\times 2=\boxed{2}$,\\
which means the area of the triangle formed by the linear function and the coordinate axes is \boxed{2}.


Question: Given the linear function $y=kx+b$ whose graph passes through points $A(0, 2)$ and point $B(-2, 0)$, for what values of $x$ is $y > 0$?

Solution: \textbf{Solution:} From the graph, it is known that when $x>-2$, $y>0$. Therefore, $\boxed{x>-2}$.


Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin. The line $y=2x-1$ intersects the line $y=\frac{3}{4}x+\frac{3}{2}$ at point A. A perpendicular line to the x-axis is drawn through point A, with point B being the foot of the perpendicular. The x-coordinate of point C is $-1$, and point C lies on the line $y=2x-1$. Connect BC. What are the coordinates of point A?

Solution:
\textbf{Solution:} Solve the system of equations:
\[
\left\{
\begin{array}{l}
y=2x-1 \\
y=\frac{3}{4}x+\frac{3}{2}
\end{array}
\right.
\]
to get
\[
\left\{
\begin{array}{l}
x=2\\
y=3
\end{array}
\right.
\]
$\therefore$ The coordinates of point A are $\boxed{(2,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system, point O is the origin. The line $y=2x-1$ and the line $y=\frac{3}{4}x+\frac{3}{2}$ intersect at point A. A perpendicular line to the x-axis is drawn from point A, and point B is the foot of the perpendicular. The x-coordinate of point C is $-1$, and point C lies on the line $y=2x-1$. Connect points B and C. What is the degree measure of $\angle CBO$?

Solution:
\textbf{Solution:} Construct $CD\perp x$-axis at D, \\
$\because A(2,3)$, \\
$\therefore B(2,0)$, \\
$\because$ point C has an x-coordinate of $-1$ and lies on the line $y=2x-1$, \\
$\therefore y=2\times(-1)-1=-3$, \\
$\therefore C(-1,-3)$, \\
$\therefore BD=3$, $CD=3$, \\
$\therefore$ triangle $CBD$ is an isosceles right triangle, \\
$\therefore \angle CBO=\boxed{45^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -\frac{1}{2}x + 5$, denoted as $l_{1}$, intersects the x-axis and y-axis at points A and B, respectively. The graph of the direct proportionality function, denoted as $l_{2}$, intersects $l_{1}$ at point C $(m, 4)$. What is the value of $m$ and the equation of $l_{2}$?

Solution:
\textbf{Solution:} Substitute $C(m,4)$ into the linear function $y=-\frac{1}{2}x+5$, we get: $4=-\frac{1}{2}m+5$, solving for $m$ gives: $m=\boxed{2}$.\\
Therefore, for $C(2,4)$, let the equation of line $l_{2}$ be $y=ax$, then we have $4=2a$, solving for $a$ gives $a=2$,\\
therefore, the equation of line $l_{2}$ is $\boxed{y=2x}$.


Question: In the Cartesian coordinate system xOy, the graph of the linear function $y = -\frac{1}{2}x + 5$, denoted by $l_1$, intersects the x and y axes at points A and B, respectively. The graph of the direct proportionality function, denoted by $l_2$, intersects $l_1$ at point C$(m, 4)$. What is the area $S_{\triangle AOC}$ of $\triangle AOC$?

Solution: \textbf{Solution:} As shown in the figure, draw $CD\perp AO$ at D, then $CD=4$. Since points A and B are on the linear function $y=-\frac{1}{2}x+5$, let $y=0$, then $x=0$, $x=10$, thus $O(0,0)$, $A(10,0)$, hence $AO=10$. Therefore, the area of $\triangle AOC$ is $\frac{1}{2}\times 10\times 4=\boxed{20}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the linear function $y = -\frac{1}{2}x + 5$, denoted as $l_1$, intersects the x and y axes at points $A$ and $B$, respectively. The graph of the direct proportion function, denoted as $l_2$, intersects $l_1$ at point $C(m, 4)$. The graph of the linear function $y = kx+1$ intersects line segment $AC$. Directly write down the value of $k$.

Solution:
\textbf{Solution:} Solve: When $y=kx+1$ exactly passes through point C, we have $4=2k+1$, and get $k=\frac{3}{2}$. When $y=kx+1$ exactly passes through point A, we have $0=10k+1$, and get $k=-\frac{1}{10}$.\\
$\therefore$ When $-\frac{1}{10}\le k\le \frac{3}{2}$, the graph of the linear function $y=kx+1$ intersects with the line segment AC.\\
Thus, the value of $k$ is $\boxed{-\frac{1}{10}\le k\le \frac{3}{2}}$.


Question: As shown in the figure, place the rectangle OABC within the Cartesian coordinate system with the x-axis and y-axis, positioning OA and OC along the positive halves of the x-axis and y-axis, respectively. Here, AB = 15, and the equation of the line containing the diagonal AC is given by $y=-\frac{5}{3}x+b$. If the rectangle OABC is folded along BE such that point A lands on point D on side OC, what are the coordinates of point B?

Solution: \textbf{Solution:} Given that $AB=15$ and the quadrilateral OABC is a rectangle, $\therefore OC=AB=15$. Thus, $C(0,15)$. By substituting into $y=-\frac{5}{3}x+b$, we find $b=15$. Therefore, the equation of line AC is $y=-\frac{5}{3}x+15$. Setting $y=0$ gives $x=9$. Therefore, $A(9,0)$ and $B=\boxed{(9,15)}$.


Question: As shown in the figure, place the rectangle OABC into the Cartesian coordinate system xO, such that OA and OC lie on the positive semi-axes of the x and y axes respectively, where AB = 15. The equation of the line containing the diagonal AC is given by $y=-\frac{5}{3}x+b$. Fold the rectangle OABC along BE such that point A falls on point D located on side OC. What is the length of EA?

Solution:
\textbf{Solution:} In $\triangle BCD$, we have $BC=9$, $BD=AB=15$,\\
$\therefore CD= \sqrt{15^2-9^2} = 12$,\\
$\therefore OD=15-12=3$. Let $DE=AE=x$,\\
In $\triangle DEO$, $\because DE^2=OD^2+OE^2$,\\
$\therefore x^2=3^2+(9-x)^2$,\\
$\therefore x=5$,\\
$\therefore AE=\boxed{5}$.


Question: As shown in the figure, place rectangle OABC in the Cartesian coordinate system with points OA and OC along the positive half-axis of x and y respectively. Here AB=15, and the equation of the line containing diagonal AC is $y=- \frac{5}{3}x+b$. Fold rectangle OABC along BE, such that point A falls on point D located on side OC. If point P is a moving point on the y-axis, there exists a point P for which the perimeter of $\triangle PBE$ is minimized. What are the coordinates of point P?

Solution:
\textbf{Solution:} Construct point $E'$, which is symmetric to point E with respect to the y-axis, and draw line $BE'$ intersecting the y-axis at point P. Under these conditions, the perimeter of $\triangle BPE$ is minimized.\\
Since $E(4,0)$,\\
it follows that $E'(-4,0)$.\\
Let the equation of the line $BE'$ be $y=kx+b$, then we have
\[
\begin{cases}
9k+b=15,\\
4k+b=0.
\end{cases}
\]
Solving these equations yields
\[
\begin{cases}
k=\frac{15}{13},\\
b=\frac{60}{13}.
\end{cases}
\]
Hence, the equation of the line $BE'$ is $y=\frac{15}{13}x+\frac{60}{13}$,\\
therefore, the coordinates of P are $\boxed{(0, \frac{60}{13})}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y = -\frac{3}{4}x + 6$ intersects the coordinate axes at points $A$ and $B$. The line $AE$ bisects the angle $\angle BAO$ and intersects the x-axis at point $E$. What is the equation of the line $AE$?

Solution:
**Solution:**

Let line $EG \perp AB$ pass through point E and intersect AB at point G,

$\because$ AE bisects $\angle BAO$,

$\therefore \angle OAE = \angle BAE$,

Also, $\because \angle AOE = \angle AGE = 90^\circ$, AE = AE,

$\therefore \triangle AOE \cong \triangle AGE$ (AAS),

$\therefore OE = GE$, AO = AG = 6,

Also, $\because$ The coordinates of point A are (0, 6), and the coordinates of point B are (8, 0),

In $\triangle AOB$, $AB= \sqrt{OA^2 + OB^2} = \sqrt{6^2 + 8^2} = 10$,

$\therefore BG = AB - AG = 4$,

Let OE = GE = x, then BE = 8 - x,

In $\triangle BEG$, $4^2 + x^2 = (8 - x)^2$,

Solving for $x$, we get: $x=3$,

$\therefore$ The coordinates of point E are (3, 0),

Let the equation of line AE be $y = kx + b$ ($k \neq 0$),

Substituting E(3, 0) and A(0, 6) into the equation, we get:

$\begin{cases} 3k + b = 0 \\ b = 6 \end{cases}$,

Solving for $k$ and $b$, we get:

$\begin{cases} k = -2 \\ b = 6 \end{cases}$,

$\therefore$ The equation of line AE is $\boxed{y = -2x + 6}$.


Question: In the Cartesian coordinate system, the graph of the linear function $y=-\frac{3}{4}x+6$ intersects with the coordinate axes at points $A$ and $B$. Segment $AE$ bisects $\angle BAO$ and intersects the x-axis at point $E$. A line is drawn from point $B$ perpendicular to $AE$, denoted as $BF$, with $F$ as the foot of the perpendicular. The segment $OF$ is then connected. Determine the shape of $\triangle OFB$ and calculate the area of $\triangle OFB$.

Solution:
\textbf{Solution:} Draw a line $FM$ perpendicular to the x-axis from point F, intersecting the x-axis at point M,\\
since $A(0,6), E(3,0)$,\\
in $\triangle AOE$, $AE= \sqrt{OA^2+OE^2}=\sqrt{6^2+3^2}=3\sqrt{5}$,\\
since $BF\perp AE$,\\
therefore $\angle AOE=\angle BFE$,\\
also, since $\angle AEO=\angle BEF$,\\
therefore $\triangle AOE \sim \triangle BEF$,\\
therefore $\frac{BF}{OA}=\frac{BE}{AE}$,\\
therefore $\frac{BF}{6}=\frac{5}{3\sqrt{5}}$,\\
solving, we find $BF= 2\sqrt{5}$,\\
since $\angle FMB=\angle BFE, \angle MBF=\angle FBE$,\\
therefore $\triangle BMF \sim \triangle BFE$,\\
therefore $\frac{BF}{BE}=\frac{BM}{BF}$,\\
therefore $\frac{2\sqrt{5}}{5}=\frac{BM}{2\sqrt{5}}$,\\
solving, we find $BM=4$,\\
in $\triangle BMF$, $FM= \sqrt{BF^2-BM^2}=2$,\\
also, since the coordinates of point B are $(8,0)$,\\
therefore $OM=BM=4$ and $FM\perp x$-axis,\\
therefore $OF=BF$,\\
therefore $\triangle OFB$ is an isosceles triangle,\\
therefore the area of $\triangle OFB$ is $\frac{1}{2}OB\cdot FM=\frac{1}{2}\times 8\times 2=\boxed{8}$.


Question: As shown in the Cartesian coordinate system, the graph of line $L_1$: $y=kx+b$ passes through point $C(-2, 7)$, and intersects the x-axis at point $B$, and intersects line $L_2$: $y=\frac{1}{2}x$ at point $A$, where the x-coordinate of point $A$ is $6$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Given the problem's statement, point $A$ lies on the line $y=\frac{1}{2}x$ and the abscissa of point $A$ is $6$,\\
$\therefore$ the ordinate of point $A$ is $3$.\\
$\because$ line $L_1$ passes through point $C(-2, 7)$, we can establish the system of equations $\left\{\begin{array}{l}-2k+b=7, \\ 6k+b=3,\end{array}\right.$ which yields $\left\{\begin{array}{l}k=-\frac{1}{2}, \\ b=6.\end{array}\right.$\\
$\therefore$ the equation of line $AB$ is $y=-\frac{1}{2}x+6$; thus, the answer is $\boxed{y=-\frac{1}{2}x+6}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the line $L_1\colon y = kx + b$ passes through point $C(-2, 7)$, intersects with the $x$-axis at point $B$, and intersects with the line $L_2\colon y = \frac{1}{2}x$ at point $A$, where the $x$-coordinate of point $A$ is $6$. Write directly the solution set of the inequality $kx + b > \frac{1}{2}x$ in terms of $x$.

Solution: \textbf{Solution:} The solution set of the inequality $kx+b>\frac{1}{2}x$ with respect to $x$ is: $\boxed{x<6}$.


Question: As shown in the figure, in the Cartesian coordinate system, line $L_1$: $y=kx+b$ passes through point $C(-2, 7)$ and intersects the $x$-axis at point $B$, and it intersects line $L_2$: $y=\frac{1}{2}x$ at point $A$, where the abscissa of point $A$ is $6$. If $D$ is a point on the $y$-axis, and $S_{\triangle AOD}=\frac{1}{3}S_{\triangle AOB}$, find the coordinates of point $D$.

Solution: \textbf{Solution:} Let $D\left(0, y\right)$. Since point $B$ is the intersection of line $L_{1}$ and the x-axis, we get $-\frac{1}{2}x+6=0$. Solving for $x$, we find $x=12$, hence $OB=12$. Since the area $S_{\triangle AOD}=\frac{1}{3}S_{\triangle AOB}$, it follows that $\frac{1}{2}\times 6\times |y|=\frac{1}{3}\times 18=6$. Solving for $y$, we get: $y=\pm 2$. Therefore, the coordinates of point $D$ are $\boxed{D\left(0, 2\right)}$ or $\boxed{\left(0, -2\right)}$.