Question: In $\triangle ABC$, $\angle C=90^\circ$. Point P is a moving point on side $AC$, and line $PQ$ is drawn parallel to $AB$ intersecting $BC$ at point Q. Point D is the midpoint of segment $PQ$, and $AD$ bisects $\angle BAC$. Given $\triangle ABC \sim \triangle PQC$ and $AB=13$, $BC=12$, find the length of $AP$.

Solution: \textbf{Solution:} Since $\angle C=90^\circ$ and $AB=13$, $BC=12$, it follows that $AC= \sqrt{AB^{2}-BC^{2}} = \sqrt{13^{2}-12^{2}} =5$. Given that $PQ\parallel AB$, it implies that $\angle BAD=\angle ADP$. Furthermore, since $AD$ bisects $\angle BAC$, we have $\angle BAD=\angle PAD$, which leads to $\angle ADP=\angle PAD$, thus $PA=PD$. Given that $D$ is the midpoint of segment $PQ$, we have $PD= \frac{1}{2}PQ$, hence $QP=2PA. Since \triangle ABC \sim \triangle PQC$, it follows that $\frac{CP}{CA}=\frac{PQ}{AB}$, which means $\frac{5-AP}{5}=\frac{2AP}{13}$. Solving this gives $AP=\boxed{\frac{65}{23}}$.


Question: As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at point O, with BE $\parallel$ AC, and CE $\parallel$ DB. Quadrilateral OBEC is a rhombus; if AD = 4 and AB = 2, find the area of the rhombus OBEC.

Solution:
\textbf{Solution:} Given $AD=4$ and $AB=2$,\\
therefore, the area of rectangle $ABCD=4\times 2=8$,\\
thus, the area of $\triangle OBC= \frac{1}{4}$ area of rectangle $ABCD=2$,\\
hence, the area of the rhombus $OBEC=2\times$ area of $\triangle OBC=\boxed{4}$.


Question: As shown in the square $ABCD$, the vertices $E$ and $F$ of $\triangle AEF$ lie on $BC$ and $CD$ respectively. $AG \perp EF$ with the foot being $G$ and $AG = AB$. Find the measure of $\angle EAF$.

Solution:
\textbf{Solution:} In right-angled triangles $Rt\triangle ABE$ and $Rt\triangle AGE$, we have $\left\{\begin{array}{l}AB=AG\\ AE=AE\end{array}\right.$, \\
$\therefore \triangle ABE\cong \triangle AGE\left(HL\right)$, \\
$\therefore \angle BAE=\angle GAE$, similarly $\angle GAF=\angle DAF$, \\
$\therefore \boxed{\angle EAF=45^\circ}$.


Question: In quadrilateral $ABCD$, $AB\parallel DC$, $AB=AD$, diagonals $AC$, $BD$ intersect at point $O$, $AC$ bisects $\angle BAD$, and through point $C$, draw $CE\parallel BD$ intersecting the extension of $AB$ at point $E$, connect $OE$. Quadrilateral $ABCD$ is a rhombus: If $\angle DAB=60^\circ$ and $AB=4$, what is the length of $OE$?

Solution: \textbf{Solution:} Given that (1) informs us that quadrilateral $ABCD$ is a rhombus, with $\angle DAB=60^\circ$, hence $\angle OAB=30^\circ$, and $\angle AOB=90^\circ$. Since $AB=4$, it follows that $OB=2$, and $AO=OC=2\sqrt{3}$. Since $CE\parallel DB$ and $DC\parallel BE$, quadrilateral $DBEC$ is a parallelogram. Thus, $CE=DB=4$, and $\angle ACE=90^\circ$. Therefore, $OE=\sqrt{OC^{2}+CE^{2}}=\sqrt{(2\sqrt{3})^{2}+4^{2}}=\boxed{2\sqrt{7}}$.


Question: As illustrated in the square $ABCD$, where $E$ is a moving point on side $CD$ (not coinciding with points $C$ or $D$), connect $AE$. By folding $\triangle ADE$ along $AE$, $\triangle AEF$ is obtained. Extend $EF$ to intersect $BC$ at point $G$, and connect $AG$. Draw $EH\perp AE$ passing through point $E$ and extending to intersect the extended line of $AG$ at point $H$, and then connect $CH$. What is the measure of $\angle EAH$ in degrees?

Solution: \textbf{Solution:} Given that quadrilateral $ABCD$ is a square, we have $AD=DC=AB$ and $\angle BAD=\angle B=\angle D=90^\circ$. By the property of folding, we get $AF=AD$ and $\angle AFE=\angle D=90^\circ$, $\angle EAF=\angle EAD$. Thus, $AB=AF$ and $\angle AFG=90^\circ$. In right triangles $Rt\triangle ABG$ and $Rt\triangle AFG$, we have $\left\{\begin{array}{l}AG=AG \\ AB=AF\end{array}\right.$, hence $Rt\triangle ABG \cong Rt\triangle AFG$ (by HL congruence), implying $\angle BAG=\angle FAG$. Consequently, $\angle EAH=\angle FAG+\angle EAF=\angle BAG+\angle EAD$. Also, since $\angle FAG+\angle EAF+\angle BAG+\angle EAD=\angle BAD=90^\circ$, we thus have $\angle EAH+\angle EAH=90^\circ$. Solving this, we find $\angle EAH=\boxed{45^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the coordinates of point $C$ are $(\sqrt{3}, -2)$, and the line $AB$ intersects the $x$ axis and $y$ axis at point $A$ and point $B$ respectively, with the coordinates of point $B$ being $(0, 3)$, and $\angle BAO = 30^\circ$. What is the analytical equation of line $AB$?

Solution: \textbf{Solution:} Since $B(0, 3)$, it follows that $OB=3$. In the right triangle $\triangle AOB$, we have $\angle AOB=90^\circ$ and $\angle BAO=30^\circ$. Therefore, $AB=2OB=6$ and $OA=\sqrt{AB^{2}-OB^{2}}=3\sqrt{3}$. Hence, $A(3\sqrt{3}, 0)$. Let the equation of line $AB$ be $y=kx+b$. Substituting points $A(3\sqrt{3}, 0)$ and $B(0, 3)$ gives $\begin{cases}3\sqrt{3}k+b=0\\ b=3\end{cases}$. Solving, we find $\begin{cases}k=-\frac{\sqrt{3}}{3}\\ b=3\end{cases}$. Thus, the equation of line $AB$ is $y=-\frac{\sqrt{3}}{3}x+3$; hence, the answer is $\boxed{y=-\frac{\sqrt{3}}{3}x+3}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the coordinate of point $C$ is ($\sqrt{3}$, $-2$), and the line $AB$ intersects the x-axis and y-axis at point $A$ and point $B$, respectively, with the coordinate of point $B$ being (0, 3), and $\angle BAO=30^\circ$. If point $D$ is a moving point on the y-axis, and point $E$ ($\sqrt{3}$, $m$) is on line $AB$, find the coordinates of points $D$ and $E$ when $CD+DE$ is minimized?

Solution:
\textbf{Solution:} As shown in the figure, construct the symmetric point $C'$ of point $C$ with respect to the $y$-axis, and connect $DC'$ and $EC'$. By the property of axial symmetry, we have $C'(-\sqrt{3}, -2)$, $DC'=CD$, $\therefore CD+DE=DC'+DE$. According to the principle that the shortest distance between two points is a straight line, when points $C'$, $D$, and $E$ are collinear, meaning point $D$ is the intersection of line $EC'$ and the $y$-axis, $DC'+DE$ achieves its minimum value.

Substituting point $E(\sqrt{3}, m)$ into $y=-\frac{\sqrt{3}}{3}x+3$ yields:
\[ m=-\frac{\sqrt{3}}{3}\times \sqrt{3}+3=2 \]
hence $E(\sqrt{3}, 2)$.

Assume the equation of line $EC'$ is $y=k_0x+b_0$. Substituting points $E(\sqrt{3}, 2)$ and $C'(-\sqrt{3}, -2)$ into it yields:
\[ \left\{\begin{array}{l}\sqrt{3}k_0+b_0=2 \\ -\sqrt{3}k_0+b_0=-2\end{array}\right. \]
solving this system of equations gives
\[ \left\{\begin{array}{l}k_0=\frac{2\sqrt{3}}{3} \\ b_0=0\end{array}\right. \]
Therefore, the equation of line $EC'$ is $y=\frac{2\sqrt{3}}{3}x$. When $x=0$, $y=0$, meaning when $CD+DE$ is minimized, the coordinates of point $D$ are $D(0, 0)$.

Thus, the answer is $D=\boxed{(0, 0)}$, $E=\boxed{(\sqrt{3}, 2)}$.


Question: As shown in the figure, in the rhombus $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. A perpendicular line is drawn from point $D$ to the extended line of $BA$ at point $E$. Quadrilateral $ACDE$ is a parallelogram; if $AC = 8$ and $BD = 6$, what is the perimeter of $\triangle BDE$?

Solution: \textbf{Solution:} From what is given in the problem, the perimeter of $\triangle BDE$ is $\boxed{24}$.


Question: As shown in the figure, in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$, and $BE \parallel AC$, $AE \parallel BD$. Connect $EO$. If $CD=6$, find the length of $OE$.

Solution:
\textbf{Solution:} \\
Since quadrilateral AEBO is a rectangle, \\
it follows that $EO = AB$, \\
Since quadrilateral ABCD is a rhombus, \\
it follows that $AB = CD$. \\
Therefore, $EO = CD = \boxed{6}$.


Question: As shown in the figure, fold the corner where vertex D of rectangle ABCD is located, so that point D falls on point F on side BC, with the crease being AE. If $\angle DAE=25^\circ$, what is the measure of $\angle EFC$?

Solution: \textbf{Solution}: As shown in the figure, since quadrilateral $ABCD$ is a rectangle, thus $AD \parallel BC$ and $\angle D=90^\circ$. By folding, it is known that $\triangle ADE \cong \triangle AFE$. Therefore, $\angle FAE=\angle DAE=25^\circ$, and $\angle AFE=\angle D=90^\circ$. Therefore, $\angle AFB=\angle DAF=2\angle DAE=50^\circ$. Therefore, $\angle EFC=180^\circ-\angle AFB-\angle AFE=180^\circ-50^\circ-90^\circ=\boxed{40^\circ}$.


Question: As shown in the figure, fold the corner where vertex D of rectangle ABCD is located, so that point D falls on point F on side BC, with the crease being AE. If $AB=8$ and $BC=10$, what is the length of $EF$?

Solution: \textbf{Solution}: Since quadrilateral $ABCD$ is a rectangle, we have $AB=CD=8$, $BC=AD=AF=10$, and $\angle B=\angle C=90^\circ$. Therefore, $BF=\sqrt{AF^2-AB^2}=\sqrt{10^2-8^2}=6$. Consequently, $FC=BC-BF=10-6=4$. Let $EF=DE=x$, then $EC=CD-DE=8-x$. In $\triangle EFC$, by the Pythagorean theorem, we have $FC^2+EC^2=EF^2$. Therefore, $4^2+(8-x)^2=x^2$. Solving this equation yields $x=\boxed{5}$. Therefore, $EF=\boxed{5}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a square. Point $M$ is on the side $BC$ (not coinciding with the endpoints $B$, $C$), and point $N$ is on the diagonal $AC$, with $MN \perp AC$. Draw line $AM$, and let point $G$ be the midpoint of $AM$. Connect $NG$ and $DN$. If $AB = 10$ and $BM = 2\sqrt{3}$, what is the length of $NG$?

Solution: \textbf{Solution:} Since quadrilateral ABCD is a square, $\angle B=90^\circ$. In $\triangle ABM$, since AB＝10 and BM＝$2\sqrt{3}$, we have $AM=\sqrt{AB^2+BM^2}=4\sqrt{7}$. Since MN$\perp$AC and point G is the midpoint of AM, we have GN＝$\frac{1}{2}AM=2\sqrt{7}$. Therefore, the length of NG is $\boxed{2\sqrt{7}}$.


Question: As shown in the figure, in square $ABCD$, $AB=\sqrt{5}$. Let $E$ be a point inside square $ABCD$, with $DE=AB$ and $\angle EDC=\alpha$ $\left(0^\circ<\alpha<90^\circ\right)$. Connect $CE$ and $AE$, draw $DF\perp AE$ with the foot being point $F$, and extend $CE$ to meet $DF$ at point $G$. Connect $AG$. When $\alpha=20^\circ$, find the measure of $\angle DAE$ in degrees.

Solution:
\textbf{Solution}: Since quadrilateral $ABCD$ is a square,\\
it follows that $\angle ADC=90^\circ$, and $AB=AD$,\\
since $\angle CDE=20^\circ$,\\
it follows that $\angle ADE=70^\circ$,\\
since $DE=AB$,\\
it follows that $DA=DE$,\\
therefore $\angle DAE=\angle DEA=\frac{1}{2}(180^\circ-70^\circ)=\boxed{55^\circ}$.


Question: As shown in the figure, in square $ABCD$, where $AB=\sqrt{5}$, $E$ is a point inside square $ABCD$, $DE=AB$, and $\angle EDC=\alpha$ $\left(0^\circ<\alpha<90^\circ\right)$. Connect $CE$ and $AE$. Through point $D$, draw $DF\perp AE$, where $F$ is the foot of the perpendicular, and it intersects the extension of $CE$ at point $G$. Connect $AG$. When $GF=1$, what is the length of $CE$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $AC$,

since the quadrilateral $ABCD$ is a square,

therefore $AC=\sqrt{2}AB=\sqrt{10}$,

since $\triangle AEG$ is an isosceles right triangle and $GF\perp AE$,

therefore $GF=AF=EF=1$,

therefore $AG=GE=\sqrt{2}$,

since $AC^{2}=AG^{2}+GC^{2}$,

therefore $10=2+(EC+\sqrt{2})^{2}$,

therefore $EC=\boxed{\sqrt{2}}$ (negative values have been disregarded).


Question: As shown in the figure, quadrilateral $ABCD$ is a parallelogram, with $AD=BD$. Through point $C$, construct $CE\parallel BD$, intersecting the extension line of $AD$ at point $E$. Quadrilateral $BDEC$ is a rhombus; connect $BE$. If $AB=2$ and $AD=4$, what is the length of $BE$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BE$ intersecting $CD$ at $O$,\\
since quadrilateral $BDEC$ is a rhombus,\\
thus $DO=CO=\frac{1}{2}CD=1$, $BO=\frac{1}{2}BE$, and $CD\perp BE$,\\
in $\triangle BDO$, $AD=BD=4$, $DO=1$,\\
thus $BO=\sqrt{BD^2-DO^2}=\sqrt{16-1}=\sqrt{15}$,\\
therefore, $BE=2BO=2\sqrt{15}$.\\
Hence, the length of $BE$ is $\boxed{2\sqrt{15}}$.


Question: As shown in the figure, the four corners of the rectangular piece of paper $ABCD$ are folded inward to form a quadrilateral $EFGH$ without gaps or overlaps. The quadrilateral $EFCH$ is a rectangle. If $AH=2$ and $HD=3$, what is the area of the rectangle $EFCH$?

Solution:
\textbf{Solution:} Method 1: Knowing that the quadrilateral EFGH is a rectangle implies that EF=GH, and $\angle B=\angle D$, $\angle BFE=\angle CGF=\angle DHG$,\\
$\therefore \triangle BFE\cong \triangle DHG$, $\therefore BF=DH$, $\therefore JF=BF=HD=HK=3$, $\therefore HF=HJ+JF=2+3=5$.\\
From folding, it is known that AE=EJ=BE, let AE=x,\\
$\because \angle HEF=90^\circ$, $\therefore EH^2+ EF^2= HF^2$,\\
$\therefore 2^2+x^2+3^2+x^2=(2+3)^2$, $\therefore x= \sqrt{6}$,\\
$S_{\text{Rectangle} EFGH}=2S_{\triangle EFH}=2\times \frac{1}{2} \times5\times \sqrt{6} = \boxed{5 \sqrt{6}}$.\\
Method 2: Similarly, we obtain JF=BF=HD=HK=3, HF=5, thus HO=FO=2.5, JO=KO=0.5,\\
$\because$ quadrilateral EFGH is a rectangle, EG=FH=5, EO=GO=2.5,\\
$\therefore$, knowing that EJ$\perp$HF, $\therefore HJ^2+EJ^2=EH^2, FJ^2+EJ^2= EF^2$,\\
$\therefore \angle EJO=90^\circ$, $\therefore EJ^2= EO^2−JO^2 =2.5^2−0.5^2= 6$,\\
$\therefore EJ= \sqrt{6}$,\\
$\therefore S_{\text{Rectangle} EFGH}=2S_{\triangle EFH}=2\times \frac{1}{2} \times5\times \sqrt{6} = \boxed{5 \sqrt{6}}$.\\
Method 3: As above, we obtain HF=5, BF=3, drawing a line HP$\perp$BC at point P, then PF=l, HP= $\sqrt{6}$,\\
$\therefore S_{\text{Rectangle} EFGH}= \frac{1}{2} S_{\triangle EFH}= \frac{1}{2} \times5\times2 \sqrt{6} = \boxed{5 \sqrt{6}}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $E$ and $F$ are points on $CB$ and $CD$ respectively, such that $\angle AEB = \angle AFD$ and $BE = DF$. Quadrilateral $ABCD$ is a rhombus. If $\triangle ABE \cong \triangle ADF$, what is the measure of $\angle B$?

Solution: \textbf{Solution:} From (1), we know that $\triangle ABE \cong \triangle ADF$, $\therefore \angle BAE = \angle DAF$, $\because \triangle ABE \cong \triangle AEF$, $\therefore \angle BAE = \angle EAF$, $AB = AE$, $\therefore \angle B = \angle AEB$, let $\angle B = \angle AEB = x$, then $\angle BAE = \angle EAF = \angle DAF = 180^\circ - 2x$, $\because$ quadrilateral $ABCD$ is a parallelogram, $\therefore AD \parallel BC$, $\therefore \angle B + \angle BAD = 180^\circ$, that is $x + 3(180^\circ - 2x) = 180^\circ$, solving yields: $x = \boxed{72^\circ}$, that is, the measure of $\angle B$ is $72^\circ$.


Question: As shown in the diagram, within square $ABCD$, $E$ is a point on diagonal $AC$, $FH \perp AC$ at point $E$, intersecting $AD$, $AB$ at points $F$, $H$ respectively. Given that $CF = CH$ and $AH = \frac{1}{3}CH$ with $AB = 4$, determine the length of $AH$.

Solution:
\textbf{Solution:} Let $AH=x$, then $CH=3x$ and $HB=4-x$.\\
Since quadrilateral $ABCD$ is a square,\\
it follows that $BC=AB=4$ and $\angle B=90^\circ$.\\
In $\triangle CBH$, by the Pythagorean theorem, we have\\
$CH^2=HB^2+BC^2$,\\
which implies $(3x)^2=(4-x)^2+4^2$,\\
solving for $x$ gives $x=\frac{-1+\sqrt{17}}{2}$ or $x=\frac{-1-\sqrt{17}}{2}$ (discard),\\
thus, $AH=\boxed{\frac{-1+\sqrt{17}}{2}}$.


Question: As shown in the figure, in quadrilateral \(ABCD\), \(AD \parallel BC\), diagonal \(BD\) bisects diagonal \(AC\) perpendicularly at point \(O\). Quadrilateral \(ABCD\) is a rhombus; if \(\angle DBC = 30^\circ\) and \(BC = 2\), find the area of quadrilateral \(ABCD\).

Solution:
Solution: Since $AC \perp BD$, \\
it follows that $\angle BOC = 30^\circ$, \\
Given that $\angle DBC = 30^\circ$, and $BC = 2$, \\
we find that $OC = 1$, and $OB = \sqrt{3}$, \\
Therefore, the area of rhombus ABCD $= \frac{1}{2} AC \cdot BD = \frac{1}{2} \times 2OC \times 2 \times OB = \frac{1}{2} \times 2 \times 2 \sqrt{3} = \boxed{2 \sqrt{3}}$.


Question: As shown in the figure, point $A(0,2)$, $B(3,3)$, and $C(7, -1)$, the quadrilateral $OBCD$ is a rectangle, and $BD$ intersects the x-axis at point $E$. Find the equation of line $AB$.

Solution:
\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting point A$(0,2)$ and B$(3,3)$ into $y=kx+b$ gives
\[
\begin{cases}
b=2\\
3k+b=3
\end{cases}
\]
Solving this, we get
\[
\begin{cases}
k=\frac{1}{3}\\
b=2
\end{cases}
\]
$\therefore \boxed{y=\frac{1}{3}x+2}$;


Question: As shown in the figure, point $A(0, 2)$, $B(3, 3)$, $C(7, -1)$, and quadrilateral $OBCD$ is a rectangle, with $BD$ intersecting the $x$ axis at point $E$. Find the length of the line segment $OE$.

Solution:
\textbf{Solution:} Connect OC to intersect BD at point F.\\
Since quadrilateral OBCD is a rectangle, with point C's coordinates at (7, -1) and point O's coordinates at (0, 0), and point F being the midpoint of OC, therefore the coordinates of point F are $\left(\frac{7}{2}, -\frac{1}{2}\right)$. Let the equation of line BD be $y=kx+b$. Substituting (3, 3) and $\left(\frac{7}{2}, -\frac{1}{2}\right)$ into $y=kx+b$ yields $\left\{\begin{array}{l}3k+b=3 \\ \frac{7}{2}k+b=-\frac{1}{2}\end{array}\right.$. Solving this system of equations gives $\left\{\begin{array}{l}k=-7 \\ b=24\end{array}\right.$, therefore $y=-7x+24$. Substituting $y=0$ into $y=-7x+24$ gives $0=-7x+24$, solving for $x$ yields $x=\frac{24}{7}$. Therefore, the coordinates of point E are $\left(\frac{24}{7}, 0\right)$, thus $OE=\boxed{\frac{24}{7}}$.


Question: As shown in the diagram, point A is located at (0, 2), B at (3, 3), and $C(7, -1)$. The quadrilateral OBCD is a rectangle, and BD intersects with the x-axis at point E. If point P is a moving point on line AB, with the area of $\triangle POB$ denoted as $S_{1}$ and the area of $\triangle OBE$ denoted as $S_{2}$, and given that $S_{1} = 2S_{2}$, find the coordinates of point P.

Solution:
Solution: By the problem statement, we have

\[\because \text{the area of } \triangle OBE \text{ is } S_{2} = \frac{1}{2}OE \cdot yB = \frac{1}{2} \times \frac{24}{7} \times 3 = \frac{36}{7},\]

\[\therefore S_{1} = 2S_{2} = \frac{72}{7}, \text{draw } BM \perp y \text{ axis at point } M,\]

(1) When point $P$ is on the left side of the y-axis, connect $OP$, construct $PN \perp y$ axis at point $N$,

\[\because S_{\triangle POB} = S_{\triangle POA} + S_{\triangle AOB},\]

\[\therefore S_{\triangle POA} = S_{\triangle POB} - S_{\triangle AOB} = S_{1} - \frac{1}{2} \cdot OA \cdot xB = \frac{72}{7} - \frac{1}{2} \times 2 \times 3 = \frac{51}{7},\]

\[\therefore \frac{1}{2}OA \cdot |xP| = \frac{1}{2} \times 2 \cdot (-xP) = \frac{51}{7},\] solving for $xP = -\frac{51}{7}$, substitute $x = -\frac{51}{7}$ into $y = \frac{1}{3}x + 2$ to get $y = -\frac{51}{7} \times \frac{1}{3} + 2 = -\frac{3}{7}$,

\[\therefore \text{the coordinates of point } P \text{ are } \boxed{\left(-\frac{51}{7}, -\frac{3}{7}\right)}.\]

(2) When point $P$ is on the right side of the y-axis, connect $OP$, construct $PN \perp y$ axis at point $N$,

\[\because S_{\triangle POB} = S_{\triangle POA} - S_{\triangle AOB} = \frac{72}{7},\]

\[\therefore S_{\triangle POA} = S_{\triangle POB} + S_{\triangle AOB} = \frac{72}{7} + \frac{1}{2} \times 2 \times 3 = \frac{93}{7},\]

\[\therefore \frac{1}{2}OA \cdot xP = \frac{93}{7},\] solving for $xP = \frac{93}{7}$, substitute $x = \frac{93}{7}$ into $y = \frac{1}{3}x + 2$ to get $y = \frac{45}{7}$,

\[\therefore \text{the coordinates of point } P \text{ are } \boxed{\left(\frac{93}{7}, \frac{45}{7}\right)}.\]

In summary, the coordinates of point $P$ are \boxed{\left(-\frac{51}{7}, -\frac{3}{7}\right)} or \boxed{\left(\frac{93}{7}, \frac{45}{7}\right)}.


Question: As shown in the figure, two lines $l_{1}$ and $l_{2}$ passing through point A (1, 0) intersect the y-axis at points B and C, respectively, where point B is above the origin and point C is below the origin. It is known that $AB = \sqrt{10}$. Find the coordinates of point B.

Solution: \textbf{Solution:} Since the coordinates of point A are $(1,0)$, it follows that $AO= 1$. Also, since $AB= \sqrt{10}$, it follows that $BO= \sqrt{AB^{2}-AO^{2}}=3$. Since point B is above the origin, the coordinates of point B are $\boxed{(0,3})$;


Question: As shown in the figure, two lines \( {l}_{1} \) and \( {l}_{2} \) passing through point \(A(1, 0)\) intersect the y-axis at points \(B\) and \(C\) respectively, where point \(B\) is above the origin and point \(C\) is below the origin. It is known that \( AB = \sqrt{10} \). If the area of \( \triangle ABC \) is 3, what is the equation of line \( {l}_{2} \)?

Solution:
\textbf{Solution:} Given that the area of $\triangle ABC$ is 3, it follows that $\frac{1}{2}BC \cdot AO = 3$. Hence, $\frac{1}{2}BC \times 1 = 3$, which implies $BC=6$. Since $BO=3$, it follows that $CO=3$. Therefore, the coordinates of point C are $(0, -3)$. Let the equation of line $l_{2}$ be $y= kx + b$ ($k \neq 0$). We then have
\[
\left\{
\begin{array}{l}
b=-3 \\
k+b=0
\end{array}
\right.
\]
Solving this system of equations, we obtain
\[
\left\{
\begin{array}{l}
k=3 \\
b=-3
\end{array}
\right.
\]
Thus, the equation of the line $l_{2}$ is $y=3x-3$. Therefore, the answer is \boxed{y=3x-3}.


Question: As shown in the figure, the line $l: y = \sqrt{3}x + \sqrt{3}$ intersects the two coordinate axes at points $A$ and $B$, respectively. Point $M$ is the midpoint of line segment $AB$. Find the coordinates of $A$, $B$, and $M$.

Solution:
\textbf{Solution:} In the equation of the line $l\colon y=\sqrt{3}x+\sqrt{3}$, \\
let $x=0$, we obtain $y=\sqrt{3}$; let $y=0$, we obtain $x=-1$, \\
thus, the coordinates of points A and B are $\boxed{(0, \sqrt{3})}$ and $\boxed{(-1, 0)}$, respectively; \\
since point M is the midpoint of segment AB, $\therefore \boxed{M\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}$.


Question: As shown in the figure, the line $l\colon y = \sqrt{3}x + \sqrt{3}$ intersects the two coordinate axes at points $A$ and $B$, respectively, and point $M$ is the midpoint of line segment $AB$. The line $l'$ is symmetrical to line $l$ with respect to the $y$-axis. Write the equation of line $l'$.

Solution:
\textbf{Solution:} Since line $l'$ is symmetric to line $l$ with respect to the $y$-axis, let line $l'$ intersect the $x$-axis at point $C$. Thus, point $C$ is symmetric to point $B$ with respect to the $y$-axis, that is $C(1, 0)$,\\
Since line $l'$ passes through point $A(0, \sqrt{3})$, we can set the equation of line $l'$ to be $y=kx+\sqrt{3}$,\\
Given that line $l'$ passes through point $C(1, 0)$, $\therefore 0=k\times 1+\sqrt{3}$, $\therefore k=-\sqrt{3}$,\\
thus, the equation of the function is: \boxed{y=-\sqrt{3}x+\sqrt{3}}.


Question: As shown in the figure, the line $l\colon y = \sqrt{3}x + \sqrt{3}$ intersects the two coordinate axes at points $A$ and $B$ respectively. Point $M$ is the midpoint of line segment $AB$. If line $l'$ intersects the $x$-axis at point $C$, and the line $MC$ intersects the $y$-axis at point $N$, connect $OM$. Find $\frac{OM}{ON}$.

Solution:
\textbf{Solution:} As shown in the diagram, draw $MD\perp x$-axis at D. In right triangle $MDO$, by the Pythagorean theorem we have:
\[OM=\sqrt{OD^{2}+DM^{2}}=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=1,\]
Also, let the equation of line $MC$ be $y=ax+b$,
then \[\left\{\begin{array}{l}
-\frac{1}{2}a+b=\frac{\sqrt{3}}{2}, \\
a+b=0.
\end{array}\right.\] Solving the system of equations, we get \[\left\{\begin{array}{l}
a=-\frac{\sqrt{3}}{3} \\
b=\frac{\sqrt{3}}{3}
\end{array}\right.,\]
\(\therefore\) The equation of line $MC$ is $y=-\frac{\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}$,
\(\therefore\) The intersection of line $MC$ with the y-axis is $N\left(0, \frac{\sqrt{3}}{3}\right)$,
\(\therefore \frac{OM}{ON}=1\div\frac{\sqrt{3}}{3}=\boxed{\sqrt{3}}.\)


Question: As shown in the figure, in the Cartesian coordinate system, the line AC passing through the point $C(0, 12)$ intersects with the line OA at the point $A(8, 4)$. What is the equation of line AC?

Solution:
\textbf{Solution:} Let the equation of the line $AC$ be $y=kx+b$, \\
Substituting $C(0, 12)$ and $A(8, 4)$ into it yields: $\begin{cases}12=b\\ 4=8k+b\end{cases}$, \\
Solving these equations, we get $\begin{cases}k=-1\\ b=12\end{cases}$, \\
$\therefore$ the equation of the line $AC$ is \boxed{y=-x+12}.


Question: As shown in the figure, in the Cartesian coordinate system, the line AC passing through the point $C(0, 12)$ intersects with the line OA at the point $A(8, 4)$. What is the area of $\triangle OAC$?

Solution:
\textbf{Solution:} Construct $AH \perp OC$ at H, as shown in the diagram:\\
$\because A\left(8, 4\right)$ and $AH \perp OC$,\\
$\therefore AH=8$,\\
$\because C\left(0, 12\right)$,\\
$\therefore OC=12$,\\
$\therefore {S}_{\triangle OAC}=\frac{1}{2}OC \cdot AH=\frac{1}{2} \times 12 \times 8=\boxed{48}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the line passing through point $C(0, 12)$ intersects with line OA at point $A(8, 4)$. Suppose a point M is moving along the ray OA. Is there any position of M that makes the area of $\triangle OBM$ equal to the area of $\triangle OAC$? If such a point exists, find the coordinates of M; if not, explain why.

Solution:
\textbf{Solution:} Exists, \\
By substituting $y=0$ into $y=-x+12$, we get $x=12$, \\
$\therefore$ point $B(12, 0)$, \\
$\therefore OB=12$, \\
Let the equation of line OA be $y=mx$, \\
Substituting $A(8, 4)$ gives $8m=4$, \\
$\therefore m=\frac{1}{2}$, \\
The equation of line OA is $y=\frac{1}{2}x$, \\
Let the coordinates of point M be $\left(x, \frac{1}{2}x\right)$, \\
$\frac{1}{2}\times OB\times \frac{1}{2}x=48$, \\
$\frac{1}{2}\times 12\times \frac{1}{2}x=48$, \\
$x=16$, \\
Thus, the coordinates of point M are $\boxed{\left(16, 8\right)}$.


Question: As shown in the figure, the line $y=x+1$ and $y=ax+4$ intersect at point $A(1, m)$. Find the value of $m$.

Solution:
\textbf{Solution:} Substitute $A\left(1, m\right)$ into $y=x+1$,\\
yielding $m=1+1=\boxed{2}$.


Question: As shown in the figure, the line $y = x + 1$ and $y = ax + 4$ intersect at point $A(1, m)$. Write the solution set for the inequality $x + 1 < ax + 4$.

Solution: \textbf{Solution:} From the figure, we can see that the solution set of the inequality $x + 1 < ax + 4$ is $\boxed{x < 1}$.


Question: As shown in the figure, line $l_1$ passes through points A$(-2, 0)$ and B$(0, 2)$, and line $l_2: y = -4x + 14$ intersects $l_1$ at point C. Find the analytical formula of line $l_1$.

Solution:
\textbf{Solution:} Let the equation of line $ {l}_{1}$ be $y=ax+b$.\\
Since line $ {l}_{1}$ passes through points A($-2,0$) and B($0,2$),\\
we have $\begin{cases} -2a+b=0\\ b=2 \end{cases}$, solving these gives us $\begin{cases} a=1\\ b=2 \end{cases}$.\\
Thus, the equation of line $ {l}_{1}$ is $\boxed{y=x+2}$.


Question: As shown in the figure, line $ {l}_{1}$ passes through points A($-2$, $0$) and B($0$, $2$), and line $ {l}_{2}: y=-4x+14$ intersects $ {l}_{1}$ at point C. Find the coordinates of point C.

Solution: \textbf{Solution}: Since line $l_1$ intersects with line $l_2$ at point C, we have $\begin{cases}y=x+2\\ y=-4x+14\end{cases}$. Solving these equations yields $\begin{cases}x=\frac{12}{5}\\ y=\frac{22}{5}\end{cases}$. Therefore, the coordinates of point C are $\boxed{\left(\frac{12}{5}, \frac{22}{5}\right)}$.


Question: As shown in the figure, the quadrilateral OABC is a rectangle, with points A and C located on the x-axis and y-axis, respectively. Triangle $ODE$ is obtained by rotating triangle $OCB$ $90^\circ$ clockwise around point O. Point D is on the x-axis, line BD intersects the y-axis at point F, and intersects OE at point H. The coordinates of point B are $(-2, 4)$. What is the equation of line BD?

Solution:
\textbf{Solution:} Given: $D(4,0)$ \\
Let the expression of line BD be: $y=kx+b$ \\
Then we have $\begin{cases} -2k+b=4 \\ 4k+b=0 \end{cases}$ \\
Solving, we get: $\begin{cases} b=\frac{8}{3} \\ k=-\frac{2}{3} \end{cases}$ \\
$\therefore$The expression of line BD is: $\boxed{y= -\frac{2}{3}x+ \frac{8}{3}}$


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle, with points $A$ and $C$ located on the $x$-axis and $y$-axis, respectively. Triangle $ODE$ is obtained by rotating triangle $OCB$ $90^\circ$ clockwise around point $O$. Point $D$ is located on the $x$-axis. The line $BD$ intersects the $y$-axis at point $F$, and intersects $OE$ at point $H$. The coordinates of point $B$ are $(-2, 4)$. Find the area of $\triangle DEH$.

Solution:
\textbf{Solution:} From the given, we know the coordinates of point E are (4, 2)\\
Let the equation of line OE be: $y=mx$, then we have $4m=2$, solving this gives: $m= \frac{1}{2}$\\
$\therefore y= \frac{1}{2} x$\\
$\because \left\{\begin{array}{l}y=\frac{1}{2}x\\ y=-\frac{2}{3}x+\frac{8}{3}\end{array}\right.$\\
$\therefore \left\{\begin{array}{l}x=\frac{16}{7}\\ y=\frac{8}{7}\end{array}\right.$\\
$\therefore H\left( \frac{16}{7}, \frac{8}{7}\right)$\\
$S_{\triangle DEH} = \frac{1}{2}\times DE\times \left(4-\frac{16}{7}\right) = \frac{1}{2}\times 2\times \left(4-\frac{16}{7}\right) = \boxed{\frac{12}{7}}$


Question: As shown in the diagram, the quadrilateral OABC is a rectangle, with points A and C located on the x-axis and y-axis, respectively. The triangle $ \triangle ODE$ is obtained by rotating $ \triangle OCB$ clockwise around point O by $90^\circ$. Point D is on the x-axis, and the line BD intersects the y-axis at point F and intersects OE at point H. The coordinate of point B is $(-2, 4)$. Point M is on the x-axis. Is there a point N in the plane such that the quadrilateral formed by points D, F, M, N is a rectangle? If yes, please directly write down the coordinates of point N; if not, please explain why.

Solution: \textbf{Solution:} The coordinates of point N are $\boxed{N(4, \frac{8}{3})}$ or $\boxed{N(\frac{20}{9}, -\frac{8}{3})}$.


Question: Given the line AB: $y = \frac{1}{2}x + 1$ intersects the x-axis and y-axis at points A and B, respectively. The line CD: $y = x + b$ intersects the x-axis and y-axis at points C and D, respectively. The line AB intersects line CD at point P, and $S_{\triangle ABD} = 4$. Find the equation of line CD.

Solution:
\textbf{Solution:} When $y=0$, that is, when $\frac{1}{2}x+1=0$, we find $x=-2$,\\
$\therefore \boxed{A(-2,0)}$,\\
when $x=0$, $y=\frac{1}{2}\times 0+1=1$,\\
$\therefore \boxed{B(0,1)}$,\\
$\because$ ${S}_{\triangle ABD} = \frac{1}{2} \text{BD} \times \text{OA} = \frac{1}{2}\times \text{BD}\times 2 = 4$,\\
$\therefore \text{BD}=4$,\\
Also, $\because$ when $x=0$ $y=0+b=b$,\\
$\therefore \boxed{D(0,b)}$,\\
$\because \text{BD}=1-b=4$,\\
$\therefore b=-3$,\\
$\therefore$ the equation of line CD is: $\boxed{y=x-3}$.


Question: Given line AB: $y = \frac{1}{2}x + 1$ intersects the x-axis and y-axis at points A and B, respectively, and line CD: $y = x + b$ intersects the x-axis and y-axis at points C and D, respectively. Line AB and line CD intersect at point P, and the area of $\triangle ABD = 4$. Please directly write the solution set for the inequality $\frac{1}{2}x + 1 \leq x + b$.

Solution: \textbf{Solution:} By the given condition, the solution set of the inequality $\frac{1}{2}x + 1 \leq x + b$ is $\boxed{x \geq 8}$.


Question: Given that the graph of the function $y=-\frac{1}{2}x+b$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the graph of the function $y=2x$ at point M(2,4); on the x-axis, there is a moving point P. A perpendicular line to the x-axis is drawn from point P, intersecting the graphs of the functions $y=-\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Find the equation of line AB and the coordinates of point A.

Solution:
\textbf{Solution:} Substituting point M$(2,4)$ into $y=-\frac{1}{2}x+b$ yields: $4=-\frac{1}{2}\times 2+b$. Solving for $b$ gives $b=5$. Thus, the equation of line AB is $y=-\frac{1}{2}x+5$.\\
Substituting $y=0$ into $y=-\frac{1}{2}x+5$ yields $x=10$, meaning the coordinates of point \boxed{A(10,0)}.


Question: Given that the graph of the function $y=-\frac{1}{2}x+b$ intersects the x-axis and the y-axis at points A and B, respectively, and intersects the graph of the function $y=2x$ at point M (2,4); there is a moving point P on the x-axis, vertical lines are drawn from point P, intersecting the graphs of $y=-\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Let point P be $(a,0)$, if $CD=\frac{1}{2}OB$, find the value of a and the coordinates of point C?

Solution:
\textbf{Solution:} Substitute $x=0$ into $y=-\frac{1}{2}x+5$ to get $y=5$,\\
$\therefore$ The coordinates of point B are $(0,5)$, that is: $OB=5$,\\
Given $CD=\frac{1}{2}OB$, we have $CD=\frac{5}{2}$,\\
Since PC$\perp x$ axis, and point P is $(a,0)$,\\
$\therefore$ The coordinates of point C are $(a,-\frac{1}{2}a+5)$, and the coordinates of point D are $(a,2a)$,\\
$\therefore$ $|2a-(-\frac{1}{2}a+5)|=\frac{5}{2}$, solving this yields: $a=\boxed{3}$ or $1$,\\
When $a=3$, $y=-\frac{1}{2}x+5=\frac{7}{2}$; When $a=1$, $y=-\frac{1}{2}x+5=\frac{9}{2}$;\\
$\therefore$ The coordinates of point C are $(1,\frac{9}{2})$ or \boxed{(3,\frac{7}{2})}.


Question: Given the graph of the function $y=-\frac{1}{2}x+b$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the graph of the function $y=2x$ at point M(2, 4); there is a moving point P on the x-axis, through which a perpendicular line to the x-axis is drawn, intersecting the graphs of the functions $y=-\frac{1}{2}x+b$ and $y=2x$ at points C and D, respectively. Is there a point E on the y-axis such that $\triangle OEM$ is an isosceles triangle? If it exists, find the coordinates of point E; if not, explain why.

Solution:
\textbf{Solution:} Let $E(0, m)$, since $M(2, 4)$. Therefore, $OM^2 = 2^2 + 4^2 = 20$, that is: $OE^2 = m^2$, $EM^2 = 2^2 + (m-4)^2 = m^2 - 8m + 20$,
\begin{enumerate}
    \item When $OE=OM$, $OE^2 = OM^2$, therefore $m^2 = 20$. Solving this gives $m = \pm\sqrt{20}$, therefore, the coordinates of point E are $(0, \sqrt{20})$ or $(0, -\sqrt{20})$;
    \item When $OE=EM$, $OE^2 = EM^2$, therefore $m^2 = m^2 - 8m + 20$. Solving this gives $m = \frac{5}{2}$, therefore, the coordinates of point E are $\left(0, \frac{5}{2}\right)$;
    \item When $OM=EM$, draw $MF \perp OE$, where point F is the midpoint of OE, \\
    $OF = \sqrt{20 - 4} = 4$, $OE = 8$. \\
    When E coincides with O (which is not according to the question, discard) \\
    (Or: When $OM=EM$, $OM^2 = EM^2$, therefore $20 = m^2 - 8m + 20$, solving this gives $m = 8$ or $0$ (discard), \\
    therefore, the coordinates of point E are $(0, 8)$);
\end{enumerate}
Hence, the coordinates of point E are $\boxed{(0, \sqrt{20})}$, $\boxed{(0, -\sqrt{20})}$, $\boxed{\left(0, \frac{5}{2}\right)}$.


Question: As shown in the diagram, the graph of the linear function $y=x+3$, denoted by ${l}_{1}$, intersects the $x$-axis at point $B$, and intersects the graph of another linear function passing through point $A(3, 0)$, denoted by ${l}_{2}$, at point $C(1, m)$. What is the equation of the linear function represented by the graph ${l}_{2}$?

Solution:
\textbf{Solution:} Since $C(1, m)$ is on the graph of $l_1\colon y=x+3$,\\
$\therefore m=1+3=4$;\\
Assume the expression for the linear function graph $l_2$ is $y=kx+b\ (k\ne 0)$,\\
Substituting the points $A(3, 0)$ and $C(1, 4)$, we get: $\left\{\begin{array}{l}3k+b=0 \\ k+b=4\end{array}\right.$,\\
Solving this, we find $\left\{\begin{array}{l}k=-2 \\ b=6\end{array}\right.$,\\
$\therefore$ The equation of $l_2$ is: \boxed{y=-2x+6}.


Question: As shown in the figure, the graph of the linear function $y=x+3$ denoted as ${l}_{1}$ intersects the $x$-axis at point $B$ and intersects with the graph of another linear function that passes through point $A(3, 0)$, denoted as ${l}_{2}$, at point $C(1, m)$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Let $CD\perp x$-axis at point $D$ passing through point $C$,\\
$\therefore B(-3, 0)$,\\
$\therefore AB=6$, $CD=4$,\\
$\therefore$ The area of $\triangle ABC = \frac{1}{2} \cdot AB \cdot CD = \frac{1}{2} \times 6 \times 4 = \boxed{12}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\frac{1}{2}x-1$ intersects with the line $y=-2x+2$ at point $P$, and intersects with the $x$-axis at points $A$ and $B$, respectively. What are the coordinates of the intersection point $P$?

Solution:
\textbf{Solution:} From the given, we have $\begin{cases}y=-\frac{1}{2}x-1 \\ y=-2x+2\end{cases}$,\\
Solving this, we get $\begin{cases}x=2 \\ y=-2\end{cases}$,\\
$\therefore \boxed{P\left(2, -2\right)}$.


Question: As shown in the figure, in a Cartesian coordinate system, the line $y=-\frac{1}{2}x-1$ intersects with the line $y=-2x+2$ at point $P$, and they intersect the $x$-axis at point A and B, respectively. What is the area of $\triangle PAB$?

Solution:
\textbf{Solution:} Let $y=-\frac{1}{2}x-1=0$,\\
then $-\frac{1}{2}x-1=0$,\\
solving, we find: $x=-2$,\\
$\therefore \boxed{A(-2, 0)}$,\\
Let $y=-2x+2=0$,\\
then $-2x+2=0$,\\
solving, we find: $x=1$,\\
$\therefore \boxed{B(1, 0)}$,\\
$\therefore AB=3$,\\
$\therefore {S}_{\triangle PAB}=\frac{1}{2}AB\cdot \left|{y}_{P}\right|=\frac{1}{2}\times 3\times 2=\boxed{3}$.


Question: As shown in Figure 1, the point $A(4, n)$ lies on the graph of the inverse proportion function $y=\frac{8}{x}$ ($x>0$). Find the value of $n$ and the equation of the line $OA$.

Solution:
\textbf{Solution:} Let's substitute point A$(4,n)$ into $y=\frac{8}{x}$, we get $n=2$;\\
Let the equation of line OA be $y=kx$. Substitute $\left\{\begin{array}{l}x=4\\ y=2\end{array}\right.$ into it, we have\\
$4k=2$, solving this yields: $k=\frac{1}{2}$,\\
$\therefore$ the equation of line OA is $y=\frac{1}{2}x$;\\
$\therefore$ the value of $n$ is $\boxed{2}$, and the equation of line $OA$ is $\boxed{y=\frac{1}{2}x}$.


Question: Given: As shown in Figure 1, point $A(4, n)$ lies on the graph of the inverse proportion function $y=\frac{8}{x}$ ($x>0$). As shown in Figure 2, after rotating the graph of the inverse proportion function $y=\frac{8}{x}$ ($x>0$) counterclockwise around the origin $O$ by $45^\circ$, it intersects the $y$-axis at point $M$. Find the length of segment $OM$.

Solution:
\textbf{Solution:} As shown in Figure 1, rotate the y-axis clockwise by $45^\circ$, intersecting the graph of $y=\frac{8}{x}(x>0)$ at point N,\\
then $OM=ON$,\\
the equation of line ON is $y = x$,\\
from $\left\{\begin{array}{l}y=\frac{8}{x}\\ y=x\end{array}\right.$, we get: $\left\{\begin{array}{l}x=2\sqrt{2}\\ y=2\sqrt{2}\end{array}\right.$ or $\left\{\begin{array}{l}x=-2\sqrt{2}\\ y=-2\sqrt{2}\end{array}\right.$ (discard)\\
$\therefore$Point N ($2\sqrt{2}, 2\sqrt{2}$)\\
$\therefore OM=ON=\sqrt{(2\sqrt{2})^{2}+(2\sqrt{2})^{2}}=\boxed{4}$;


Question: Given: As shown in Figure 1, the point $A(4, n)$ lies on the graph of the inverse proportion function $y=\frac{8}{x}(x>0)$. As shown in Figure 3, rotate line $OA$ counterclockwise by $45^\circ$ around the origin $O$, and it intersects the graph of the inverse proportion function $y=\frac{8}{x}(x>0)$ at point $B$. Find the coordinates of point $B$.

Solution:
\textbf{Solution:} As shown in Figure 2, construct the symmetric point $A_1$ of point A with respect to the line OB, \\
then $OA=OA_1$, $AA_1 \perp OB$, \\
construct $A_1C \perp y$-axis at point C, construct $AD \perp x$-axis at point D, \\
it is easy to prove $\triangle A_1OC \cong \triangle AOD$, \\
$\therefore OC=OD$, $A_1C=AD$, \\
$\because$ The coordinates of A are $(4,2)$, \\
$\therefore$ The coordinates of $A_1$ are $(-2, 4)$, \\
$\therefore$ The equation of line $AA_1$ is: $y=-\frac{1}{3}x+\frac{10}{3}$, \\
$\therefore$ The equation of line $OB$ is: $y=3x$, \\
From $\begin{cases}y=\frac{9}{x}\\ y=3x\end{cases}$, we get $\begin{cases}x=\frac{2\sqrt{6}}{3}\\ y=2\sqrt{6}\end{cases}$ or $\begin{cases}x=-\frac{2\sqrt{6}}{3}\\ y=-2\sqrt{6}\end{cases}$ (negative solution is discarded) \\
$\therefore$ The coordinates of point $B\left(\frac{2\sqrt{6}}{3}, 2\sqrt{6}\right)$ are $\boxed{\left(\frac{2\sqrt{6}}{3}, 2\sqrt{6}\right)}$.


Question: As illustrated in the Cartesian coordinate system, the line \( l: y=-\frac{\sqrt{3}}{3}x+4 \) intersects the x-axis and y-axis at points M and N, respectively. An equilateral triangle \(\triangle ABC\) with a height of 3 units, and side \(BC\) lying on the x-axis, is translated in the positive direction of the x-axis. During this translation, \(\triangle A_1B_1C_1\) is obtained. When point \(B_1\) coincides with the origin, determine the coordinates of point \(A_1\) and ascertain whether \(A_1\) lies on line \(l\)?

Solution:
\textbf{Solution:} As shown in the figure, draw $A_1D\perp OM$ through point $A_1$, with D being the foot of the perpendicular.\\
Since $\triangle A_1B_1C_1$ is an equilateral triangle, and $A_1D\perp OM$,\\
therefore, $\angle B_1A_1D=30^\circ$,\\
therefore, in $Rt\triangle A_1DB_1$, $B_1D=\frac{1}{2}A_1B_1$.\\
Since $A_1D=3$,\\
therefore, $A_1B_1^2-B_1D^2=3B_1D^2=A_1D^2=9$,\\
therefore, $B_1D=\sqrt{3}, A_1B_1=2\sqrt{3}$.\\
therefore, the coordinates of point $A_1$ are \(\boxed{(\sqrt{3}, 3)}\).\\
From the equation of line l, we get\\
when $x=\sqrt{3}$, $y=-\frac{\sqrt{3}}{3}\times \sqrt{3}+4=3$\\
therefore, point \(\boxed{A_1}\) lies on the line l.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $l\colon y=-\frac{\sqrt{3}}{3}x+4$ intersects the x-axis and y-axis at points M and N, respectively. An equilateral triangle ABC with a height of 3, where side BC is on the x-axis, is translated in the positive direction of the x-axis. During the translation process, $\triangle A_1B_1C_1$ is obtained. When point $B_1$ coincides with the origin, find a point P on the right side of line $A_1C_1$ such that the quadrilateral formed by vertices P, $A_1$, $C_1$, and M is a parallelogram. Find the coordinates of point P.

Solution:
\textbf{Solution:}

(1) If we consider $\angle A_1C_1M$ as an internal angle of the parallelogram, then the desired parallelogram is parallelogram $A_1C_1MP$.

As shown in Figure (1), draw $A_1E\perp ON$ from point $A_1$, with E being the foot of the perpendicular.

From the analytic equation of line $l$, we get

when $y=0$, $-\frac{\sqrt{3}}{3}x+4=0$,

therefore $x=4\sqrt{3}$.

Therefore, the coordinates of point M are $(4\sqrt{3}, 0)$

Therefore, $OM=4\sqrt{3}$.

Because $B_1C_1=A_1B_1=2\sqrt{3}$.

Therefore, $C_1M=OM-B_1C_1=4\sqrt{3}-2\sqrt{3}=2\sqrt{3}$

Therefore, $A_1P=C_1M=2\sqrt{3}$

Therefore, $PE=A_1E+A_1P=\sqrt{3}+2\sqrt{3}=3\sqrt{3}$.

Therefore, the coordinates of point P are $\boxed{(3\sqrt{3}, 3)}$;

(2) If we consider $\angle C_1A_1M$ as an internal angle of the parallelogram, then the desired parallelogram is $A_1C_1PM$. As shown in Figure (2), draw $PG\perp OM$ from point P, with G being the foot of the perpendicular.

Because $\triangle A_1B_1C_1$ is an equilateral triangle,

therefore $\angle A_1C_1B_1=60^\circ$,

therefore $\angle AC_1G=120^\circ$,

because parallelogram $A_1C_1PM$ is a parallelogram,

therefore $A_1C_1\parallel PM$, $PM=A_1C_1=A_1B_1=2\sqrt{3}$,

therefore $\angle PMC_1=120^\circ$,

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

therefore, in $\triangle PMG$, $\angle MPG=90^\circ-\angle PMG=90^\circ-60^\circ=30^\circ$.

Therefore, $MG=\frac{1}{2}PM=\frac{1}{2}\times 2\sqrt{3}=\sqrt{3}$,

$PG=\sqrt{PM^2-MG^2}=\sqrt{(2\sqrt{3})^2-(\sqrt{3})^2}=3$,

therefore $OG=OM+MG=4\sqrt{3}+\sqrt{3}=5\sqrt{3}$.

Therefore, the coordinates of point P are $\boxed{(5\sqrt{3}, -3)}$

In conclusion, the coordinates of point P are either $\boxed{(3\sqrt{3}, 3)}$ or $\boxed{(5\sqrt{3}, -3)}$.


Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} According to the problem, we can establish the following system of equations:
\[
\begin{cases}
y = -2x + 8\\
y = \frac{1}{2}x + 3
\end{cases}
\]
Solving this, we obtain:
\[
\begin{cases}
x = 2\\
y = 4
\end{cases}
\]
Therefore, the coordinates of point $D$ are $\boxed{(2, 4)}$.


Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. As shown in figure 2, draw $AE\parallel y$-axis from point $A$ to intersect the line $y=\frac{1}{2}x+3$ at point $E$, and connect $AC$, $BE$. Prove that quadrilateral $ACBE$ is a rhombus.

Solution:
\textbf{Solution:} Proof: \\
$ \because $ the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, \\
$ \therefore $ point $B(0,8)$, and point $A(4, 0)$, \\
$ \because $ the line $y=\frac{1}{2}x+3$ intersects the $y$-axis at point $C$, \\
$ \therefore $ point $C(0, 3)$, \\
$ \because $AE$\parallel$y-axis and intersects the line $y=\frac{1}{2}x+3$ at point $E$, \\
$ \therefore $ point $E(4, 5)$ \\
$ \because $ point $B(0,8)$, point $A(4, 0)$, point $C(0, 3)$, and point $E(4, 5)$, \\
$ \therefore $BC=5, $AE=5$, $AC=5$, $BE=5$, \\
$ \therefore $BC=AE=AC=BE, \\
$ \therefore $ the quadrilateral $ACBE$ is a rhombus $\boxed{}$;


Question: As shown in the figure, the line $y=-2x+8$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $y=\frac{1}{2}x+3$ intersects the $y$-axis at point $C$. The two lines intersect at point $D$. As shown in Figure 3, under the condition of (2), point $F$ is on the line segment $BC$, and point $G$ is on the line segment $AB$. Connect $CG$ and $FG$. When $CG=FG$ and $\angle CGF=\angle ABC$, what are the coordinates of point $G$?

Solution:
\textbf{Solution:} Given $BC=AC$,\\
it follows that $\angle ABC=\angle CAB$,\\
since $\angle CGF=\angle ABC$ and $\angle AGF=\angle ABC+\angle BFG=\angle AGC+\angle CGF$,\\
it implies $\angle AGC=\angle BFG$ and $FG=CG$, $\angle ABC=\angle CAB$,\\
thus $\triangle ACG\cong \triangle BGF$ (By the AAS criterion),\\
implying $BG=AC=5$,\\
let point $G(a, -2a+8)$,\\
hence $(-2a+8-8)^2+(a-0)^2=5^2$,\\
leading to $a=\pm \sqrt{5}$,\\
since point $G$ lies on segment $AB$,\\
it must be $a=\sqrt{5}$,\\
therefore, the coordinates of point $G$ are $\boxed{(\sqrt{5}, 8-2\sqrt{5})}$.


Question: Given the figure, in the Cartesian coordinate system, the graph of the function $y=kx+2$ passes through the point $A(3,0)$. After moving the graph upwards by 2 units, it intersects the $x$ axis at point $B$ and the $y$ axis at point $C$. What is the linear equation represented by the graph passing through points $B$ and $C$?

Solution:
\textbf{Solution:}  Substituting $A(3, 0)$ into $y=kx+2$ gives: $3k+2=0$, \\
Solving for $k$, we find $k=-\frac{2}{3}$, \\
$\therefore y=-\frac{2}{3}x+2$. \\
$\because$ Shifting the graph of the function $y=-\frac{2}{3}x+2$ up by 2 units results in $y=-\frac{2}{3}x+4$, \\
$\therefore$ The linear equation passing through points B and C is $\boxed{y=-\frac{2}{3}x+4}$;


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the function $y=kx+2$ passes through point $A(3,0)$. After the graph is translated upwards by $2$ units, it intersects the $x$ axis at point $B$ and the $y$ axis at point $C$. Find the area of $\triangle OBC$.

Solution:
\textbf{Solution:} In the line $y=-\frac{2}{3}x+4$, when $x=0$, then $y=4$; when $y=0$, then $x=6$,\\
$\therefore$ the coordinates of $B$ are $(6, 0)$, and the coordinates of $C$ are $(0, 4)$\\
$\therefore$ $OB=6, OC=4$,\\
$\therefore$ the area of $\triangle OBC$ is $\frac{1}{2}\times 6\times 4=\boxed{12}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ intersects the x-axis at point A $(-3,0)$, intersects the y-axis at point B, and intersects with line $l_{2}: y=\frac{4}{3}x$ at point C $(a,4)$. What is the equation of line $l_{1}$?

Solution:
\textbf{Solution:} Given point C(3, 4), let the equation of line $l_{1}$ be $y=kx+b$, \\
$\left\{\begin{array}{l}
3k+b=4\\
-3k+b=0
\end{array}\right.$ Solving this system, we get $\left\{\begin{array}{l}
k=\frac{2}{3}\\
b=2
\end{array}\right.$\\
$\therefore$ The equation of line $l_{1}$ is: $\boxed{y=\frac{2}{3}x+2}$.


Question: As illustrated in the Cartesian coordinate system, the line $l_{1}$ intersects with the x-axis at point A$(-3, 0)$ and with the y-axis at point B, while its intersection with line $l_{2}\colon y=\frac{4}{3}x$ is point C$(a, 4)$. If the quadrilateral formed by points O, D, B, C is a parallelogram, write down the coordinates of point D directly.

Solution: \textbf{Solution:} Given that $B(0,2)$ and $C(3,4)$. When $OB$ is a side, by drawing a parallel line to $OB$ through point $C$ and setting $CD=OB$, point $C$ can be translated up or down by 2 units. Thus, the positions of point $D$ are $D_1(3,2)$ and $D_2(3,6)$. When $OB$ is a diagonal, by constructing point $D_1$'s symmetric point $D_3$ with respect to point $O$ and given that $D_1(3,2)$, we find that $D_3(-3,-2)$. Therefore, the coordinates of point $D$ are $\boxed{(3,2)}$, $\boxed{(3,6)}$, or $\boxed{(-3,-2)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}$ intersects the x-axis at point $A(-3,0)$ and intersects the y-axis at point $B$. Additionally, it intersects line $l_{2}\colon y=\frac{4}{3}x$ at point $C(a,4)$. The line $l_{1}$ is translated 3 units downward along the y-axis to obtain line $l_{3}$. Point $P(m,n)$ is a moving point on line $l_{2}$, from which a perpendicular to the x-axis is drawn, intersecting lines $l_{1}$ and $l_{3}$ at points $M$ and $N$ respectively. When point $P$ is on the segment $MN$, determine the range of values for $m$.

Solution:
\textbf{Solution:} Since the line $l_{1}$ is translated 3 units downwards along the y-axis to obtain the line $l_{3}$,\\
therefore, the equation of the line $l_{3}$ is $y=\frac{2}{3}x+2-3=\frac{2}{3}x-1$\\
Solving for: $\left\{\begin{array}{l}
y=\frac{2}{3}x-1\\
y=\frac{4}{3}x
\end{array}\right.$\\
Yields: $\left\{\begin{array}{l}
x=-\frac{3}{2}\\
y=-2
\end{array}\right.$\\
Therefore, the intersection point of line $l_{2}$ and line $l_{3}$ is $\left(-\frac{3}{2}, -2\right)$;\\
Since the intersection point of line $l_{2}$ and line $l_{3}$ is given as $(3, 4)$,\\
therefore, the range of values for $m$ is: \boxed{-\frac{3}{2}\le m \le 3}


Question: As shown in the figure, point $N(0, 6)$ lies on the plane, and point $M$ is on the negative $x$-axis with $ON = 3OM$. $A$ is a point on line segment $MN$, $AB \perp x$-axis with $B$ as the foot of the perpendicular, and $AC \perp y$-axis with $C$ as the foot of the perpendicular. Write down the coordinates of point $M$.

Solution:
\textbf{Solution:} Given, $\because N(0,6)$,\\
$\therefore ON=6$,\\
$\because ON=3OM$,\\
$\therefore OM=2$,\\
$\therefore$ the coordinates of point M are $\boxed{(-2,0)}$;


Question: As shown in the figure, point $N(0,6)$ and point $M$ is on the negative half axis of the $x$-axis, with $ON=3OM$. $A$ is a point on the line segment $MN$. $AB \perp x$-axis with the foot of the perpendicular as point $B$, and $AC \perp y$-axis with the foot of the perpendicular as point $C$. What is the equation of line $MN$?

Solution:
\textbf{Solution:} To find the equation of line $MN$, we utilize the equation $y=kx+b$. Substituting $M(-2,0)$ and $N(0,6)$ into it, we get
\[
\left\{
\begin{array}{l}
-2k+b=0 \\
b=6
\end{array}
\right.
\]
Solving this system yields
\[
\left\{
\begin{array}{l}
k=3 \\
b=6
\end{array}
\right.
\]
$\therefore$ The equation of line $MN$ is $\boxed{y=3x+6}$.


Question: As shown in the diagram, point N(0,6) and point M is on the negative half-axis of the x-axis, with ON = 3OM. A is a point on segment MN, AB$\perp$x-axis with B as its foot, and AC$\perp$y-axis with C as its foot. If the x-coordinate of point A is $-1$, what is the area of rectangle ABOC?

Solution:
\textbf{Solution:} In the equation $y=3x+6$, when $x=-1$, we have $y=3$,\\
$\therefore OB=1$, $AB=3$,\\
$\therefore$ the area of the rectangle ABOC is $1\times 3=\boxed{3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line \( y=\frac{3}{4}x+6\) intersects the x-axis at point A and the y-axis at point B. Triangle \(AOB\) is folded along BC, with point O exactly falling on point D on side AB, where BC is the fold line. What is the length of segment AB?

Solution:
\textbf{Solution:} Let $x=0$, then $y=6$,\\
Let $y=0$, then $\frac{3}{4}x+6=0$, solving this gives $x=-8$,\\
$\therefore$ the coordinates of point A are $(-8,0)$, and the coordinates of point B are $(0,6)$,\\
$\therefore$ OA=8, OB=6,\\
$\therefore$ $AB=\sqrt{OA^2+OB^2}=\boxed{10}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{3}{4}x+6$ intersects the x-axis at point A and the y-axis at point B. Triangle $\triangle AOB$ is folded along BC, where point O precisely lands on point D on side AB, with BC being the crease. Find the analytical expression of line BC.

Solution:
\textbf{Solution:} According to the problem statement, we have: $\angle ADC=\angle BDC=\angle BOC=90^\circ$, $BD=OB=6$,\\
$\therefore AD=AB-BD=10-6=4$,\\
let $OC=x$, then $AC=OA-OC=8-x$,\\
in $\triangle ACD$, $AD^2+CD^2=AC^2$,\\
$\therefore 4^2+x^2=(8-x)^2$, solving for $x$ yields: $x=3$, that is $OC=3$,\\
$\therefore$ the coordinates of point C are $(-3,0)$,\\
let the equation of line BC be $y=kx+b\ (k\neq 0)$,\\
substituting points B$(0,6)$, C$(-3,0)$ into the equation gives:\\
$\begin{cases} b=6 \\ -3k+b=0 \end{cases}$, solving for $k$ and $b$ gives: $\begin{cases} k=2 \\ b=6 \end{cases}$,\\
$\therefore$ the equation of line BC is $y=2x+6$. Hence, the answer is $\boxed{y=2x+6}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the line \( l_{1}: y_{1}=2x \) intersects with the line \( l_{2}: y_{2}=-x+3 \) at point A. What are the coordinates of point A?

Solution:
\textbf{Solution:} From the given, we have
\[
\begin{cases}
y=2x \\
y=-x+3
\end{cases}
\]
Solving these equations, we get:
\[
\begin{cases}
x=1 \\
y=2
\end{cases}
\]
$\therefore$ The coordinates of point A are $\boxed{(1, 2)}$.


Question: As shown in the graph, in the Cartesian coordinate system \(xOy\), the line \(l_1: y_1 = 2x\) intersects with the line \(l_2: y_2 = -x + 3\) at point A. When \(y_1 < y_2\), directly write the range of values for \(x\).

Solution:
\textbf{Solution:} From (1), it is known that point A is at $(1, 2)$,\\
According to the graph, when ${y}_{1}<{y}_{2}$, the range of x can be directly written as $\boxed{x<1}$;


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the graph of the inverse proportion function $y=\frac{m}{x}$ where $x>0$, at points $A(1, 4)$ and $B(4, n)$. What are the expressions for the linear function and the inverse proportion function?

Solution: \textbf{Solution:} Substituting point A into the inverse proportion function $y=\frac{m}{x}\left(x>0\right)$, we have $m=xy=4$. Therefore, the expression of the inverse proportion function is: $y=\frac{4}{x}$. Hence, $4n=4$, solving this equation, we find $n=1$. Therefore, point $B(4, 1)$ is determined. This leads to the system $\left\{\begin{array}{l}1=4k+b\\ 4=k+b\end{array}\right.$, from which we solve $\left\{\begin{array}{l}k=-1\\ b=5\end{array}\right.$. Therefore, the equation of the linear function is: \boxed{y=-x+5};


Question: As shown in the figure, the graph of the linear function $y=kx+b$ and the graph of the inverse proportion function $y=\frac{m}{x}$ (for $x>0$) intersect at points $A(1, 4)$ and $B(4, n)$. Based on the graph, directly write out the solution set for the inequality $kx+b-\frac{m}{x}>0$ in terms of $x$.

Solution: \textbf{Solution:} According to the graph, we can directly write the solution set of the inequality $kx + b - \frac{m}{x} > 0$ with respect to $x$ as $\boxed{1<x<4}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}\ (x>0)$ at points $A(1, 4)$ and $B(4, n)$. Find the area of $\triangle AOB$.

Solution:
Solution: As shown in the figure, let the graph of the linear function intersect the coordinate axis at points C and D, and draw AE perpendicular to the y-axis at E and BF perpendicular to the x-axis at F, \\
$\because A\left(1, 4\right)$ , $B(4, 1)$,\\
$\therefore$AE=BF=1,\\
$\because$ the equation of the linear function is: $y=-x+5$,\\
$\therefore$OC=OD=5,\\
$\therefore$ $ {S}_{\triangle AOC}={S}_{\triangle BOD}=\frac{1}{2}OA\times AE=\frac{5}{2}$,\\
$ {S}_{\triangle COD}=\frac{1}{2}OC\times OD=\frac{25}{2}$,\\
$\therefore$ $ {S}_{\triangle AOB}={S}_{\triangle COD}-{S}_{\triangle AOC}-{S}_{\triangle BOD}=\frac{25}{2}-\frac{5}{2}-\frac{5}{2}=\boxed{\frac{15}{2}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+10$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Another line passing through point $B$ intersects the positive half of the $x$ axis at point $C$, and the area of $\triangle ABC$ is $60$. What are the coordinates of point $C$ and the equation of line $BC$?

Solution: \textbf{Solution:} Since point B is on the line $y=2x+10$, we have B(0,10). When $y=0$, $x=-5$, therefore OA=5. The area of $\triangle ABC$, $S_{\triangle ABC}=\frac{1}{2}AC\cdot OB=60$, solving this gives $AC=12$. Therefore, OC=AC−OA=7, so C(7,0). Let the equation of line BC be $y=kx+10$, thus $0=7k+10$. Solving for $k$ yields $k=-\frac{10}{7}$, hence $y=-\frac{10}{7}x+10$. Therefore, the coordinates of point C are $\boxed{C(7,0)}$, and the equation of line BC is $\boxed{y=-\frac{10}{7}x+10}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=2x+10$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Another line passing through point $B$ intersects the positive half of the $x$-axis at point $C$, and the area of $\triangle ABC$ is $60$. If $M$ is a point on segment $BC$ and the line $AM$ divides the area of $\triangle ABC$ into two parts with an area ratio of $1\colon 2$, what are the coordinates of $M$?

Solution:
\textbf{Solution:} Let $M(m,-\frac{10}{7}m+10)$,\\
(1) When $\frac{S_{\Delta ACM}}{S_{\Delta ABM}} = \frac{1}{2}$, that is $S_{\Delta ACM} = \frac{1}{3}S_{\Delta ABC}$,\\
$\therefore \frac{1}{2}AC \cdot y_{M} = \frac{1}{3} \times 60 = 20$,\\
$\because AC = 12$,\\
$\therefore y_{M} = \frac{10}{3}$,\\
$\therefore \frac{10}{3} = -\frac{10}{7}m + 10$,\\
Solving, we get $m = \frac{14}{3}$,\\
$\therefore M\left(\frac{14}{3}, \frac{10}{3}\right)$;\\
(2) When $\frac{S_{\Delta ABM}}{S_{\Delta ACM}} = \frac{1}{2}$, that is $S_{\Delta ACM} = \frac{2}{3}S_{\Delta ABC}$,\\
$\therefore \frac{1}{2}AC \cdot y_{M} = \frac{2}{3} \times 60 = 40$,\\
$\because AC = 12$,\\
$\therefore y_{M} = \frac{20}{3}$,\\
$\therefore \frac{20}{3} = -\frac{10}{7}m + 10$,\\
Solving, we get $m = \frac{7}{3}$,\\
$\therefore M\left(\frac{7}{3}, \frac{20}{3}\right)$,\\
In conclusion, the coordinates of point M are $\boxed{\left(\frac{14}{3}, \frac{10}{3}\right)}$, $\boxed{\left(\frac{7}{3}, \frac{20}{3}\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+10$ intersects with the $x$ axis at point $A$ and with the $y$ axis at point $B$. Another line passing through point $B$ intersects the positive $x$ axis at point $C$, and the area of $\triangle ABC$ is $60$. When the area of $\triangle ABM$ is $20$, point $E$ is a moving point on line $AM$. Does there exist a point $D$ on the $x$ axis such that the quadrilateral formed by points $D$, $E$, $B$, and $C$ is a parallelogram? If it exists, write down the coordinates of point $D$ directly; if not, please explain the reason.

Solution:
\textbf{Solution:} There exist coordinates for point $D$ as $(13,0)$, $(-23,0)$, or $(1,0)$, for the following reasons:\\
$\because S_{\triangle ACM}=S_{\triangle ABC}-S_{\triangle ABM}=60-20=40$,\\
$\therefore \frac{1}{2}AC \cdot y_M=40$,\\
$\because AC=12$,\\
$\therefore y_M=\frac{20}{3}$,\\
$\therefore \frac{20}{3}=-\frac{10}{7}m+10$,\\
Solving this yields: $m=\frac{7}{3}$,\\
$\therefore M\left(\frac{7}{3}, \frac{10}{3}\right)$,\\
Let the equation of line AM be $y=kx+b$, then $\left\{\begin{array}{l}\frac{7}{3}m+n=\frac{20}{3}\\ -5m+n=0\end{array}\right.$,\\
$\therefore \left\{\begin{array}{l}m=\frac{10}{11}\\ n=\frac{50}{11}\end{array}\right.$,\\
$\therefore$ The equation of line AM is $y=\frac{10}{11}x+\frac{50}{11}$,\\
(1) When BC is a side of a parallelogram, and quadrilateral BCDE forms a parallelogram,\\
$\because BE\parallel CD$, $B(0,10)$,\\
$\therefore y_E=10$,\\
$\therefore 10=\frac{10}{11}x+\frac{50}{11}$,\\
Solving for $x$ gives $x=6$,\\
$\therefore E(6,10)$,\\
$\because$ quadrilateral BCDE forms a parallelogram,\\
$\therefore CD=BE=6$,\\
$\therefore OD=OC+CD=7+6=13$,\\
$\therefore \boxed{D(13,0)}$;\\
(2) When BC is a side of a parallelogram, and quadrilateral BDEC forms a parallelogram, by drawing EF$\perp$CD at point F,\\
$\because$ quadrilateral BDCE forms a parallelogram,\\
$\therefore BC=DE$, $BC\parallel DE$,\\
$\therefore \angle EDF=\angle BCO$,\\
And $\because \angle EFD=\angle BOC$,\\
$\therefore \triangle BDC\cong \triangle ECD$ (AAS),\\
$\therefore EF=OB=10$, $DF=OC=7$,\\
$\therefore y_E=-10$,\\
$\therefore -10=\frac{10}{11}x+\frac{50}{11}$,\\
Solving for $x$ gives $x=-16$,\\
$\therefore E(-16,10)$,\\
$\therefore OF=16$,\\
$\therefore OD=OF+FD=16+7=23$,\\
$\therefore \boxed{D(-23,0)}$;\\
(3) When BC is a diagonal of a parallelogram, and quadrilateral BDEC forms a parallelogram,\\
$\because BE\parallel CD$, $B(0,10)$,\\
$\therefore y_E=10$,\\
$\therefore 10=\frac{10}{11}x+\frac{50}{11}$,\\
Solving for $x$ gives $x=6$,\\
$\therefore E(6,10)$,\\
$\because$ quadrilateral BCDE forms a parallelogram,\\
$\therefore CD=BE=6$,\\
$\because OC=7$,\\
$\therefore OD=OC-CD=1$,\\
$\therefore \boxed{D(1,0)}$;\\
In summary, the coordinates of point $D$ are: \boxed{(13,0)}, \boxed{(-23,0)}, or \boxed{(1,0)}.


Question: As shown in the figure, the graph of the direct proportion function $y=kx$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ ($x>0$) at point $A(2, 2)$. By translating the line $y=kx$ downwards, we obtain line $l$. If line $l$ intersects with the graph of the inverse proportion function at point $B(3, n)$. What are the values of $m$ and $n$?

Solution:
\textbf{Solution}: From the given information, substituting point A$(2,2)$ into the inverse proportion function $y=\frac{m}{x}$, we get: $m=2\times 2=4$,\\
$\therefore y=\frac{4}{x}$. Then, substituting B$(3,n)$ into $y=\frac{4}{x}$, we find: $n=\frac{4}{3}$;\\
Hence, $m=\boxed{4}$, $n=\boxed{\frac{4}{3}}$;


Question: As shown in the figure, the graph of the direct proportion function $y=kx$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ (where $x>0$) at point $A(2, 2)$. By translating the line $y=kx$ downward, we obtain line $l$. If line $l$ intersects with the graph of the inverse proportion function at point $B(3, n)$, connect $AB$ and $OB$. What is the area of $\triangle AOB$?

Solution: \textbf{Solution:} Start by substituting point A(2,2) into $y=kx$, we get: $2=2k$, thus $k=1$, and therefore $y=x$. Since the line $y=kx$ is translated downwards to form line $l$, we set the equation of line $l$ as $y=x+b$. Given it intersects the x-axis at point C, substituting point B(3,$\frac{4}{3}$) yields: $b=-\frac{5}{3}$. Hence, the equation of line $l$ is $y=x-\frac{5}{3}$. When $y=0$, we have $x=\frac{5}{3}$, thus $OC=\frac{5}{3}$. Connecting AC and knowing that $OA\parallel BC$, we conclude that ${S}_{\triangle AOB}={S}_{\triangle AOC}=\frac{1}{2}\times \frac{5}{3}\times 2=\boxed{\frac{5}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1\colon y_1=kx+b$ that passes through point $A(-6, 0)$ intersects with the line $l_2\colon y_2=2x$ at point $B(m, 4)$. What is the analytical expression of line $l_1$?

Solution:
\[ \textbf{Solution:} \]
Since point B $(m, 4)$ lies on the line $y = 2x$,\\
we have $2m = 4$. Solving this gives $m = 2$.\\
Therefore, point B is $(2, 4)$;\\
Since the line $l_1$ passes through point A $(-6, 0)$ and also goes through point B $(2, 4)$,\\
we get
\[\left\{\begin{aligned}
-6k + b &= 0\\
2k + b &= 4
\end{aligned}\right.\]
Solving these equations, we find
\[\left\{\begin{aligned}
k &= \frac{1}{2}\\
b &= 3
\end{aligned}\right.,\]
Therefore, $y_1 = \frac{1}{2}x + 3 = \boxed{\frac{1}{2}x + 3}$.


Question: As illustrated in the Cartesian coordinate system, the line $l_1\colon y_1=kx+b$, passing through point $A(-6,0)$, intersects with the line $l_2\colon y_2=2x$ at point $B(m,4)$. Utilize the graphical representation of functions to directly determine the range of $x$ values for which $y_2\leq y_1$.

Solution: \textbf{Solution:} By using the function graph, we can directly write the range of $x$ when $y_2 \leq y_1$ as $\boxed{x \leq 2}$.


Question: As shown in the figure, the line $l\colon y = \frac{4}{3}x+b$ passes through the point $A(-3,0)$ and intersects with the $y$-axis at point $B$. The bisector of $\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular is drawn from point $C$ to line $AB$, intersecting the $x$-axis at point $E$, with the foot of the perpendicular being point $D$. What are the coordinates of points $B$ and $C$?

Solution:
\textbf{Solution}: Substituting point $A(-3,0)$ into $y=\frac{4}{3}x+b$, we find $b=4$,\\
$\therefore B(0,4)$,\\
$\therefore OB=4$,\\
$\because A(-3,0)$,\\
$\therefore OA=3$,\\
In $\triangle AOB$, $\angle AOB=90^\circ$,\\
$\therefore AB=\sqrt{OA^2+OB^2}=5$.\\
$\because AC$ bisects $\angle OAB$, $CD\perp AB$, $CO\perp OA$,\\
$\therefore CD=CO$, $\angle ACD=\angle ACO$,\\
$\because AC=AC$,\\
$\therefore \triangle ACD\cong \triangle ACO$ (SAS),\\
$\therefore AD=AO=3$, $BD=AB-AD=2$.\\
Let $CD=CO=m$, then $BC=4-m$,\\
In $\triangle BDC$,\\
by the Pythagorean theorem, $CD^2+BD^2=BC^2$,\\
$\therefore m^2+2^2=(4-m)^2$,\\
Solving, we get $m=\frac{3}{2}$,\\
$\therefore C(0,\frac{3}{2})$;\\
$\therefore$ The coordinates of $B$ and $C$ are $B(0,4)$ and $C(0,\frac{3}{2})$ respectively.\\
Therefore, the coordinates of points $B$ and $C$ are $B=\boxed{(0,4)}$ and $C=\boxed{(0,\frac{3}{2})}$.


Question: As shown in the diagram, the line $l: y = \frac{4}{3}x + b$ passes through the point $A(-3, 0)$ and intersects the $y$-axis at point $B$. The bisector of $\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular line to $AB$ is drawn through point $C$, intersecting the $x$-axis at point $E$, with the foot of the perpendicular being point $D$. What is the equation of the line $DE$?

Solution:
\textbf{Solution:} Since $CD\perp AB$ and $CO\perp OA$, \\
it follows that $\angle CDB = \angle COE = 90^\circ$, \\
Since $CD=CO$ and $\angle BCD = \angle ECO$, \\
it follows that $\triangle BCD \cong \triangle ECO$ (ASA), \\
OE=BD=2, \\
Thus, the coordinates of E are $(2,0)$, \\
Given that $C(0, \frac{3}{2})$, \\
let the equation of line DE be $y=kx+\frac{3}{2}$, \\
Thus, $0=2k+\frac{3}{2}$, solving this gives $k=-\frac{3}{4}$, \\
Therefore, the equation of line DE is $\boxed{y=-\frac{3}{4}x+\frac{3}{2}}$.


Question: As shown in the figure, the line $l\colon y=\frac{4}{3}x+b$ passes through point $A(-3,0)$ and intersects the $y$-axis at point $B$. The bisector of $\angle OAB$ intersects the $y$-axis at point $C$. A perpendicular line to line $AB$ passing through point $C$ intersects the $x$-axis at point $E$, and the foot of the perpendicular is point $D$. Let point $P$ be a moving point on the $y$-axis. When the value of $PA+PD$ is minimized, please write down the coordinates of point $P$ directly.

Solution: \textbf{Solution:} The coordinate of point $P$ is $\boxed{\left(0, \frac{12}{7}\right)}$.


Question: As shown in the figure, the line $y=ax+1$ intersects with the x-axis and y-axis at points $A$ and $B$, respectively, and intersects with the hyperbola $y=\frac{k}{x}$ ($x>0$) at point $P$. $PC\perp x$-axis at point $C$, and $PC=2$. The coordinates of point $A$ are $(-2, 0)$. Find the expressions of line $AP$ and the hyperbola?

Solution:
\textbf{Solution:} Substitute $A(-2,0)$ into $y=\frac{1}{2}x+1$ and obtain $a=\frac{1}{2}$,\\
$\therefore y=\frac{1}{2}x+1$,\\
Given $PC=2$, substitute $y=2$ into $y=\frac{1}{2}x+1$ to find $x=2$,\\
that is, $P(2,2)$,\\
Substituting $P(2,2)$ into $y=\frac{k}{x}$ yields $k=4$,\\
$\therefore$ the expression for the hyperbola is $\boxed{y=\frac{4}{x}}$;


Question: As shown in the figure, the straight line $y=ax+1$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$ respectively, and intersects the hyperbola $y=\frac{k}{x}$ (where $x>0$) at point $P$, $PC\perp x$-axis at point $C$, and $PC=2$, the coordinates of point $A$ are $(-2, 0)$. If point $Q$ is a point on the right side of point $P$ on the hyperbola, and $QH\perp x$-axis at $H$, when the triangle formed by points $Q$, $C$, and $H$ is similar to $\triangle AOB$, what are the coordinates of point $Q$?

Solution:
\textbf{Solution:} Given the diagram, draw $QH\perp$ x-axis and let $Q(m,n)$, \\
since $Q(m,n)$ lies on $y= \frac{4}{x}$, \\
thus, $n= \frac{4}{m}$.\\
When $\triangle QCH\sim \triangle BAO$, it follows that $\frac{CH}{AO}=\frac{QH}{BO}$,\\
which means $\frac{m-2}{2}=\frac{n}{1}$,\\
therefore $m-2=2n$,\\
which implies $m-2=\frac{8}{m}$,\\
leading to $m^{2}-2m-8=0$,\\
solving this, we find $m=4$ or $m=-2$ (discard),\\
hence, $Q\left(4, 1\right)$,\\
When $\triangle QCH\sim \triangle ABO$, it follows that $\frac{CH}{BO}=\frac{QH}{AO}$,\\
which means $\frac{m-2}{1}=\frac{n}{2}$,\\
rearranging gives $2m-4=\frac{8}{m}$,\\
leading to $2m^{2}-4m-8=0$,\\
solving this, we find $m=1+\sqrt{3}$ or $m=1-\sqrt{3}$ (discard),\\
hence, $Q\left(1+\sqrt{3}, 2\sqrt{3}-2\right)$,\\
In conclusion, $Q\left(4, 1\right)$ or $Q\left(1+\sqrt{3}, 2\sqrt{3}-2\right)$.\\
Therefore, the coordinates of point $Q$ are $\boxed{Q\left(4, 1\right)}$ or $\boxed{Q\left(1+\sqrt{3}, 2\sqrt{3}-2\right)}$.


Question: As shown in the figure, the line $y=-\frac{3}{2}x-2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects with the hyperbola $y=\frac{m}{x}$ (where $m \neq 0$) at point $C$ in the second quadrant. $CD$ is perpendicular to the $y$-axis at point $D$, and $CD=4$. What is the equation of the hyperbola?

Solution:
\textbf{Solution:} Given that the point of intersection in the second quadrant is C, and CD is perpendicular to the y-axis at point D, with CD equal to 4,\\
thus the x-coordinate of point C is $-4$,\\
substituting $x = -4$ into $y = -\frac{3}{2}x - 2$, we get $y = -\frac{3}{2} \times (-4) - 2 = 4$,\\
therefore, C$(-4,4)$,\\
substituting C$(-4,4)$ into $y = \frac{m}{x}$, we get $4 = \frac{m}{-4}$,\\
hence, $m = -16$,\\
therefore, the equation of the hyperbola is: \boxed{y = -\frac{16}{x}}.


Question: As illustrated, the line $y=-\frac{3}{2}x-2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the hyperbola $y=\frac{m}{x}$ (where $m\neq 0$) at point $C$ in the second quadrant. Line segment $CD$ is perpendicular to the $y$-axis at point $D$, and $CD=4$. Let point $Q$ be a point on the hyperbola, and the area of triangle $QOB$ is twice the area of triangle $AOB$. Find the coordinates of point $Q$.

Solution:
\textbf{Solution:} Insert $x=0$ into $y=-\frac{3}{2}x-2$, we get $y=-\frac{3}{2}\times0-2=-2$,\\
$\therefore B(0,-2)$,\\
Insert $y=0$ into $y=-\frac{3}{2}x-2$, we get $0=-\frac{3}{2}x-2$,\\
$\therefore x=-\frac{4}{3}$,\\
$\therefore A\left(-\frac{4}{3},0\right)$,\\
$\because S_{\triangle QOB}=2S_{\triangle AOB}$,\\
$\therefore \frac{1}{2}OB\cdot |x_{Q}|=2\times \frac{1}{2}OA\cdot OB$,\\
$\therefore \frac{1}{2}\times 2|x_{Q}|=2\times \frac{1}{2}\times \left|-\frac{4}{3}\right|\times 2$,\\
solving this gives $x_Q=\pm \frac{8}{3}$,\\
Insert $x=\frac{8}{3}$ into $y=-\frac{16}{x}$, we get $y=-6$,\\
Insert $x=-\frac{8}{3}$ into $y=-\frac{16}{x}$, we get $y=6$,\\
$\therefore Q\left(\frac{8}{3},-6\right)$ or $\boxed{Q\left(-\frac{8}{3},6\right)}$;


Question: As shown in the figure, the line $y=-\frac{3}{2}x-2$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. It intersects with the hyperbola $y=\frac{m}{x}$ ($m\neq 0$) at point $C$ in the second quadrant. $CD\perp y$-axis at point $D$, and $CD=4$. There exists a point $P$ on the $y$-axis such that $PA+PC$ is the shortest. Please directly write the coordinates of point $P$.

Solution: \textbf{Solution:} The coordinates of point $P$ are $\boxed{(0,1)}$.


Question: As shown in the figure, in the Cartesian coordinate system, point $O$ is the origin. The line $BC$ intersects the y-axis at point $B$ and the x-axis at point $C$. The coordinates of point $B$ are $(0,4)$, and $OB=OC$. What is the analytical expression of line $BC$?

Solution:
\textbf{Solution:} Since $B(0, 4)$, \\
it follows that $OB=4$, \\
thus $OC=OB=4$, \\
implying $C(4, 0)$ \\
Let the equation of line $BC$ be $y=kx+b\ (k \neq 0)$, \\
therefore $\left\{\begin{array}{l}b = 4 \\ 4k + b = 0 \end{array}\right.$ \\
hence $\left\{\begin{array}{l}k = -1 \\ b = 4 \end{array}\right.$ \\
Thus, the equation of line $BC$ is $y = -x + 4$, so the answer is \boxed{y = -x + 4}.


Question: In the Cartesian coordinate system, point O is the origin, and the line $BC$ intersects the y-axis at point B and the x-axis at point C. The coordinates of point B are $\left(0,4\right)$, and $OB=OC$. If the area of $\triangle ABO$ is 2, find the coordinates of point A.

Solution:
\textbf{Solution:} Construct $AD \perp y$-axis at point D, \\
$\therefore {S}_{\triangle ABO} = \frac{1}{2}BO \times AD = \frac{1}{2} \times 4AD = 2AD$, \\
$\because {S}_{\triangle ABO} = 2$, \\
$\therefore 2AD = 2$, \\
$\therefore AD = 1$, \\
$\therefore$ The x-coordinate of point A is 1, \\
When $x=1$, $y = -1 + 4 = 3$, \\
$\therefore A\left(1, 3\right)$. Therefore, the coordinates of point A are $\boxed{A(1, 3)}$.


Question: As shown in the figure, the quadrilateral OABC is a rectangle, where points A and C are on the coordinate axes, and $\triangle ODE$ is obtained by rotating $\triangle OCB$ $90^\circ$ clockwise around point O. Point D is on the x-axis, and line BD intersects the y-axis at point F and intersects OE at point H. The lengths of segments BC and OC are the solutions to the system of equations $\begin{cases} 2x+3y=14 \\ 4x-5y=6 \end{cases}$, with OC > BC. What is the analytical expression of line BD?

Solution:
\textbf{Solution}: From the given problem, we have
\[
\begin{cases}
2x+3y=14\\
4x-5y=6
\end{cases}
\]
Solving these equations, we obtain
\[
\begin{cases}
x=4\\
y=2
\end{cases},
\]
Since $OC>BC$,\\
Hence, $CO=4$ and $BC=2$,\\
Thus, point B is $(-2,4)$,\\
Since $\triangle ODE$ is obtained by rotating $\triangle OCB$ $90^\circ$ clockwise around point O,\\
Therefore, $\triangle ODE \cong \triangle OCB$,\\
Hence, $OD=OC$ and $DE=BC$,\\
Thus, point D is $(4,0)$ and point E is $(4,2)$,\\
Let the equation of line BD be $y=kx+b$,\\
Substituting points B and D, we get
\[
\begin{cases}
-2k+b=4\\
4k+b=0
\end{cases},
\]
Solving these, we find
\[
\begin{cases}
k=-\frac{2}{3}\\
b=\frac{8}{3}
\end{cases},
\]
Therefore, the equation of line BD is $y=-\frac{2}{3}x+\frac{8}{3}$; hence, the equation of line BD is $\boxed{y=-\frac{2}{3}x+\frac{8}{3}}$.


Question: As shown in the figure, quadrilateral $OABC$ is a rectangle with points $A$ and $C$ on the coordinate axes. $\triangle ODE$ is obtained by rotating $\triangle OCB$ $90^\circ$ clockwise about point $O$. Point $D$ is on the $x$ axis, and line $BD$ intersects the $y$ axis at point $F$ and $OE$ at point $H$. The lengths of segments $BC$ and $OC$ are the solutions of the system of equations $\left\{\begin{array}{l}2x+3y=14 \\ 4x-5y=6\end{array}\right.$. Given that $OC > BC$, find the area of $\triangle OFH$.

Solution:
\textbf{Solution:} Given $y=-\frac{2}{3}x+\frac{8}{3}$, by setting $x=0$, we obtain $y=\frac{8}{3}$\\
$\therefore F\left(0, \frac{8}{3}\right)$\\
Let the equation of line OE be $y=k_1x$,\\
Substituting point E yields $k_1=\frac{1}{2}$,\\
$\therefore y=\frac{1}{2}x$,\\
$\begin{cases} y=\frac{1}{2}x \\ y=-\frac{2}{3}x+\frac{8}{3} \end{cases}$,\\
Solving this system gives $\begin{cases} x=\frac{16}{7} \\ y=\frac{8}{7} \end{cases}$,\\
$\therefore H\left(\frac{16}{7}, \frac{8}{7}\right)$,\\
Therefore, the area of $\triangle OFH$ $= \frac{1}{2} \times \frac{8}{3} \times \frac{16}{7} = \boxed{\frac{64}{21}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=x+b$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the graph of the direct proportionality function $y=-2x$ at point C. The ordinate of point C is 4. Find the coordinates of points A, B, and C?

Solution:
\textbf{Solution:} \\
Since point C lies on the graph of $y=-2x$, when $y=4$, $-2x=4$, thus $x=-2$, \\
hence the coordinates of point C are $\left(-2, 4\right)$, \\
furthermore, since $y=x+b$ passes through point $\left(-2, 4\right)$, \\
it follows that $-2+b=4$, solving this gives $b=6$, \\
hence $y=x+6$, when $x=0$, $y=6$, thus the coordinates of point A are $\left(0, 6\right)$, \\
when $y=0$, $x=-6$, thus the coordinates of point B are $\left(-6, 0\right)$. \\
$\therefore \boxed{A\left(0, 6\right)}, \boxed{B\left(-6, 0\right)}, \boxed{C\left(-2, 4\right)}$.


Question: In the Cartesian coordinate system, the graph of the linear function \(y = x + b\) intersects the x-axis and the y-axis at points \(A\) and \(B\) respectively, and intersects the graph of the direct proportionality function \(y = -2x\) at point \(C\), where the y-coordinate of point \(C\) is \(4\). If a moving point \(P\) is on the ray \(CA\), and the area of triangle \(OAP\) is \(\frac{1}{8}\) of the area of triangle \(OBC\), what are the coordinates of point \(P\)?

Solution:
\textbf{Solution:} Let $P\left(x, x+6\right)$,\\
when $S_{\triangle OAP}=\frac{1}{8}S_{\triangle OBC}$, then $\frac{1}{2}AO\cdot |x|=\frac{1}{8}\times \frac{1}{2}\times BO\cdot |y_c|$,\\
$\therefore \frac{1}{2}\times 6|x|=\frac{1}{8}\times \frac{1}{2}\times 6\times 4$, hence $x=\pm \frac{1}{2}$,\\
Moreover, since the moving point P is on the ray $CA$,\\
$\therefore x\ge -2$,\\
$\therefore$ the coordinates of point P are $\boxed{P_{1}\left(-\frac{1}{2}, \frac{11}{2}\right)}$, $\boxed{P_{2}\left(\frac{1}{2}, \frac{13}{2}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=x+b$ intersects with the $x$ and $y$ axes at points A and B, respectively, and intersects with the graph of the direct proportion function $y=-2x$ at point C, where the ordinate of point C is 4. If the point $Q(m, 2)$ is inside $\triangle OBC$ (excluding the boundary), please write down the range of the values of $m$ directly.

Solution: \textbf{Solution:} The range of $-4<m<-1$ is $m \in \boxed{(-4, -1)}$.


Question: As shown in the figure, the line $y=ax+6$ intersects with the line $y=2x$ at point A $(m, 4)$, and intersects the x-axis at point B and the y-axis at point C. Find the values of $a$ and $m$?

Solution:
\textbf{Solution:} Substitute point A$(m,4)$ into $y=2x$, we get $4=2m$\\
$\therefore m=2$\\
$\therefore$ the coordinates of point A are $(2,4)$,\\
Substituting point A$(2,4)$ into $y=ax+6$, we get $4=2a+6$\\
Solving this, we find $a=-1$,\\
$\therefore m=\boxed{2}$, $a=\boxed{-1}$;


Question: As shown in the diagram, the line $y=ax+6$ intersects the line $y=2x$ at point A ($m, 4$), and it intersects the x-axis at point B, and the y-axis at point C. Calculate the area of $\triangle AOC$.

Solution:
\textbf{Solution:} As shown in the diagram, draw a perpendicular line from A to the y-axis with the foot at D, thus $AD=2$,\\
when $x=0$, we have $y=-x+6=6$,\\
$\therefore$ the coordinates of point C are $(0,6)$\\
$\therefore$ the area of $\triangle AOC$ is $\frac{1}{2}\times 6\times 2=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}\colon y=kx+b$ and the line $l_{2}\colon y=mx+n$ intersect at point $A(1,2)$, the line $l_{2}$ intersects with the $y$-axis at point $B(0,3)$, and the line $l_{1}$ intersects with the $x$-axis at point $C(-1,0)$. What are the equations of the lines $l_{1}$ and $l_{2}$?

Solution:
\textbf{Solution:} Let's consider the line $ l_{1}: y=kx+b$. It intersects the x-axis at point C ($-1$, 0) and intersects at point A ($1, 2$), then
\[
\begin{cases}
0=-k+b \\
2=k+b
\end{cases}
\]
Solving these, we get
\[
\begin{cases}
k=1 \\
b=1
\end{cases}
\]
Therefore, the equation of the line $ l_{1}$ is: $\boxed{y=x+1}$.

For line $ l_{2}: y=mx+n$ which intersects at point A ($1, 2$) and intersects the y-axis at point B ($0, 3$), we have
\[
\begin{cases}
2=m+n \\
3=n
\end{cases}
\]
Solving this gives
\[
\begin{cases}
m=-1 \\
n=3
\end{cases}
\]
Therefore, the equation of the line $ l_{2}$ is $\boxed{y=-x+3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1: y=kx+b$ intersects with the line $l_2: y=mx+n$ at point A$(1,2)$, the line $l_2$ intersects with the $y$-axis at point B$(0,3)$, and the line $l_1$ intersects with the $x$-axis at point C$(-1,0)$. Connect $BC$. What is the area of triangle ABC?

Solution:
\textbf{Solution:} As shown in the diagram: \\
When $y=0$, we have $0=-x+3$, which means $x=3$, \\
$\therefore$ the coordinates of point D are: $(3, 0)$, \\
$\therefore$ CD=4, \\
$\therefore$ the area of the triangle ABC is $S_{\triangle ABC}=S_{\triangle BCD}-S_{\triangle ACD}=\frac{1}{2}\times 4\times 3-\frac{1}{2}\times 4\times 2=\boxed{2}$.


Question: As shown in the figure, in the Cartesian coordinate system, point B is in the first quadrant. $BA \perp x$-axis at A, $BC \perp y$-axis at C, $BA = 3$, and $BC = 5$. There is an inverse proportion function graph that exactly passes through point B. Find the expressions of the inverse proportion function that passes through point B and the linear function that passes through points A and C, respectively.

Solution:
\textbf{Solution:} Given $\because$ BA $\perp$ x-axis at A, and BC $\perp$ y-axis at C, with BA $= 3$ and BC $=5$, \\
$\therefore$ the coordinates of point B are $(5,3)$, point A $(5,0)$, and point C $(0,3)$.\\
Let the inverse proportional function be $y = \frac{k}{x}$ ($x \neq 0$),\\
$\because$ point B lies on the graph of the inverse proportional function,\\
$\therefore k=3 \times 5 = 15$\\
$\therefore$ the equation of the inverse proportional function is $y = \frac{15}{x}$;\\
Let the linear function be $y = ax + b$,\\
Substituting A$(5, 0)$ and C$(0, 3)$, we get\\
$\left\{\begin{array}{l}
b=3\\
5a+b=0
\end{array}\right.$\\
Solving gives:\\
$\left\{\begin{array}{l}
a=-\frac{3}{5}\\
b=3
\end{array}\right.$\\
$\therefore$ the equation of the linear function is $\boxed{y = -\frac{3}{5}x + 3}$.


Question: As shown in the figure, in the Cartesian coordinate system, point $B$ is in the first quadrant. $BA\perp x$-axis at point $A$, $BC\perp y$-axis at point $C$, with $BA=3$ and $BC=5$. There is an inverse proportion function graph that exactly passes through point $B$. A moving point $P$ is on the ray $CA$ (excluding point $C$), and a line $l\perp x$-axis is drawn through point $P$, intersecting the graph of the inverse proportion function at point $D$. Is there such a point $Q$, such that the quadrilateral with vertices $B$, $D$, $P$, and $Q$ forms a rhombus? Determine the coordinates of point $Q$.

Solution:
\textbf{Solution:} Let point P be $ \left(m, -\frac{3}{5}m+3\right)$ and point D be $ \left(m, \frac{15}{m}\right)$.\\
$\therefore$PD$= \frac{15}{m}-\left(-\frac{3}{5}m+3\right)$,\\
$\because$The quadrilateral formed by points B, D, P, Q is a rhombus, and when point Q lies on line BA and PD=BD=BQ,\\
$ \frac{15}{m}-\left(-\frac{3}{5}m+3\right)=\sqrt{(m-5)^{2}+\left(3-\frac{15}{m}\right)^{2}}$\\
Simplifying gives $ \frac{16}{25}m^{2}-\frac{32}{5}m+7=0$\\
Solving this yields $ m_{1}=\frac{5}{4}, m_{2}=\frac{35}{4}$.\\
When $ m=\frac{5}{4}$, $ PD=BQ=\frac{15}{m}-\left(-\frac{3}{5}m+3\right)=\frac{39}{4}$,\\
$\because$ BA＝3,\\
$\therefore$the y-coordinate of point Q is $ 3-\frac{39}{4}=-\frac{27}{4}$,\\
$\therefore$Point Q is $ \left(5, -\frac{27}{4}\right)$;\\
When $ m=\frac{35}{4}$, $ PD=BQ=\frac{15}{m}-\left(-\frac{3}{5}m+3\right)=\frac{111}{28}$,\\
$\because$ BA＝3,\\
$\therefore$the y-coordinate of point Q is $ 3-\frac{111}{28}=-\frac{27}{28}$,\\
$\therefore$Point Q is $ \left(5, -\frac{27}{28}\right)$;\\
When point Q lies on line AC, PD and BQ bisect each other perpendicularly,\\
$\therefore$BQ$\parallel$x-axis,\\
$\therefore$the y-coordinate of point Q is 3,\\
$\therefore$ $ \frac{\frac{3}{5}m+3+\frac{15}{m}}{2}=3$\\
Solving this yields $ m=\frac{-5\pm 5\sqrt{3}}{2}$\\
$\because$m＞0,\\
$\therefore$ $ m=\frac{-5+5\sqrt{3}}{2}$\\
$\therefore$Point Q is $ \left(\frac{-5+5\sqrt{3}}{2}, 3\right)$,\\
$\therefore$The coordinates of point Q are $\boxed{\left(5, -\frac{27}{4}\right)}$ or $\boxed{\left(\frac{-5+5\sqrt{3}}{2}, 3\right)}$ or $\boxed{\left(5, -\frac{27}{28}\right)}$.


Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis respectively at points A and B. The line $y_2=kx+b$ passes through point D(3,0) and intersects with the line $y_1=3x+6$ at point C(m,3). What is the analytical equation of line CD?

Solution:
\textbf{Solution:}
Since line $y_2$ intersects line $y_1$ at point C ($m$, 3), \\
substituting point C ($m$, 3) into $y_1=3x+6$, we get $3m+6=3$, and solve for $m=-1$, \\
thus, point C is ($-1$, 3). Substituting points C ($-1$, 3) and D (3, 0) into $y_2=kx+b$, \\
we get $\begin{cases}
-k+b=3 \\
3k+b=0
\end{cases}$, \\
solving which we find $\begin{cases}
k=-\frac{3}{4} \\
b=\frac{9}{4}
\end{cases}$, \\
therefore, the equation of line CD is $\boxed{y_2=-\frac{3}{4}x+\frac{9}{4}}$.


Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis at points A and B respectively. The line $y_2=kx+b$ passes through point D (3, 0) and intersects with the line $y_1=3x+6$ at point C (m, 3). Based on the diagram, directly write the solution set for the inequality $3x+6>kx+b$ in terms of x.

Solution:
\textbf{Solution:} From equation (1), we have point C at $(-1,3)$,\\
$3x+6>kx+b$ implies $y_1>y_2$,\\
meaning the graph of $y_1=3x+6$ is above that of $y_2=-\frac{3}{4}x+\frac{9}{4}$,\\
Based on the graph above, the solution set is $\boxed{x>-1}$.


Question: As shown in the figure, the line $y_1=3x+6$ intersects the x-axis and y-axis at points A and B, respectively. The line $y_2=kx+b$ passes through point D(3, 0) and intersects line $y_1=3x+6$ at point C(m, 3). There exists a point P on line AB, and through point P, a line PQ is drawn parallel to the y-axis, intersecting line CD at point Q. If the length of segment PQ is 5, what are the coordinates of point P?

Solution:
\textbf{Solution:} Given that point $P$ lies on line $AB$ and point $Q$ lies on line $CD$, and $PQ \parallel y$-axis, \\
let the coordinates of point $P$ be $(n, 3n+6)$, then the coordinates of point $Q$ are $\left(n, -\frac{3}{4}n+\frac{9}{4}\right)$, \\
since the length of segment $PQ$ is $3$, \\
$\therefore \left|3n+6+\frac{3}{4}n-\frac{9}{4}\right|=5$, \\
solving this gives $n=\frac{1}{3}$ or $n=-\frac{7}{3}$, \\
when $n=\frac{1}{3}$, $3n+6=7$; \\
when $n=-\frac{7}{3}$, $3n+6=-1$, \\
$\therefore$ the coordinates of point P are $\boxed{\left(\frac{1}{3},7\right)}$ or $\boxed{\left(-\frac{7}{3},-1\right)}$.


Question: As shown in the figure, the graph of the linear function $y=(m-2)x-4$ intersects the negative x-axis at point A and intersects the y-axis at point B, and the area of $\triangle OAB$ is 8. Find the value of $m$ and the coordinates of point A.

Solution:
\textbf{Solution:} When $x=0$, we have $y=(m-2)x-4=-4$, thus $B(0, -4)$,\\
$\therefore OB=4$,\\
$\because$ The area of $\triangle OAB$ is 8,\\
$\therefore \frac{1}{2}\times OA\times 4=8$, solving this gives: $OA=4$,\\
$\therefore \boxed{A(-4,0)}$.\\
Substituting point $A(-4,0)$ into $y=(m-2)x-4$, gives: $-4m+8-4=0$,\\
$\therefore m=\boxed{1}$.


Question: As shown in the figure, the graph of the linear function $y=(m-2)x-4$ intersects the negative half-axis of the x-axis at point A, and intersects the y-axis at point B, and the area of $\triangle OAB$ is 8. A line $CD \parallel AB$ is drawn through point C (0, 3), intersecting the x-axis at point D. Find the equation of line BD.

Solution:
\textbf{Solution:} Given that $C(0,3)$ and $AB\parallel CD$, thus the equation of line CD is: $y=-x+3$. Hence, the coordinate of point D is $(3,0)$. Let the equation of line BD be $y=kx+b$. Substituting points B$(0,-4)$ and D$(3,0)$ into the equation, we get:
\[b=-4\\
3k+b=0\],
solving this, we obtain:
\[k=\frac{4}{3}\\
b=-4\].\\
Therefore, the equation of line BD is $\boxed{y=\frac{4}{3}x-4}$.


Question: As shown in the figure, in rectangle OABC, point A is on the x-axis, and point C is on the y-axis. The coordinates of point B are $\left(6, 8\right)$. Rectangle OABC is folded along line BD so that point C lands exactly on point E on the diagonal OB. The line of the fold BD intersects the y-axis and the x-axis at points D and F, respectively. What is the length of line segment OE?

Solution:
\textbf{Solution:} Since in rectangle OABC, point A is on the x-axis and point C is on the y-axis, with point B having the coordinates (6, 8),\\
it follows that OA = 6, AB = 8, and $\angle$OAB$=90^\circ$,\\
thus OB $= \sqrt{OA^{2} + AB^{2}} = \sqrt{6^{2} + 8^{2}} = 10$,\\
from folding, we know BE = BC = 6,\\
therefore, OE = OB - BE = 10 - 6 = \boxed{4};


Question: As shown in the figure, in rectangle OABC, point A is on the x-axis, point C is on the y-axis, and the coordinates of point B are $(6, 8)$. The rectangle OABC is folded along the line BD, making point C exactly coincide with point E on the diagonal OB. The fold line BD intersects the y-axis and x-axis at points D and F, respectively. What are the coordinates of point F?

Solution:
\textbf{Solution:} Let the coordinates of point D be $(0,a)$, so $OD=a$, $CD=8-a$,\\
since $BC=6$, $CD=DE=8-a$, $OB=10$,\\
since $\frac{S_{\triangle ODB}}{2}=\frac{OD\cdot BC}{2}=\frac{OB\cdot DE}{2}$,\\
thus $\frac{a\times 6}{2}=\frac{10(8-a)}{2}$,\\
solving, we get $a=5$,\\
which means the coordinates of point D are $(0,5)$,\\
Let the equation of the line containing the crease BD be $y=kx+b$,\\
since points D $(0,5)$ and B $(6,8)$ lie on line BD,\\
thus $\left\{\begin{array}{l}b=5 \\ 6k+b=8\end{array}\right.$, gives $\left\{\begin{array}{l}k=0.5 \\ b=5\end{array}\right.$,\\
hence, the equation of the line containing the crease BD is $y=0.5x+5$,\\
when $y=0$, $0.5x+5=0$,\\
solving, we get $x=-10$,\\
thus, the coordinates of point F are $\boxed{(-10,0)}$;


Question: As shown in the figure, in rectangle $OABC$, point $A$ is on the $x$-axis, point $C$ is on the $y$-axis, and the coordinates of point $B$ are $\left(6, 8\right)$. The rectangle $OABC$ is folded along the line $BD$ so that point $C$ exactly falls on point $E$ on the diagonal $OB$. The fold line $BD$ intersects with the $y$-axis and the $x$-axis at points $D$ and $F$, respectively. If point $M$ is on the line $y=-\frac{1}{2}x$, then is there a point $P$ on the line $BD$ such that the quadrilateral with vertices $C$, $D$, $M$, and $P$ is a parallelogram? Find the coordinates of the point $P$ that satisfy the condition.

Solution:
\textbf{Solution}: Let there exist a point $P$ on line $BD$ such that the quadrilateral constituted by vertices $C$, $D$, $M$, and $P$ forms a parallelogram,\\
$\because$ from (2) we know the equation of line $BD$ is $y=0.5x+5$,\\
$\therefore D(0,5)$,\\
and $\because C(0,8)$,\\
$\therefore CD=3$,\\
Point $M$ lies on the line $y=-0.5x$, and point $P$ lies on line $BD$,\\
To make the quadrilateral with vertices $C$, $D$, $M$, and $P$ a parallelogram,\\
it requires either $CD$ to be parallel and equal to $MP$ or $CP$ to be parallel and equal to $MD$,\\
When $CD$ is parallel and equal to $MP$, let the coordinate of point $P$ be $(m,0.5m+5)$, then $M(m,-0.5m)$,\\
$\therefore MP=|\left(0.5m+5\right)-\left(-0.5m\right)|=3$,\\
From which, $m_{1}=-2$, $m_{2}=-8$,\\
$\therefore P_{1}(-2,4)$, $P_{2}(-8,1)$\\
When $CP$ and $MD$ are parallel and equal, let the coordinate of point $P$ be $(m,0.5m+5)$, then $M(-m,0.5m)$,\\
$\therefore |8-(0.5m+5)|=|0.5m-5|$,\\
From which, $m=8$,\\
$\therefore P_{3}(8,9)$\\
Thus, the coordinates of point $P$ satisfying the requirements are \boxed{P_{1}(-2,4)}, \boxed{P_{2}(-8,1)}, and \boxed{P_{3}(8,9)}.


Question: As shown in the figure, in $\triangle AOB$, $\angle ABO=90^\circ$, $OB=4$, $AB=8$. The line $y=-x+b$ intersects $OA$ and $AB$ at points $C$ and $D$, respectively, and the area of $\triangle BOD$ is $4$. What is the equation of line $AO$?

Solution:
\textbf{Solution:} Given that $OB=4$, $AB=8$, and $\angle ABO=90^\circ$,\\
thus, the coordinates of point A are (4, 8).\\
Assume the equation of line $AO$ is $y=kx$, then $4k=8$,\\
solving this gives $k=2$, thus the equation of line $AO$ is $\boxed{y=2x}$.


Question: As shown in the diagram, in $\triangle AOB$, $\angle ABO=90^\circ$, $OB=4$, and $AB=8$. The line $y=-x+b$ intersects $OA$ and $AB$ at points $C$ and $D$, respectively, and the area of $\triangle BOD$ is $4$. Find the equation of line $CD$.

Solution:
\textbf{Solution:} Given that $OB=4$ and $\angle ABO=90^\circ$, the area of $\triangle BOD$ is $S_{\triangle BOD} = \frac{1}{2} \times OB \times BD = 4$, \\
thus $DB=2$, \\
therefore, the coordinates of point D are $(4,2)$, \\
substituting D$(4,2)$ into $y=-x+b$ yields: $b=6$, \\
therefore, the equation of line CD is $\boxed{y=-x+6}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ ($k\neq 0$) intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ ($m\neq 0$) at points $A(-3, 2)$ and $B(n, -\frac{3}{2})$. Find the expressions of the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given the inverse proportion function $y_{2}=\frac{m}{x}$ (where $m\neq 0$) passes through point A ($-3$, $2$),

$\therefore m=-3\times 2=-6$,

$\therefore$ the expression of the inverse proportion function is $y=-\frac{6}{x}$.

Given point B ($n$, $-\frac{3}{2}$) lies on the graph of the inverse proportion function $y=-\frac{6}{x}$,

$\therefore n=4$.

$\therefore$ the coordinates of point B are ($4$, $-\frac{3}{2}$).

Given the linear function $y_{1}=kx+b$ (where $k\neq 0$) passes through points A ($-3$, $2$) and B ($4$, $-\frac{3}{2}$),

$\therefore \left\{\begin{array}{l}-3k+b=2 \\ 4k+b=-\frac{3}{2}\end{array}\right.$,

solving this gives $\left\{\begin{array}{l}k=-\frac{1}{2} \\ b=\frac{1}{2}\end{array}\right.$,

$\therefore$ the linear function is $\boxed{y=-\frac{1}{2}x+\frac{1}{2}}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ ($k\neq 0$) intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ ($m\neq 0$) at points $A(-3, 2)$ and $B(n, -\frac{3}{2})$. Please directly write down the range of $x$ when $y_{1}>y_{2}$.

Solution: \textbf{Solution:} When $y_{1}>y_{2}$, the range of $x$ is $x<-3$ or $0<x<4$. Therefore, the answer is $\boxed{x \in (-\infty, -3) \cup (0, 4)}$.


Question: As shown in the figure, the graph of the linear function $y_1=kx+2$ intersects with the graph of the inverse proportion function $y_2=-\frac{8}{x}$ at points $A(a, -2a)$ and $B(4, -2)$. What are the values of $a$ and $k$?

Solution:
\textbf{Solution:} Since the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the reciprocal function $y_{2}=-\frac{8}{x}$ at point A$(a, -2a)$ and B$(4, -2)$,\\
it follows that $-2=4k+2$,\\
which implies $k=-1$,\\
thus $y_{1}=-x+2$,\\
and therefore $-2a=-a+2$,\\
leading to $a=\boxed{-2}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the inverse proportion function $y_{2}=-\frac{8}{x}$ at points $A(a，-2a)$ and $B(4，-2)$. Considering the graphs, directly write down the solution set of the inequality $kx+2+\frac{8}{x}<0$:

Solution:
\textbf{Solution:} From (1) it is known that: Points A(-2, 4) and B(4, -2),\\
since $kx+2+\frac{8}{x}<0$,\\
therefore $kx+2<-\frac{8}{x}$, which means when $y_1<y_2$, the graph of the linear function is below that of the hyperbolic function,\\
thus $-2<x<0$ or $x>4$. Hence, the solution set is $x\in\boxed{(-2,0)\cup(4,+\infty)}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+2$ intersects with the graph of the inverse proportion function $y_{2}=-\frac{8}{x}$ at points $A(a，-2a)$ and $B(4，-2)$. Connect $OA$ and $OB$. What is the area of $\triangle AOB$?

Solution: \textbf{Solution:} As shown in the figure, AB intersects the x-axis at point C. According to (1), we have $y_1 = -x + 2$, $\therefore$ the coordinates of point C are $(2, 0)$, $\therefore OC = 2$, $\therefore$ the area of $\triangle AOB$, $S_{\triangle AOB} = \frac{1}{2} \times OC \times (y_A - y_B) = \frac{1}{2} \times 2 \times (4 + 2) = \boxed{6}$.


Question: As shown in the figure, line $y_{1} = -x + 3$ and line $y_{2} = kx - 2$ intersect the $y$-axis at points $A$ and $B$, respectively, and intersect at point $C(2, m)$. What are the values of $m$ and $k$?

Solution:
\textbf{Solution:} Substitute $C(2, m)$ into $y_1 = -x + 3$, \\
we get: $m = -2 + 3 = 1$, \\
$\therefore$ the coordinates of point $C$ are $(2, 1)$, \\
substitute $C(2, 1)$ into $y_2 = kx - 2$, \\
we get: $2k - 2 = 1$, \\
solving for $k$, we obtain $k = \boxed{1.5}$. \\
In conclusion, $m = \boxed{1}$ and $k = \boxed{1.5}$.


Question: The lines $y_1 = -x + 3$ and $y_2 = kx - 2$ intersect the y-axis at points $A$ and $B$, respectively, and intersect at point $C(2, m)$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} When $x=0$, $y_1=3$,\\
then $A(0,3)$;\\
When $x=0$, $y_2=-2$,\\
then $B(0,-2)$,\\
$\therefore$ The area of $\triangle ABC = \frac{1}{2} \times (3+2) \times 2 = \boxed{5}$.


Question: The line $y_{1} = -x + 3$ and the line $y_{2} = kx - 2$ respectively intersect the $y$-axis at points $A$ and $B$, and intersect each other at point $C(2, m)$. Based on the graph, directly write the range of the independent variable $x$ when $y_{1} > y_{2}$.

Solution: \textbf{Solution:} As shown in the figure, when $\boxed{x<2}$, $y_{1}>y_{2}$.


Question: It seems like there was a mistake in your question. You've asked for a translation of a text, but you haven't provided any text to translate. Could you please provide the text you'd like translated into English?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=kx+b$ ($k\neq 0$) passes through the points $A(-2,-4)$ and $B(2,0)$,\\
we have $\begin{cases} -2k+b=-4\\ 2k+b=0 \end{cases}$,\\
solving this system of equations gives $k=1$ and $b=-2$,\\
therefore, the expression of the function is: \boxed{y=x-2}.


Question: It is known that the graph of the linear function $y=kx+b\left(k\ne 0\right)$ passes through points $A\left(-2, -4\right)$ and $B\left(2, 0\right)$. If point $P$ lies on the $x$-axis and the area of $\triangle ABP$ is 10, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Since point P is on the x-axis,\\
let P be $(x,0)$,\\
then $BP=|x-2|$,\\
since the area of $\triangle ABP$ is 10,\\
$\frac{1}{2} \times 4 \times |x-2|=10$,\\
thus, $|x-2|=5$,\\
so, $x-2=5$ or $x-2=-5$,\\
solving this gives $x_{1}=-3$ or $x_{2}=7$,\\
therefore, the coordinates of point P are \boxed{(-3,0)} or \boxed{(7,0)}.