Question: As shown in the figure, it is known that the parabola $y=\frac{k}{8}(x-2)(x+4)$ (where $k$ is a constant and $k>0$) intersects the $x$-axis at points $A$ and $B$ from left to right, respectively, and intersects the $y$-axis at point $C$. Point $D$ lies on the graph of the parabola in the first quadrant, and $\angle BAD=30^\circ$. Under these conditions, let $F$ be a point on segment $AD$ (excluding the endpoints), and connect $BF$. A moving point $M$ starts from point $B$, moves along segment $BF$ at a speed of 1 unit per second to point $F$, then moves along segment $FD$ at a speed of 2 units per second to point $D$ and stops. When the coordinates of point $F$ are what values, will point $M$ spend the least time in the entire movement process, and what is the minimum time spent?

Solution:
\textbf{Solution:} From (1), we know $D(5,3\sqrt{3})$. As shown in the diagram, draw $DN\perp x$-axis at point N, then $DN=3\sqrt{3}$, $ON=5$, $AN=4+5=9$. Draw $DK\parallel x$-axis through point D, then $\angle KDF=\angle DAB=30^\circ$. Draw $FG\perp DK$ at point G through point F, then $FG=\frac{1}{2}DF$. According to the problem statement, the path of the moving point M is the polyline $BF+DF$, with the movement time: $t=BF+\frac{1}{2}DF$, $\therefore t=BF+FG$, that is, the time value of the movement is equal to the length value of the polyline $BF+FG$. It is known from the shortest vertical segment that the minimum length of the polyline $BF+FG$ is the vertical segment between $DK$ and the $x$-axis. Draw $BH\perp DK$ at point B through point H, then $t_{\text{minimum}}=BH$, and the intersection of line segment $BH$ with line $AD$ is the desired point F. Suppose the equation of line $AD$ is $y=kx+b$, substituting $(-4,0)$ and $(5,3\sqrt{3})$, we get
\[
\begin{cases}
-4k+b=0\\
5k+b=3\sqrt{3}
\end{cases}
\]
Solving, we find
\[
\begin{cases}
k=\frac{\sqrt{3}}{3}\\
b=\frac{4\sqrt{3}}{3}
\end{cases}
\]
$\therefore$ The equation of line $AD$ is: $y=\frac{\sqrt{3}}{3}x+\frac{4\sqrt{3}}{3}$, $\because$ the $x$-coordinate of point B is 2, $\therefore y=\frac{2\sqrt{3}}{3}+\frac{4\sqrt{3}}{3}=2\sqrt{3}$, $\therefore F(2,2\sqrt{3})$. $\therefore HF=BH-BF=3\sqrt{3}-2\sqrt{3}=\sqrt{3}$, $\therefore FD=2\sqrt{3}$.

In summary, when the coordinates of point F are $(2,2\sqrt{3})$, point M takes the least amount of time in the entire motion, which is $2\sqrt{3}\div 1+2\sqrt{3}\div 2=2\sqrt{3}+\sqrt{3}=\boxed{3\sqrt{3}}$ (seconds).


Question: As shown in the diagram, $O$ is the intersection point of the diagonals of $\square ABCD$, $BE\perp OC$ at point $E$, $BE$ is extended to point $F$ such that $EF=BE$, and $DF$ is connected. $\angle F=90^\circ$. When $\square ABCD$ is a rectangle, $AC=6$, and $BF= 2\sqrt{5}$, find the lengths of $DF$ and $CE$.

Solution:
\textbf{Solution:} Since $ABCD$ is a rectangle, $BD=AC=6$, thus $OB=\frac{1}{2}BD=3$. Given $EF=BE$ and $BF=2\sqrt{5}$, it follows that $BE=\sqrt{5}$. Since $BE\perp OC$, $\angle BEO=90^\circ$. In $\triangle BOE$, $OE=\sqrt{BO^{2}-BE^{2}}=2$,\\
thus $DF=2OE=\boxed{4}$,\\
and since $CO=\frac{1}{2}AC=3$,\\
it follows that $CE=CO-OE=3-2=\boxed{1}$.


Question: As shown in the diagram, the vertices A and B of the rhombus ABCD are located on the y-axis and the positive half of the x-axis, respectively, while C and D are in the first quadrant, with $AC \parallel x$-axis. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through vertex D. Given $A(0, 2)$ and $B(1, 0)$, find the equation of the inverse proportion function.

Solution:
**Solution:**

(1) Draw a perpendicular line from point D to the y-axis, intersecting at point F,

$\because ABCD$ is a rhombus,

$\therefore BD\perp AC$, $BE=DE$, $AE=EC$

It is easy to prove that quadrilaterals AOBE and AEDF are rectangles.

$\therefore AO=BE=DE=2$, $\therefore D\left(1, 4\right)$, $\therefore y=\frac{4}{x}$

(2) Draw a perpendicular line from point C to the x-axis, intersecting at point G,

It is easy to prove that quadrilaterals AEBO and ACGO are rectangles.

$\therefore AO=CG=2$, $AC=2AE=2$ $\therefore C\left(2, 2\right)$, $\therefore$C lies on the graph of the inverse proportional function $\boxed{y=\frac{4}{x}}$


Question: As shown in the figure, the vertices A and B of rhombus ABCD are located on the positive y-axis and positive x-axis, respectively, while vertices C and D are located in the first quadrant. $AC$ is parallel to the x-axis, and the graph of the inverse proportion function $y=\frac{k}{x}$ passes through vertex D. Given $k=18\sqrt{3}$ and $\angle ABD=30^\circ$, what is the length of the sides of rhombus ABCD?

Solution:
\textbf{Solution:} Given that $\angle ABD=30^\circ$ and $AC\perp BD$. Let $AE=a$, $BE=\sqrt{3}a$, and $AB=2a$.\\
Thus, $D\left(a, 2\sqrt{3}a\right)$. Since D lies on the inverse proportional function,\\
we have $2\sqrt{3}a^2=18\sqrt{3}$, thus $a=3\left(a>0\right)$.\\
Therefore, $AB=2a=\boxed{6}$.


Question: Fold a parallelogram sheet $ABCD$ as shown in Diagram 1, so that vertices $A$ and $B$ coincide at point $M$ on segment $HF$, and vertices $C$ and $D$ coincide at point $N$ on segment $HF$, where the length of $AD$ is $6$, and the length of $AE$ is $x$. Quadrilateral $EFGH$ is a rectangle; please express the length of $HF$ with an algebraic expression involving $x$; if you can't, simply write out the length of $HF$.

Solution:
\textbf{Solution:} By folding, we get: $\angle B=\angle FME$, $\angle D=\angle HNG$, $AH=MH$, $FC=FN$, $BF=FM$, $HD=HN$,\\
$\therefore HF=MH+FM=AH+BF$,\\
$\because$ Rectangle EFGH,\\
$\therefore EH=FG$, $EH\parallel FG$,\\
$\therefore \angle EHM=\angle NFG$,\\
Furthermore, $\because$ parallelogram ABCD,\\
$\therefore \angle B=\angle D$,\\
$\therefore \angle FME=\angle HNG$,\\
$\therefore \angle EMH=\angle GNF$,\\
$\therefore \triangle EMH \cong \triangle GNF$,\\
$\therefore MH=FN$,\\
$\therefore BF=MF=NH=DH$,\\
$\therefore HF=AH+DH=AD$,\\
$\because AD=6$,\\
$\therefore HF=\boxed{6}$,\\
Hence, the length of $HF$ remains unchanged.


Question: Folding the parallelogram-shaped paper ABCD as shown in Figure 1, so that the vertices A, B both fall on point M on the segment HF, and the vertices C, D both fall on point N on the segment HF, where $AD=6$, $AE=x$. If $AE=2$ and $AF$ is connected such that $AF\perp BC$ (as shown in Figure 2), find the value of $\frac{BF}{FC}$.

Solution:
\textbf{Solution:} Solution Method (1): Given $AE=EM=BE$, \\
$\therefore AB=2AE=4$, \\
$\because AF\perp BC$, $\therefore EF=\frac{1}{2}AB=2$, $HG=EF=2$, \\
Similarly, $CG=DG=2$, \\
$\therefore CH\perp AD$, $CH\perp BC$, $CH=AF$, \\
Let $BF=\alpha$, then $CF=6-\alpha$, \\
$\therefore AF^{2}=AB^{2}-BF^{2}=16-\alpha^{2}$, $CH^{2}=FH^{2}-FC^{2}=36-(6-\alpha)^{2}$, \\
$\therefore 16-\alpha^{2}=36-(6-\alpha)^{2}$, \\
$\therefore \alpha=\frac{4}{3}$, $6-\alpha=\frac{14}{3}$, $\therefore \frac{BF}{CF}=\boxed{\frac{2}{7}}$; \\
Method (2): As shown in the diagram, connect EG intersecting AF at point P, \\
Given: $AE=EM=BE$, \\
$\therefore AB=2AE=4$, \\
$\because AP\perp BC$, $\therefore EF=\frac{1}{2}AB=2$, $EG=FH=6$, \\
$\therefore FG=4\sqrt{2}$, \\
$\because EG\parallel BC$, $AF\perp BC$, $\therefore EG\perp AF$, \\
$\because AE=EF$, $\therefore AP=PF=\frac{EF\cdot FG}{EG}=\frac{4\sqrt{2}}{3}$, $AF=\frac{8\sqrt{2}}{3}$, \\
In $\triangle ABF$, $AB^{2}=AF^{2}+BF^{2}$, \\
$\therefore BF=\frac{4}{3}$, $FC=\frac{14}{3}$, $\therefore \frac{BF}{FC}=\boxed{\frac{2}{7}}$.


Question: As shown in the figure, within rectangle $ABCD$, $E$ and $F$ are two points on $BC$, and $BE=CF$. Lines $AF$ and $DE$ intersect at point $O$. $\triangle AOD$ is an isosceles triangle; if $AF\perp DE$, $OF= \frac{1}{3} OA=1$, calculate the perimeter of rectangle $ABCD$.

Solution:
Solution: Since $OF = \frac{1}{3} OA = 1$,\\
it follows that $OA = 3, AF = 4$,\\
which means $OA = OD = 3$,\\
Given $AF \perp DE$,\\
it follows that $\angle AOD = 90^\circ$,\\
therefore, $\angle OAD = \angle ODA = 45^\circ$,\\
$AD = \sqrt{3^2 + 3^2} = 3\sqrt{2}$,\\
In rectangle ABCD, $\angle BAD = \angle ADC = 90^\circ$,\\
it follows that $\angle BAF = \angle CDE = 45^\circ$,\\
thus $\angle AFB = 45^\circ = \angle BAF$,\\
which leads to $AB = BF$,\\
Given $AB^2 + BF^2 = AF^2$,\\
it follows that $AB = BF = 2\sqrt{2}$,\\
Hence, the perimeter of rectangle ABCD is $2 \times (2\sqrt{2} + 3\sqrt{2}) = \boxed{10\sqrt{2}}$.


Question: As shown in the figure, in rectangle $ABCD$, $M$ is the midpoint of side $AD$, and $P$ is a moving point on side $BC$, such that $PE\perp MC$, $PF\perp BM$, with the perpendicular feet being $E$ and $F$, respectively. If quadrilateral $PEMF$ is a rectangle, when does point $P$ move to a position that makes quadrilateral $PEMF$ a square? Prove your conclusion.

Solution:
\textbf{Solution:} When point $P$ moves to the midpoint of side $BC$, the rectangle $PEMF$ becomes a square, at which point $BP=CP$. \\
\textit{Proof:} Since quadrilateral $PEMF$ is a rectangle, \\
$\therefore$ $\angle PFM=\angle PFB=\angle PEC=90^\circ$. \\
In $\triangle BFP$ and $\triangle CEP$, $\left\{\begin{array}{l}\angle FBP=\angle ECP=45^\circ \\ \angle PFB=\angle PEC \\ BP=CP \end{array}\right.$, \\
$\therefore$ $\triangle BFP\cong \triangle CEP$ (AAS), \\
$\therefore$ $PF=PE$. \\
Moreover, since quadrilateral $PEMF$ is a rectangle, \\
$\therefore$ rectangular $PEMF$ is a square, \\
that is, when point $P$ moves to the midpoint of $BC$, quadrilateral $PEMF$ is a \boxed{square}.


Question: As shown in the figure, in quadrilateral ABCD, AC and BD intersect at point O, where O is the midpoint of AC, AB$\parallel$DC, AC=20, and BD=10. Quadrilateral ABCD is a parallelogram; if AC$\perp$BD, find the area of parallelogram ABCD.

Solution:
\textbf{Solution:} It follows from equation (1) that quadrilateral ABCD is a parallelogram. Furthermore, since $AC \perp BD$, the parallelogram ABCD is thereby a rhombus. Given that $AC=20$, and $BD=10$, \\
the area of quadrilateral ABCD is $=\frac{1}{2}AC \times BD=\frac{1}{2} \times 20 \times 10=\boxed{100}$.


Question: As shown in the diagram, let $ABCD$ be a square, and let points $E$ and $F$ be on the diagonal $AC$. Given that $\angle EBF = 45^\circ$, when lines $BE$ and $BF$ are connected, triangles $ABE$ and $GBE$ are symmetric about line $BE$. Point $G$ is on $BD$, and line $FG$ is connected. What is the measure of $\angle FBC$?

Solution:
\textbf{Solution:} Let the intersection point of the diagonals be $O$.\\
$\because \angle ABO=\angle EBF=45^\circ$\\
$\therefore \angle ABE+\angle EBO=\angle EBO+\angle DBF$\\
$\therefore \angle ABE=\angle DBF$\\
$\because \triangle ABE$ and $\triangle GBE$ are symmetric with respect to the line $BE$\\
$\therefore \angle ABE=\angle EBO$\\
$\therefore \angle EBO=\angle OBF=\frac{1}{2}\angle EBF=22.5^\circ$\\
$\therefore \angle FBC=45^\circ-\angle OBF=\boxed{22.5^\circ}$


Question: As shown in the figure, for the square $ABCD$, with points $E$ and $F$ located on the diagonal $AC$, $\angle EBF=45^\circ$. The lines $BE$ and $BF$ are drawn. The triangles $ABE$ and $GBE$ are symmetrical about the line $BE$. Point $G$ is on $BD$, and the line $FG$ is drawn. As shown in the auxiliary figure, extend $BF$ to intersect $CD$ at point $H$. Draw the line $HG$. Find the value of $\frac{CD}{CH}$.

Solution:
\textbf{Solution:}

Given that\\
\begin{equation}
    \because \triangle ABE \text{ and } \triangle GBE \text{ are symmetrical about the line } BE
\end{equation}
\begin{equation}
    \therefore AB=BG=BC
\end{equation}
\begin{equation}
    \therefore \left\{\begin{array}{l}BG=BC \\ \angle DBF=\angle FBC \\ BF=BF \end{array}\right.
\end{equation}
\begin{equation}
    \therefore \triangle GBF \cong \triangle CBF(SAS)
\end{equation}
\begin{equation}
    \therefore GF=FC, \angle BGF=\angle BCF=45^\circ
\end{equation}
\begin{equation}
    \because \left\{\begin{array}{l}BG=BC \\ \angle DBF=\angle FBC \\ BH=BH \end{array}\right.
\end{equation}
\begin{equation}
    \therefore \triangle HGB \cong \triangle HCB(SAS)
\end{equation}
\begin{equation}
    \therefore \angle BGH=\angle BCH=90^\circ
\end{equation}
\begin{equation}
    \because \text{Square } ABCD
\end{equation}
\begin{equation}
    \therefore AC \perp BD
\end{equation}
\begin{equation}
    \therefore \angle BOC=90^\circ
\end{equation}
\begin{equation}
    \therefore \angle BOC=\angle BGH=90^\circ
\end{equation}
\begin{equation}
    \therefore OF \parallel GH
\end{equation}
\begin{equation}
    \therefore OC \parallel GH
\end{equation}
\begin{equation}
    \therefore CF \parallel GH
\end{equation}
\begin{equation}
    \because \angle BGF=45^\circ, \angle BOC=90^\circ
\end{equation}
\begin{equation}
    \therefore \angle OFG=45^\circ
\end{equation}
\begin{equation}
    \therefore \angle OFG=\angle ACD=45^\circ
\end{equation}
\begin{equation}
    \therefore GF \parallel CD
\end{equation}
\begin{equation}
    \therefore GF \parallel CH
\end{equation}
\begin{equation}
    \because CF \parallel GH
\end{equation}
\begin{equation}
    \therefore \text{Quadrilateral } GHCF \text{ is a parallelogram}
\end{equation}
\begin{equation}
    \because GF=FC
\end{equation}
\begin{equation}
    \therefore \text{Quadrilateral } GHCF \text{ is a rhombus;}
\end{equation}
(2) Given that $\angle BGH=90^\circ$
\begin{equation}
    \therefore \angle HGD=90^\circ
\end{equation}
\begin{equation}
    \because \angle BDC=45^\circ
\end{equation}
\begin{equation}
    \therefore \angle GHD=45^\circ
\end{equation}
\begin{equation}
    \therefore \triangle DGH \text{ is an isosceles right triangle}
\end{equation}
\begin{equation}
    \therefore DG=GH
\end{equation}
\begin{equation}
    \therefore DH=\sqrt{2}GH
\end{equation}
\begin{equation}
    \because \text{Quadrilateral } GHCF \text{ is a rhombus}
\end{equation}
\begin{equation}
    \therefore GH=CH
\end{equation}
\begin{equation}
    \therefore CD=CH+DH=GH+DH=GH+\sqrt{2}GH=\left(1+\sqrt{2}\right)GH
\end{equation}
\begin{equation}
    \therefore \frac{CD}{CH}=\frac{\left(1+\sqrt{2}\right)CH}{CH}=1+\sqrt{2}
\end{equation}
Thus, $\frac{CD}{CH}=\boxed{1+\sqrt{2}}$.


Question: Quadrilateral $ABCD$ is a rectangle, and quadrilateral $AEFG$ is a square, with $AD > AE$, point $E$ is on the left side of segment $AD$. Connect $DE$ and $BG$. As shown in Figure 1, if point $F$ is on side $AD$ and $AD=3$, $AE= \sqrt{2}$, find the length of $DE$.

Solution:
\textbf{Solution:} Let the intersection of EG and AF be point H, as shown in the figure,\\
$\because$ Quadrilateral AEFG is a square\\
$\therefore EG\perp AF$, $EH=HG$, $FH=AF$, $AF=EG$\\
$\therefore FH=EH=HG=AH=\frac{1}{2}EG=\frac{1}{2}FA$\\
In $\triangle AEF$,\\
$AF^{2}=EF^{2}+AE^{2}$\\
$\therefore AF=\sqrt{2^{2}+2^{2}}=2$\\
$\therefore EH=AH=1$\\
In $\triangle DHE$,\\
$DE^{2}=EH^{2}+DH^{2}$\\
That is, $DE^{2}=EH^{2}+(AD-AH)^{2}$\\
$\therefore DE=\sqrt{1^{2}+(3-1)^{2}}=\sqrt{5}=\boxed{\sqrt{5}}$


Question: As shown in the figure, the vertex of the parabola is $P(1, 4)$, it intersects with the $y$-axis at point $C(0, 3)$, and intersects with the $x$-axis at points $A$ and $B$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Let $y=-a(x-1)^2+4$, where $a\neq 0$.

Substituting point $C(0, 3)$, we get: $-a+4=3$, which means $a=-1$.

$\therefore$ The equation of the parabola is $\boxed{y=-(x-1)^2+4=-x^2+2x+3}$.


Question: Given that the vertex of the parabola is $P(1, 4)$, it intersects the y-axis at point $C(0, 3)$, and it intersects the x-axis at points $A$ and $B$. If the parabola has only one intersection point with the line $y = x + m$, what is the value of $m$?

Solution:
\textbf{Solution:} Since the parabola and the line $y=x+m$ intersect at only one point,\\
it follows that $-{x}^{2}+2x+3=x+m$, which simplifies to ${x}^{2}-x+(m-3)=0$,\\
hence, $\Delta=1-4(m-3)=0$,\\
solving for $m$, we get $m=\boxed{\frac{13}{4}}$;


Question: As shown in the figure, the vertex of the parabola is $P(1, 4)$, it intersects the y-axis at point $C(0, 3)$, and intersects the x-axis at points $A$ and $B$. Let $Q$ be a point on the parabola other than point $P$, and the area of $\triangle BCQ$ is equal to the area of $\triangle BCP$, find the coordinates of point $Q$.

Solution:
\textbf{Solution:} Given the parabola's equation $y=-x^2+2x+3$, by setting $y=0$, we find $x_1=-1, x_2=3$,\\
$\therefore$ points $A(-1, 0), B(3, 0)$,\\
Let the equation of line BC be $y=kx+b$, then:\\
$\begin{cases}
3k+b=0\\
b=3
\end{cases}$, solving gives: $\begin{cases}
k=-1\\
b=3
\end{cases}$,\\
$\therefore$ the equation of line BC is $y=-x+3$,\\
Draw $PQ_1 \parallel BC$ through $P$, intersecting the parabola at point $Q_1$,\\
Let the equation of line $PQ_1$ be $y=-x+n$,\\
$\because P(1, 4)$,\\
$\therefore$ equation of line $PQ_1$ is $y=-x+5$,\\
By solving: $\begin{cases}
y=-x+5\\
y=-x^2+2x+3
\end{cases}$,\\
We get: $\begin{cases}
x=1\\
y=4
\end{cases}$ or $\begin{cases}
x=2\\
y=3
\end{cases}$, that is, $(1, 4)$ coincides with $P$, $Q_1(2, 3)$;\\
Through point $P$, draw a perpendicular to the $x$-axis intersecting BC at $G$, take $PG=GH$ on line $PG$,\\
$\because$ the equation of line BC is $y=-x+3$, $P(1, 4)$,\\
$\therefore G(1, 2)$,\\
$\therefore PG=GH=2$,\\
$\therefore H(1, 0)$,\\
Draw line $Q_2Q_3 \parallel BC$ through $H$, intersecting the parabola at points $Q_2, Q_3$,\\
Similarly, we find the equation of line $Q_2Q_3$ to be $y=-x+1$,\\
By solving: $\begin{cases}
y=-x+1\\
y=-x^2+2x+3
\end{cases}$,\\
We get: $\begin{cases}
x=\frac{3+\sqrt{17}}{2}\\
y=\frac{-1-\sqrt{17}}{2}
\end{cases}$ or $\begin{cases}
x=\frac{3-\sqrt{17}}{2}\\
y=\frac{-1+\sqrt{17}}{2}
\end{cases}$,\\
$\therefore Q_2\left(\frac{3-\sqrt{17}}{2}, \frac{-1+\sqrt{17}}{2}\right)$, $Q_3\left(\frac{3+\sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2}\right)$;\\
Therefore, the coordinates of point $Q$ are $\boxed{Q_2\left(\frac{3-\sqrt{17}}{2}, \frac{-1+\sqrt{17}}{2}\right)}$ or $\boxed{Q_3\left(\frac{3+\sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2}\right)}$.


Question: As shown in the figure, in rhombus $ABCD$, the diagonals intersect at point $O$, with $DE \parallel AC$ and $CE \parallel BD$. Quadrilateral $OCED$ is a rectangle; if $AD = 5$ and $BD = 8$, calculate the value of $DE$.

Solution:
Solution: Since quadrilateral ABCD is a rhombus, and $BD=8$, \\
therefore, $OD=\frac{1}{2}BD=4$, $OC=OA$, and $AD=CD$, \\
since $AD=5$, \\
thus, $OC=AO=\sqrt{AD^{2}-OD^{2}}=3$, \\
since quadrilateral OCED is a rectangle, \\
therefore, $DE=OC=\boxed{3}$.


Question: Let $A$ and $B$ be two fixed points on the side $OP$ of $\angle POQ$, and let $E$ be a moving point on the side $OQ$. Construct squares $ABCD$ and $AEFG$ on the same side above $OP$, with $AB$ and $AE$ as their respective sides. As shown in the figure, $\angle POQ=45^\circ$, $OB=6$, and $OA=2$. Connect $BG$ and $DE$. When point $E$ moves along side $OQ$, the length of segment $BG$ has a minimum value. Determine this minimum value.

Solution:
\textbf{Solution:} As shown in the diagram, let $OQ$ intersect $AD$ at point $M$, and $CD$ at point $N$. Draw $DE' \perp OQ$ at point $E'$,\\
$\because \angle OAD=90^\circ$, $\angle QOA=45^\circ$,\\
$\therefore \angle QOA=\angle AMO=45^\circ$,\\
$\therefore OA=AM=2$,\\
$\because OB=6$,\\
$\therefore AD=AB=OB-OA=4$,\\
$\therefore DM=AD-AM=2$,\\
$\because \angle DMN=\angle AMO=45^\circ$,\\
$\therefore DE'=DM\sin\angle DMN=2 \times \frac{\sqrt{2}}{2}=\sqrt{2}$,\\
$\therefore$ when point $E$ coincides with $E'$, the value of $DE$ is at its minimum,\\
$\because BG=DE$,\\
$\therefore$ the minimum value of $BG$ is $\boxed{\sqrt{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, AD is the altitude, $AE\parallel BC$, $BC=2AE$. The quadrilateral ADCE is a rectangle, point F is the midpoint of AB, connect DF, EF, if $\angle DFE=90^\circ$, $AB=4$, find the length of EF.

Solution:
\textbf{Solution:} As shown in the figure, connect DE.\\
Since quadrilateral ADCE is a rectangle,\\
it follows that $DE=AC=AB=4$.\\
Since point F is the midpoint of AB, and $\triangle ABD$ is a right-angled triangle,\\
it follows that $DF=\frac{1}{2}AB=2$.\\
In $\triangle DEF$, by the Pythagorean theorem, we have $EF=\sqrt{DE^{2}-DF^{2}}=2\sqrt{3}$.\\
Therefore, the length of EF is $\boxed{2\sqrt{3}}$.


Question: As shown in the figure, O is the intersection point of the diagonals of rectangle ABCD, DE $\parallel$ AC, and CE $\parallel$ BD. Quadrilateral OCED is a rhombus. If AB = 6 and BC = 8, find the perimeter of the quadrilateral OCED.

Solution:
\textbf{Solution:} Since quadrilateral ABCD is a rectangle,\\
it follows that $\angle ABC=90^\circ$,\\
Since AB=6, and BC=8,\\
then in $\triangle ABC$, by the Pythagorean theorem, we have: AC=10,\\
which means OC$=\frac{1}{2}$AC=5,\\
Since quadrilateral OCED is a rhombus,\\
it follows that OC=OD=DE=CE=5,\\
therefore, the perimeter of quadrilateral OCED is $5+5+5+5=\boxed{20}$.


Question: As shown in the figure, within square $ABCD$, diagonals $AC$ and $BD$ intersect at point $O$, with $E$ and $F$ located on $OB$ and $OC$ respectively. The extension of line $AE$ intersects $BF$ at point $M$, and $OE=OF$. $\triangle AOE \cong \triangle BOF$; $AM \perp BF$; Given $AE=\sqrt{5}$ and $OE=1$, calculate the length of $EM$.

Solution:
Solution: Since $AE=\sqrt{5}$ and $OE=1$, $\angle AOE=90^\circ$, \\
it follows that $OA=\sqrt{AE^2-OE^2}=2$, \\
Since $\triangle AOE \cong \triangle BOF$, \\
it follows that $S_{\triangle AOE}=S_{\triangle BOF}=S_{\triangle EOF}+S_{\triangle BEF}$, with $BF=AE=\sqrt{5}$, \\
Therefore, $\frac{1}{2}OE \cdot AO=\frac{1}{2}OE \cdot OF+\frac{1}{2}BF \cdot EM$, \\
Thus, $\frac{1}{2} \times 1 \times 2=\frac{1}{2} \times 1 \times 1+\frac{\sqrt{5}}{2}EM$, \\
Hence, $EM=\boxed{\frac{2}{\sqrt{5}}}$.


Question: As shown in the figure, it is known that in the rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. Point $E$ is on the extension of side $AB$, and $AB=BE$. Connect $CE$. Given that $BD \parallel EC$, if $AD=5$ and $CE=6$, what is the area of the rhombus $ABCD$?

Solution: \textbf{Solution:} Given that rhombus $ABCD$ with $AD=5$, hence $AB=5$. The diagonals $AC$ and $BD$ are perpendicular and bisect each other, hence $\angle AOB=90^\circ$, and $AO=OC$. Also, since $AB=BE$ and $CE=6$, we have $BO= \frac{1}{2}CE=3$ and $BD=2BO=6$. Thus, $AO= \sqrt{5^2-3^2}=4$ and $AC=2AO=8$. Therefore, the area of rhombus $ABCD$ is $S_{\text{rhombus }ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 8\times 6=\boxed{24}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $AB\parallel CD$, $AD\perp CD$, $\angle B=45^\circ$. Extend $CD$ to point $E$ such that $DE=DA$, and connect $AE$. Quadrilateral $ABCE$ is a parallelogram. If $AB=6$, $CD=2$, find the area of quadrilateral $ABCE$.

Solution:
\textbf{Solution:} Given from (1) $AB=CE$,\\
since $CD=2$, $AB=6$,\\
thus $DE=4$,\\
since $AD=DE$,\\
thus $AD=4$,\\
therefore, the area of quadrilateral $ABCE=AB\times AD=6\times 4=\boxed{24}$.


Question: As shown in Figure 1, there is a rectangular piece of paper $ABCD$, where $AD=8\text{cm}$ and $AB=6\text{cm}$. First, fold along the diagonal $BD$, and point $C$ falls onto the position of point $C'$, with $BC'$ intersecting $AD$ at point $G$. $BG=DG$; what is the length of $C'G$ in centimeters?

Solution:
\textbf{Solution:} Since $\triangle GAB \cong \triangle GC'D$,\\
$\therefore AG=C'G$,\\
Let $C'G=x$, then $GD=BG=8-x$,\\
$\therefore x^2+6^2=(8-x)^2$,\\
Solving this gives: $x=\frac{7}{4}$,\\
$\therefore C'G=\boxed{\frac{7}{4}\text{cm}}$;


Question: As shown in Figure 1, there is a rectangular piece of paper $ABCD$, where $AD=8\text{cm}$ and $AB=6\text{cm}$. First, fold along the diagonal $BD$, so that point $C$ falls on the position of point $C'$, and $BC'$ intersects $AD$ at point $G$. It is given that $BG=DG$; as shown in Figure 2, fold it again such that point $D$ coincides with point $A$. The crease $EN$ intersects $AD$ at $M$. Find the length of $EM$ in cm.

Solution:
\textbf{Solution:} Since point $D$ coincides with point $A$, we get the crease $EN$,\\
$\therefore DM=4\,\text{cm}$,\\
$\because AD=8\,\text{cm}$, $AB=6\,\text{cm}$,\\
$\therefore$ in $\triangle ABD$, $BD=10\,\text{cm}$,\\
$\because EN\perp AD$, $AB\perp AD$,\\
$\therefore EN\parallel AB$,\\
$\therefore MN$ is the midline of $\triangle ABD$,\\
$\therefore DN=\frac{1}{2}BD=5\,\text{cm}$,\\
In $\triangle MND$, $MN=\sqrt{5^2-4^2}=3\,\text{cm}$,\\
By the properties of folding, it is known that $\angle NDE=\angle NDC$,\\
$\because EN\parallel CD$,\\
$\therefore \angle END=\angle NDC$,\\
$\therefore \angle END=\angle NDE$,\\
$\therefore EN=ED$,\\
Let $EM=x$, then $ED=EN=x+3$,\\
By the Pythagorean theorem, $ED^2=EM^2+DM^2$, that is, $(x+3)^2=x^2+4^2$,\\
Solving for $x$, we get $x=\frac{7}{6}$, thus $EM=\boxed{\frac{7}{6}\,\text{cm}}$.


Question: As shown in the figure, in an equilateral triangle $\triangle ABC$, points $D$ and $E$ are the midpoints of sides $AB$ and $AC$ respectively. Extend $BC$ to point $F$ such that $CF = \frac{1}{2}BC$, and connect $DE$, $CD$, and $EF$. Quadrilateral $DCFE$ is a parallelogram. If $AB=6$, what is the perimeter of parallelogram $DCFE$?

Solution: \textbf{Solution:} Given that $\triangle ABC$ is an equilateral triangle and $AB=6$, with $D$ being the midpoint of $AB$, it follows that $BC=6$ and $CD\perp AB$ with $BD=\frac{1}{2}AB=3$. Therefore, $DC=\sqrt{BC^2-BD^2}=\sqrt{6^2-3^2}=3\sqrt{3}$. From statement (1), we know that $DE$ is the median of $\triangle ABC$, and quadrilateral $DCFE$ is a parallelogram, thus $DE=CF=\frac{1}{2}BC=3$ and $EF=DC=3\sqrt{3}$. Therefore, the perimeter of the parallelogram $DCFE$ is $2(DE+DC)=2(3+3\sqrt{3})=\boxed{6+6\sqrt{3}}$.


Question: As shown in the figure, extend the side BC of the rectangle ABCD to point E, such that CE = BD, and connect AE. If $\angle ADB = 40^\circ$, what is the degree measure of $\angle E$?

Solution:
\textbf{Solution:} As shown in the figure, connect AC and let the intersection of AC and BD be O,\\
$\because$ Quadrilateral ABCD is a rectangle,\\
$\therefore$ AC=BD, AD$\parallel$BC, OA=OB=OC=OD,\\
Also, $\because$ CE=BD,\\
$\therefore$ AC=CE,\\
$\therefore$ $\angle$CAE=$\angle$E,\\
$\because$ AD$\parallel$BC,\\
$\therefore$ $\angle$ADB=$\angle$CBD=40$^\circ$,\\
$\because$ OB=OC,\\
$\therefore$ $\angle$ACB=$\angle$CBD=40$^\circ$,\\
$\therefore$ $\angle$ACB=$\angle$CAE+$\angle$E=2$\angle$E=40$^\circ$,\\
$\therefore$ $\angle$E=\boxed{20^\circ}.


Question: As shown in the figure, extend the side $BC$ of the rectangle $ABCD$ to the point $E$ such that $CE = BD$, and connect $AE$. If $AB = 3$ and $CE = 5$, what is the length of $AE$?

Solution:
\textbf{Solution}: From (1), we know $AC=CE=5$, \\
$\therefore BC= \sqrt{AC^{2}-AB^{2}} = \sqrt{5^{2}-3^{2}}=4$, \\
$\therefore BE=BC+CE=4+5=9$, \\
$\because \angle ABE=90^\circ$, \\
$\therefore AE=\sqrt{AB^{2}+BE^{2}}=\sqrt{3^{2}+9^{2}}=\boxed{3\sqrt{10}}$.


Question: As shown in the diagram, in parallelogram $ABCD$, $AC \perp AD$. Let $\angle ECA = \angle ACD$, with $CE$ intersecting $AB$ at point $O$ and the extension of $DA$ at point $E$, and connect $BE$. Quadrilateral $ACBE$ is a rectangle; connect $OD$. If $AB=4$ and $\angle ACD=60^\circ$, what is the length of $OD$?

Solution:
\textbf{Solution:} As shown in the figure, draw OF$\perp$DE at F through point O,\\
from (1) it is known that quadrilateral ACBE is a rectangle,\\
$\therefore$ the diagonals AB and CE are equal and bisect each other, AO=$ \frac{1}{2}AB=2$,\\
$\therefore$OA=OC,\\
$\because$$\angle$ACD=$\angle$ACO=60^\circ,\\
$\therefore$$\triangle$AOC is an equilateral triangle,\\
$\therefore$$\angle$OAC=60^\circ,\\
$\because$$\angle$EAC=90^\circ,\\
$\therefore$$\angle$FAO=90^\circ−60^\circ=30^\circ,\\
In right $\triangle$AFO,\\
OF=$ \frac{1}{2}AO=1$, $ AF=\sqrt{3}$,\\
In right $\triangle$AEB, $ BE=\frac{1}{2}AB=2$,\\
AD=AE=$ \sqrt{{4}^{2}−{2}^{2}}=2\sqrt{3}$,\\
$\therefore$DF=AF+AD=$ \sqrt{3}+2\sqrt{3}=3\sqrt{3}$,\\
$\therefore$OD=$ \sqrt{DF^{2}+OF^{2}}=\boxed{2\sqrt{7}}$.


Question: As shown in the figure, the graph of the function $y=kx+b$ (where $k$ and $b$ are constants) is illustrated. What is the value of $k$?

Solution:
\textbf{Solution:} From the given problem, we see that the slope $k$ of the function $y=kx+b$ is $-\frac{1}{2}$,

$\therefore k=\boxed{-\frac{1}{2}}$.


Question: As shown in the figure, the graph of the function $y=kx+b$ (where $k$ and $b$ are constants) is depicted. What is the range of $y$ when $0<x<2$?

Solution:
\textbf{Solution:} When $0<x<2$, the range of $y$ is $0<y<1$, \\
$\therefore$ the range of $y$ is $\boxed{0<y<1}$.


Question: As shown in the figure, the line $y=-\frac{4}{3}x+4$ intersects the x-axis and y-axis at points A and B, respectively. A new line is obtained by shifting the line $y=-\frac{4}{3}x+4$ to the left, which intersects the x-axis and y-axis at points C and D, respectively. If $BD-AB=3$, then the equation of line CD is $y=$?

Solution: \textbf{Solution:} Hence, the equation of the line CD is $y=-\frac{4}{3}x-4$. Therefore, $\boxed{y=-\frac{4}{3}x-4}$.


Question: As shown in the figure, it is known that the line $l\colon y=x+4$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ ($x<0$) at point $A(-1, n)$, the line $l'$ passes through point $A$ and is symmetric to $l$ with respect to the line $x=-1$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since point A$(-1, 3)$ is on the line $l\colon y=x+4$,\\
the coordinates of point A are $(-1, 3)$.\\
Since point A is also on the graph of the inverse proportion function $y= \frac{k}{x}$ $(x<0)$,\\
it follows that $k=-3$,\\
therefore, the expression for the inverse proportion function is $y= \boxed{-\frac{3}{x}}$.


Question: As shown in the figure, it is known that the line $l\colon y=x+4$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ ($x<0$) at point $A(-1,n)$, and the line $l'$ passes through point $A$ and is symmetrical to $l$ with respect to the line $x=-1$. What is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Let the line $l: y=x+4$ intersect the x-axis and y-axis at points B and C, respectively,\\
then B$(-4,0)$, C$(0,4)$.\\
Since line $l'$ passes through point A and is symmetric to $l$ with respect to the line $x=-1$,\\
let the intersection of line $l'$ with the x-axis be point E, then E$(2,0)$.\\
Let line $l': y=ax+b$, then $\left\{\begin{array}{l}
3=-a+b\\
0=2a+b
\end{array}\right.$ solves to $\left\{\begin{array}{l}
a=-1,\\
b=2,
\end{array}\right.$\\
therefore $y=-x+2$. Let the intersection of line $l'$ with the y-axis be point D, then D$(0,2)$,\\
therefore, the area of the shaded part = area of $\triangle BOC$ - area of $\triangle ACD$ = $\frac{1}{2}\times4\times4-\frac{1}{2}\times2\times1=\boxed{7}$.


Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What are the coordinates of point $A$? What are the coordinates of point $B$?

Solution:
\textbf{Solution:} The coordinates of point $A$ are $\boxed{\left(\frac{3}{2}, 0\right)}$.\\
The coordinate information of point $B$ is not provided, so an answer cannot be given.


Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What are the coordinates of the intersection point $C$ of these two lines?

Solution: \textbf{Solution:} The coordinates of the intersection point $C$ of these two lines is $\boxed{\left(\frac{2}{3}, -\frac{10}{3}\right)}$.


Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the $x$-axis at point $A$, and the line $y=-2x-2$ intersects the $x$-axis at point $B$, and these two lines intersect at point $C$. What is the area of $\triangle ABC$?

Solution:
Solution: Since $A\left(\frac{3}{2}, 0\right)$ and $B\left(-1, 0\right)$,

$\therefore AB=\frac{3}{2}-(-1)=\frac{5}{2}$,

$\therefore {S}_{\triangle ABC}=\frac{1}{2}AB\cdot\left|{y}_{c}\right|=\frac{1}{2}\times \frac{5}{2}\times \frac{10}{3}=\boxed{\frac{25}{6}}$.


Question: As shown in the figure, it is known that the line $y=4x-6$ intersects the x-axis at point $A$, and the line $y=-2x-2$ intersects the x-axis at point $B$, and these two lines intersect at point $C$. Write the range of the independent variable $x$ for which the value of the function $y=4x-6$ is greater than the value of $y=-2x-2$.

Solution: \textbf{Solution:} From the graph, it is known that the range of the independent variable $x$ that makes the value of the function $y=4x-6$ greater than the value of the function $y=-2x-2$ is $\boxed{x>\frac{2}{3}}$.


Question: As shown in the figure, the line $y_{1}=2x-2$ intersects the $y$ axis at point $A$, and the line $y_{2}=-2x+6$ intersects the $y$ axis at point $B$. The two lines intersect at point $C$. Solve the system of equations $\left\{\begin{array}{l}2x-y=2 \\ 2x+y=6 \end{array}\right.$.

Solution: \textbf{Solution:} The solution to the system of equations $\left\{\begin{array}{l}2x-y=2 \\ 2x+y=6 \end{array}\right.$ is: $\left\{\begin{array}{l}x=\boxed{2} \\ \boxed{y=2} \end{array}\right.$.


Question: As shown in the figure, the line $y_{1}=2x-2$ intersects the y-axis at point $A$, and the line $y_{2}=-2x+6$ intersects the y-axis at point $B$. The two lines intersect at point $C$. What is the range of values for $x$ when both $2x-2>0$ and $-2x+6>0$ are true?

Solution:
\textbf{Solution:} When both $y_1>0$ and $y_2>0$ hold, \\
$\therefore$ the range of values for $x$ is: $x\in\boxed{(1,3)}$.


Question: As shown in the figure, the line $y_{1} = 2x - 2$ intersects the y-axis at point $A$, and the line $y_{2} = -2x + 6$ intersects the y-axis at point $B$, with the two lines intersecting at point $C$. Find the area of $\triangle ABC$.

Solution:
Solution: Since let $x=0$, then ${y}_{1}=-2$, ${y}_{2}=6$\\
Therefore, $A(0, 2)$, $B(0, 6)$.\\
Therefore, $AB=8$\\
Therefore, the area of $\triangle ABC$ is $\frac{1}{2}\times 8\times 2=\boxed{8}.


Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the solution to the system of equations $\begin{cases}2x-y=2, \\ 2x+y=6?\end{cases}$

Solution:
\textbf{Solution:} Given the system of equations $\begin{cases}2x-y=2, \\ 2x+y=6\end{cases}$, we find the solution $x=2$, $y=2$.

$\therefore$ The solution to the system is $x=\boxed{2}$, $\boxed{y=2}$.


Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the range of $x$ when both $y_{1}>0$ and $y_{2}>0$ are satisfied?

Solution: \textbf{Solution:} When both $y_{1}>0$ and $y_{2}>0$ hold, the range of $x$ is $x=\boxed{1<x<3}$.


Question: As shown in the figure, the graph of the line $y_{1}=2x-2$ intersects the y-axis at point A, and the graph of the line $y_{2}=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Let $x=0$, then $y_1=-2$, and $y_2=6$,\\
Therefore $A(0, -2)$ and $B(0, 6)$.\\
Therefore $AB=8$.\\
Therefore, the area of $\triangle ABC$ is $\frac{1}{2} \times 8 \times 2 = \boxed{8}$;


Question: As shown in the figure, the graph of the line $y_1=2x-2$ intersects the y-axis at point A, and the graph of the line $y_2=-2x+6$ intersects the y-axis at point B. The two lines intersect at point C. There exists another point $P$ on the graph of the line $y_1=2x-2$, different from point $C$, such that the area of $\triangle ABC$ is equal to the area of $\triangle ABP$. Find the coordinates of point $P$.

Solution:
\[ \textbf{Solution:} \text{ Let } P(x_0, 2x_0-2), \text{ then } S_{\triangle ABP}=\frac{1}{2}\times 8\times |x_0|=8, \]
\[ \therefore x_0=\pm 2. \]
\[ \because \text{ Point } P \text{ is different from point } C, \]
\[ \therefore x_0=-2, 2x_0-2=-6. \]
\[ \therefore P(\boxed{-2, -6}). \]


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects with the x-axis at point $A(-3, 0)$ and with the y-axis at point B. It also intersects the graph of the directly proportional function $y=\frac{4}{3}x$ at point $C(a, 4)$. Find the value of $a$ and the equation of the linear function $y=kx+b$?

Solution: \textbf{Solution:} Since point C lies on the graph of a direct proportion function, $\therefore \frac{4}{3}a=4$. Solving this gives: $a=\boxed{3}$. Since points C(3, 4) and A(-3, 0) lie on the graph of a linear function, $\therefore \left\{\begin{array}{l} -3k+b=0 \\ 3k+b=4 \end{array}\right.$. Solving this system of equations yields $\left\{\begin{array}{l} k=\frac{2}{3} \\ b=2 \end{array}\right.$, therefore the equation of the linear function is $\boxed{y=\frac{2}{3}x+2}$.


Question: As shown in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the x-axis at point $A(-3, 0)$ and the y-axis at point $B$, and its intersection point with the direct proportion function $y=\frac{4}{3}x$ is $C(a, 4)$. What is the area of $\triangle BOC$?

Solution: \textbf{Solution:} In $y=\frac{2}{3}x+2$, let $x=0$, we get $y=2$, hence $B(0,2)$ and $OB=2$. Since point $C(3,4)$, we have $x_C=3$. Therefore, the area of $\triangle BOC$ is $\frac{1}{2}OB\times x_C=\frac{1}{2}\times 2\times 3=\boxed{3}$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with known coordinates $A(4, 0)$ and $B(0, 3)$. Directly write down the coordinates of points $C$ and $D$.

Solution: \textbf{Solution:} Since $A(4,0)$ and $B(0,3)$, it follows that $OA=4, OB=3$. Therefore, $AB=\sqrt{AO^{2}+BO^{2}}=\sqrt{4^{2}+3^{2}}=5$. Since quadrilateral ABCD is a rhombus, it follows that $BC=AD=AB=5$. Consequently, $C(\boxed{-5,3})$ and $D(\boxed{-1,0})$.


Question: As shown in the figure, the quadrilateral $ABCD$ is a rhombus, with $A(4, 0)$ and $B(0, 3)$. What is the analytical expression of the line passing through points $B$ and $D$?

Solution:
\textbf{Solution:} Let the equation of the line passing through points $B$ and $D$ be $y=mx+n$. Substituting $B(0,3)$, $D(-1,0)$ into it, we get:
\[
\left\{
\begin{array}{l}
n=3 \\
-m+n=0
\end{array}
\right.
\]
Solving this, we find:
\[
\left\{
\begin{array}{l}
m=3 \\
n=3
\end{array}
\right.
\]
$\therefore$ The equation of the line passing through points $B$ and $D$ is \boxed{y=3x+3}.


Question: As shown in the figure, quadrilateral $ABCD$ is a rhombus, with $A(4, 0)$ and $B(0, 3)$. Draw the center $M$ of the rhombus, and write down the coordinates of point $M$.

Solution:
\textbf{Solution:} Connect $AC$ and $BD$ intersecting at $M$, then $M$ is the center of the rhombus $ABCD$.\\
Since quadrilateral $ABCD$ is a rhombus,\\
therefore, $M$ is the midpoint of $BD$,\\
since $B(0,3)$, $D(-1,0)$,\\
therefore, \boxed{M\left(-\frac{1}{2}, \frac{3}{2}\right)}.


Question: As shown in the figure, the quadrilateral $ABCD$ is a diamond. It is known that $A(4, 0)$, $B(0, 3)$. Move the line from (2) up and down along the $y$-axis by $b$. If the line always intersects with the diamond, directly write out the range of values of $b$.

Solution: \textbf{Solution:} Given from equation (2) the equation of the line passing through points B and D is $y=3x+3$. Translating this line along the y-axis by b units, we get the equation $y=3x+3+b$ (where $b>0$ indicates an upward shift, and $b<0$ indicates a downward shift). When the translated line passes through point A (4,0), we have $0=12+3+b$, hence $b=\boxed{-15}$. When the translated line passes through point C (-5,3), we have $3=-15+3+b$, thus $b=\boxed{15}$. Therefore, the line always intersects with the rhombus, so the range of $b$ is $\boxed{-15\leq b\leq 15}$.


Question: As shown in the figure, point P $(x, y)$ is a moving point in the first quadrant, and it lies on the line $y=-2x+8$. The line intersects the x-axis at point A. What is the area of $\triangle APO$ when the x-coordinate of point P is 3?

Solution:
\textbf{Solution:} Let $y=0$, then $-2x+8=0$, solving this we find $x=4$,\\
$\therefore OA=4$.\\
$\because$ Point P$(x,y)$ is a moving point in the first quadrant, and lies on the line $y=-2x+8$,\\
$\therefore$ when $x=3$, $y=-2\times 3+8=2$.\\
$\therefore S_{\triangle APO}=\frac{1}{2}\times 4\times 2=\boxed{4}$.


Question: As shown in the figure, point \(P(x, y)\) is a moving point in the first quadrant and is on the line \(y = -2x + 8\), where the line intersects the x-axis at point \(A\). Let the area of \(\triangle APO\) be \(S\), express \(S\) with an algebraic formula involving \(x\), and write down the range of values for \(x\).

Solution:
\textbf{Solution:} Given that point P lies on the line $y=-2x+8$,\\
thus P$(x, -2x+8)$.\\
Therefore, $S_{\triangle APO}=OA \times (-2x+8) = 4 \times (-2x+8) = -4x+32$,\\
where the range of $x$ is $0<x<4$,\\
hence $S_{\triangle APO}=\boxed{-4x+32}$.


Question: As shown in the figure, the graphs of the functions $y = 3x$ and $y = kx + 5$ intersect at point $A(a, 3)$. What are the values of $a$ and $k$?

Solution: \textbf{Solution:} Substituting $A\left(1, 3\right)$ into $y=3x$, we have $3a=3$. Solving this gives $a=\boxed{1}$. Therefore, the coordinates of point A are $\left(1, 3\right)$. Since the graph of the function $y=kx+5$ passes through point A, we have $k+5=3$. Solving this yields $k=\boxed{-2}$.


Question: As shown in the figure, the graphs of the functions \(y=3x\) and \(y=kx+5\) intersect at point \(A(a, 3)\). Based on the graph, directly write the solution set of the inequality \(3x<kx+5\).

Solution: \textbf{Solution:} From the graph, it can be concluded that the solution set of the inequality $3x<kx+5$ is $\boxed{x<1}$.


Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects line $AB$ at point $C$. What are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given the line $y=kx+5$ passes through the point $A(5,0)$, we have $5k+5=0$. Solving this equation gives $k=-1$. Therefore, the equation of line $AB$ is $y=-x+5$. By solving the system of equations of lines $AB$ and $CD$:
\[
\begin{cases}
y=-x+5 \\
y=2x-4
\end{cases}
\]
we find:
\[
\begin{cases}
x=3 \\
y=2
\end{cases}
\]
Therefore, the coordinates of point $C$ are $\boxed{(3,2)}$.


Question: Given the line $y=kx+5$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, and the coordinates of point $A$ are $(5,0)$, the line $y=2x-4$ intersects the $x$ axis at point $D$, and intersects the line $AB$ at point $C$. Based on the graph, write the solution set of the inequality $2x-4>kx+5$ with respect to $x$.

Solution:
Observe the graph of the function, we know that for $x>3$, the line $y=2x-4$ is above the line $y=kx+5$, \\
$\therefore$ the solution set of the inequality $2x-4>kx+5$ is $\boxed{x>3}$.


Question: As shown in the figure, it is known that the line $y=kx+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$, and the coordinates of point $A$ are $(5,0)$. The line $y=2x-4$ intersects the $x$-axis at point $D$ and intersects the line $AB$ at point $C$. What is the area of $\triangle ADC$?

Solution: \textbf{Solution:} Substituting $y=0$ into $y=2x-4$, we get $2x-4=0$. Solving this yields $x=2$. Thus, the coordinates of point D are $(2,0)$. Since A$(5,0)$, it follows that $AD=3$. Given the coordinates of point C as $C(3,2)$, it follows that the area of $\triangle ADC$ is $\frac{1}{2} \cdot 3 \cdot 2 = \boxed{3}$.


Question: As shown in the figure, the analytical expression of the line $l_1$ is: $y = -3x + 3$, and the line $l_1$ intersects the x-axis at point D. The line $l_2$ passes through points A$(4,0)$ and B$\left(3, -\frac{3}{2}\right)$. The lines $l_1$ and $l_2$ intersect at point C. What are the coordinates of point D?

Solution:
\textbf{Solution}: Since point D is on the x-axis, the equation of line $l_{1}$ is $y=-3x+3$. \\
Therefore, setting $y=0$ yields $-3x+3=0$, solving for $x$ gives us $x=1$, \\
Therefore, the coordinates of point D are $\boxed{(1, 0)}$.


Question: As shown in the figure, the analytical expression of line $l_1$ is $y=-3x+3$, and line $l_1$ intersects the x-axis at point D. Line $l_2$ passes through points A$(4,0)$ and $B\left(3, -\frac{3}{2}\right)$. Lines $l_1$ and $l_2$ intersect at point C. What is the analytical expression of line $l_2$?

Solution:
\textbf{Solution:}
Let the analytical expression of line $l_2$ be: $y=kx+b (k\ne 0)$. Since line $l_2$ passes through points $A(4, 0)$ and $B\left(3, -\frac{3}{2}\right)$, substituting into the expression, we get:
\[
\begin{cases}
4k+b=0\\
3k+b=-\frac{3}{2}
\end{cases}
\]
Solving this system of equations yields:
\[
\begin{cases}
k=\frac{3}{2}\\
b=-6
\end{cases}
\]
Therefore, the analytical expression of line $l_2$ is: $\boxed{y=\frac{3}{2}x-6}$.


Question: As shown in the figure, the analytic expression of line $ l_{1}$ is: $ y=-3x+3$, and line $ l_{1}$ intersects the x-axis at point D, line $ l_{2}$ passes through point A$(4,0)$ and point $B\left(3, -\frac{3}{2}\right)$, lines $ l_{1}$ and $ l_{2}$ intersect at point C. What is the area of $\triangle ADC$?

Solution:
Solution: By solving the system of equations of lines $l_1$ and $l_2$
\[
\begin{cases}
y=-3x+3\\
y=\frac{3}{2}x-6
\end{cases}
\]
we obtain
\[
\begin{cases}
x=2\\
y=-3
\end{cases}
\]
$\therefore C(2, -3)$, $\because$ in $\triangle ADC$, $AD=3$, and the height from $AD$ is $|-3|=3$, $\therefore$ the area of $\triangle ADC$ is $\frac{1}{2} \times 3 \times 3 = \boxed{\frac{9}{2}}$.


Question: As shown in the figure, the analytical expression of the line $l_1$ is: $y=-3x+3$, and the line $l_1$ intersects with the x-axis at point D. Line $l_2$ passes through points A$(4,0)$ and B$\left(3, -\frac{3}{2}\right)$. Line $l_1$ and line $l_2$ intersect at point C. On line $l_2$, there exists another point P different from point C, such that the area of $\triangle ADP$ is equal to the area of $\triangle ADC$. Please directly write out the coordinates of point P.

Solution: \textbf{Solution:} The coordinates of point P are $\boxed{(6,3)}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, point O is the origin, and the quadrilateral $ABCO$ is a rhombus. The coordinates of point A are $(-5, 12)$, and point C is on the positive half-axis of the x-axis. The line AC intersects the y-axis at point M, and the side AB intersects the y-axis at point H. What is the side length of the rhombus $ABCO$?

Solution:
Solution: Since the coordinates of point A are $(-5, 12)$, \\
thus $AH = 5$, and $OH = 12$, \\
By the Pythagorean theorem, we have: $OA = \sqrt{AH^{2}+OH^{2}} = \sqrt{5^{2}+12^{2}} = \boxed{13}$, \\
Therefore, the side length of the rhombus $ABCO$ is \boxed{13}.


Question: As shown in Figure 1, in the Cartesian coordinate system, point $O$ is the origin, and quadrilateral $ABCO$ is a rhombus. The coordinates of point $A$ are $(-5, 12)$, and point $C$ is located on the positive half of the x-axis. Line $AC$ intersects the y-axis at point $M$, and side $AB$ intersects the y-axis at $H$. Find the analytical expression of line $AC$.

Solution:
\textbf{Solution:} Given that the side length of the rhombus $ABCO$ is $13$,\\
hence $OC=13$,\\
thus $C(13,0)$,\\
let the equation of line $AC$ be: $y=kx+b$ (where $k\neq 0$),\\
substituting $A(-5,12)$ and $C(13,0)$ into the equation yields: $\begin{cases}-5k+b=12\\ 13k+b=0\end{cases}$,\\
solving this system gives: $\begin{cases}k=-\frac{2}{3}\\ b=\frac{26}{3}\end{cases}$,\\
therefore, the equation of line $AC$ is: \boxed{y=-\frac{2}{3}x+\frac{26}{3}}.


Question: As shown in Figure 1, in the Cartesian coordinate system, point O is the origin, and quadrilateral $ABCO$ is a rhombus. The coordinate of point A is $(-5, 12)$, and point C is on the positive half-axis of the x-axis. The line AC intersects the y-axis at point M, and the side AB intersects the y-axis at point H. As shown in Figure 2, a moving point P starts from point A, moving along the polyline $A-B-C$ towards the endpoint C. A line $PQ\parallel y$-axis is drawn through point P intersecting AC at point Q, let the x-coordinate of point P be $a$, and the length of the segment PQ be $l$.

Solution:
\textbf{Solution}: Since $PQ\parallel y$-axis and the x-coordinate of point P is a,\\
it follows that $Q(a, -\frac{2}{3}a+\frac{26}{3})$,\\
(1) When $-5\le a\le 8$, point P is on AB, P$(a, 12)$,\\
thus $l=12-\left(-\frac{2}{3}a+\frac{26}{3}\right)=\frac{2}{3}a+\frac{10}{3}$;\\
When $8<a\le 13$, point P is on BC,\\
since the coordinates of point B are $(8, 12)$ and the coordinates of point C are $(13, 0)$,\\
let the equation of line BC be: $y=mx+n$ (where $m\neq 0$),\\
then $\begin{cases}8m+n=12\\ 13m+n=0\end{cases}$, solving this yields: $\begin{cases}m=-\frac{12}{5}\\ n=\frac{156}{5}\end{cases}$,\\
thus the equation of line BC is: $y=-\frac{12}{5}x+\frac{156}{5}$,\\
thus at this time P$(a, -\frac{12}{5}a+\frac{156}{5})$,\\
thus $l=-\frac{12}{5}a+\frac{156}{5}-\left(-\frac{2}{3}a+\frac{26}{3}\right)=-\frac{26}{15}a+\frac{338}{15}$,\\
summarizing, $l=\begin{cases}\frac{2}{3}a+\frac{10}{3} & (-5\le a\le 8)\\ -\frac{26}{15}a+\frac{338}{15} & (8<a\le 13)\end{cases}$;\\
(2) Possible;\\
In the equation $y=-\frac{2}{3}x+\frac{26}{3}$, let $x=0$, yielding $y=\frac{26}{3}$,\\
thus $OM=\frac{26}{3}$,\\
thus $MN=\frac{1}{2}OM=\frac{13}{3}$,\\
since $PQ\parallel MN$,\\
thus when $PQ=MN$, quadrilateral PQNM is a parallelogram,\\
when $-5\le a\le 8$, we have $\frac{2}{3}a+\frac{10}{3}=\frac{13}{3}$,\\
solving yields: $a=\frac{3}{2}$,\\
thus $P\left(\frac{3}{2}, 12\right), Q\left(\frac{3}{2}, \frac{23}{3}\right)$;\\
when $8<a\le 13$, we have $-\frac{26}{15}a+\frac{338}{15}=\frac{13}{3}$,\\
solving yields: $a=\frac{21}{2}$,\\
thus $P\left(\frac{21}{2}, 6\right), Q\left(\frac{21}{2}, \frac{5}{3}\right)$,\\
summarizing, \boxed{P\left(\frac{3}{2}, 12\right), Q\left(\frac{3}{2}, \frac{23}{3}\right)} or \boxed{P\left(\frac{21}{2}, 6\right), Q\left(\frac{21}{2}, \frac{5}{3}\right)}.


Question: As shown in the figure, the function expression of line ${l}_{1}$ is: $y = x - 3$, which intersects the x-axis at point B. Line ${l}_{2}$ passes through point $A(-2, 0)$ and intersects line ${l}_{1}$ at point $C(-1, a)$. What is the analytical expression of line ${l}_{2}$?

Solution:
\textbf{Solution:} Let's substitute point C $(-1, a)$ into $y = x - 3$, yielding $a = -4$,\\
$\therefore$ the coordinates of point C are $(-1, -4)$,\\
Let the equation of line $l_2$ be $y = kx + b$,\\
$\therefore$ $\begin{cases} -2k+b=0 \\ -k+b=-4 \end{cases}$,\\
Solving these, we get $\begin{cases} k=-4 \\ b=-8 \end{cases}$,\\
$\therefore$ the equation of line $l_2$ is $\boxed{y = -4x - 8}$.


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

Solution:
\textbf{Solution:} Solve: In the line $l_1\colon y=x-3$, let $y=0$, we get $x=3$,\\
$\therefore B(3,0)$,\\
$\therefore AB=3-(-2)=5$,\\
Let $P(b,b-3)$, since $PQ\parallel x$-axis, we have $Q(b,-4b-8)$,\\
$PQ=|b-3-(-4b-8)|=AB=5$,\\
Solving this, we get $b=0$ or $b=-2$,\\
$\therefore P(0,-3)$ or $P(-2,-5)$,\\
Hence, the coordinates of point $P$ are $\boxed{(0,-3)}$ or $\boxed{(-2,-5)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, perpendicular lines are drawn through points $C(1, 4)$ and $D(4, 1)$ to the x-axis, meeting the x-axis at points A and B, respectively. Find the analytical expressions for the lines CD and OD.

Solution:
\textbf{Solution:} Let the equation of line CD be $y=kx+b$,\\
Substituting $C\left(1, 4\right)$ and $D\left(4, 1\right)$ into the equation,\\
we get $\begin{cases}k+b=4\\ 4k+b=1\end{cases}$,\\
Solving this system of equations, we find $\begin{cases}k=-1\\ b=5\end{cases}$,\\
Therefore, the equation of line CD is $\boxed{y=-x+5}$,\\
Let the equation of line OD be $y=mx$,\\
Substituting $D\left(4, 1\right)$ into the equation, we get $m=\frac{1}{4}$,\\
Hence, the equation of line OD is $\boxed{y=\frac{1}{4}x}$.


Question: As shown in the diagram, in the Cartesian coordinate system, perpendicular lines to the $x$-axis are drawn through points $C(1, 4)$ and $D(4, 1)$, with the feet of the perpendiculars being $A$ and $B$, respectively. Point $M$ is a moving point on line $OD$, and a perpendicular line to the $x$-axis is drawn through point $M$, intersecting the $x$-axis at point $P$ and line $CD$ at point $N$.

Solution:
Solution:
(1) Since PM is the midline of $\triangle OBD$ with $D(4, 1)$,
it follows that $M\left(2, \frac{1}{2}\right)$.
Substituting $x=2$ into $y=-x+5$ yields $N\left(2, 3\right)$,
therefore $MN=\frac{5}{2}$.

(2) Possible;
Reason: Assume the coordinates of point M are $\left(a, \frac{1}{4}a\right)$, and the coordinates of point N are $\left(a, -a+5\right)$.
When $AC=MN$, the quadrilateral formed by the vertices A, C, N, M is a parallelogram.
That is, $\left|-a+5-\frac{1}{4}a\right|=4$, solving this gives $a=\frac{4}{5}$ or $\frac{36}{5}$. Thus, the satisfying x-coordinate of point M is $\boxed{\frac{4}{5}}$ or $\boxed{\frac{36}{5}}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=x+2$ is the line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the line $l_{2}$. The two lines intersect at point $A(1,a)$. It is known that the points where lines $l_{1}$ and $l_{2}$ intersect the $x$-axis are points $B$ and $C$, respectively, and line $l_{2}$ intersects the $y$-axis at point $E(0,4)$. What are the coordinates of point $A$? What are the coordinates of point $B$?

Solution: \textbf{Solution:} The coordinates of point $A$ are $\boxed{(1,3)}$, while the coordinate information for point $B$ is not provided.


Question: As shown in the diagram, the graph of the linear function $y_{1}=x+2$ is the straight line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the straight line $l_{2}$. The two lines intersect at point $A(1, a)$. It is known that the intersections of the lines $l_{1}$ and $l_{2}$ with the $x$-axis are points $B$ and $C$, respectively, and the line $l_{2}$ intersects the $y$-axis at point $E(0, 4)$. What is the equation of the line $l_{2}$?

Solution:
\textbf{Solution:} Since the linear function $y_{2}=kx+b$ passes through the points $E(0,4)$ and $A(1, 3)$,\\
we have $\left\{\begin{array}{l}
3=k+b \\
4=b
\end{array}\right.$,\\
solving this, we get $\left\{\begin{array}{l}
k=-1 \\
b=4
\end{array}\right.$,\\
$\therefore$ the equation of line $l_{2}$ is $\boxed{y_{2}=-x+4}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=x+2$ is the line $l_{1}$, and the graph of the linear function $y_{2}=kx+b$ is the line $l_{2}$. The two lines intersect at point $A(1,a)$. It is known that lines $l_{1}$ and $l_{2}$ intersect the $x$-axis at points $B$ and $C$ respectively, and line $l_{2}$ intersects the $y$-axis at point $E(0,4)$. Find the area of $\triangle ABC$.

Solution:
\textbf{Solution:} Let $y_2=0$, that is, $-x+4=0$ \\
Solving this, we get $x=4$ \\
Hence, $C(4, 0)$ \\
Therefore, the area of $\triangle ABC = \frac{1}{2} \times BC \times y_A = \frac{1}{2}[4-(-2)]\times 3 = \boxed{9}$


Question: As shown in the figure, the line $x=t$ intersects with the line $y=-x$ and the line $y=\frac{1}{2}x+2$ at points E and D (where D is above E), respectively. The line $y=-x$ and the line $y=\frac{1}{2}x+2$ intersect at point M. What are the coordinates of point M?

Solution: \textbf{Solution:} The coordinates of point M are $\boxed{\left(-\frac{4}{3}, \frac{4}{3}\right)}$.


Question: As shown in the diagram, the line $x=t$ intersects with the line $y=-x$ and the line $y=\frac{1}{2}x+2$ at points E and D (where D is above E), respectively. What is the length of the line segment DE (expressed as an algebraic expression containing $t$)?

Solution:
\textbf{Solution:} When $x=t$, $y=-t$; when $x=t$, $y=\frac{1}{2}t+2$.\\
Therefore, the coordinates of point D are $\left(t, \frac{1}{2}t+2\right)$, and the coordinates of point E are $\left(t, -t\right)$.\\
Since D is above E,\\
therefore the length of segment DE is $DE=\frac{3}{2}t+2$, and $t>-\frac{4}{3}$.\\
Hence, the length of the line segment DE is $DE=\boxed{\frac{3}{2}t+2}$.


Question: As shown in the figure, the line $x=t$ intersects the line $y=-x$ and the line $y=\frac{1}{2}x+2$ at points E and D (D is above E), respectively. Point N is a moving point on the y-axis, and $\triangle NDE$ is an isosceles right triangle. Directly write the value of $t$ and the coordinates of point N.

Solution:
\textbf{Solution:} Write directly the value of $t$ and the coordinates of point $N$.\\
When $t=-\frac{4}{5}$, the coordinates of point N are $\left(0, \frac{8}{5}\right)$ or $\left(0, \frac{4}{5}\right)$;\\
When $t=-\frac{4}{7}$, the coordinates of point N are $\left(0, \frac{8}{7}\right)$;\\
When $t=4$, $\triangle NDE$ is an isosceles right triangle, at this point the coordinates of N are \boxed{\left(0, 0\right)}.


Question: As shown in the diagram, it's known that the line $l_1\colon y=kx+b$ ($k\ne 0$), which passes through the point $B(1, 0)$, intersects with the line $l_2\colon y=2x+4$ at point $P(-1, a)$. Find the equation of the line $l_1$.

Solution:
\textbf{Solution}: Since point P ($-1$, $2$) lies on the line $l_2: y=2x+4$, \\
it follows that $2\times(-1)+4=2$, which means $a=2$, so the coordinates of P are ($-1$, $2$).\\
According to the problem, we have
\[
\left\{
\begin{array}{l}
k+b=0\\
-k+b=2
\end{array}
\right.,
\]
from which we obtain:
\[
\left\{
\begin{array}{l}
k=-1\\
b=1
\end{array}
\right.,
\]
Therefore, the equation for $l_1$ is: \boxed{y=-x+1};


Question: As shown in the figure, it is known that the line $l_1: y=kx+b$ ($k\ne 0$) passing through point $B(1, 0)$ intersects with line $l_2: y=2x+4$ at point $P(-1, a)$. What is the area of $\triangle ABP$?

Solution: Solution: Since the line $l_2\colon y=2x+4$ intersects the x-axis at point A, let $y=0$, then $2x+4=0$. Solving this gives: $x=-2$, therefore, the coordinates of point A are $(-2,0)$. Given that $AB=3$, and since $P(-1, 2)$, the area of triangle $ABP$ is $\frac{1}{2}\times3\times2=\boxed{3}


Question: As shown in Figure 1, in the Cartesian coordinate system, the line $l_{1}$: $y=kx+b\ (k \neq 0)$ intersects the x-axis at point $A(a, 0)$ and the y-axis at point $B(0, b)$. The line $l_{2}$, which passes through the origin, intersects the line $l_{1}$ at point C. If $a=2$ and $b=4$, what is the equation of the line $l_{1}$?

Solution:
\textbf{Solution:} Let's suppose $a=2$, $b=4$, then point $A(2, 0)$ and point $B(0, 4)$ are given. Substituting into the equation $y=kx+b$, we get:
\begin{align*}
\begin{cases}
2k+b=0\\
b=4
\end{cases}
\end{align*},
which solves to
\begin{align*}
\begin{cases}
k=-2\\
b=4
\end{cases}
\end{align*},
$\therefore$ the equation for the line ${l}_{1}$ is $\boxed{y=-2x+4}$.


Question: As shown in Figure 1, in the Cartesian coordinate system, the line ${l}_{1}$: $y=kx+b\ (k\ne 0)$ intersects the x-axis at point $A(a, 0)$, and it intersects the y-axis at point $B(0, b)$. The line ${l}_{2}$ passing through the origin intersects line ${l}_{1}$ at point C. Given condition (1), when the x-coordinate of point C is 1, what is the area of $\triangle AOC$?

Solution:
\textbf{Solution:} Substitute $x=1$ into $y=-2x+4$, we get $y=-2\times 1+4=2$,\\
$\therefore C(1, 2)$,\\
$\because A(2, 0)$,\\
$\therefore OA=2$,\\
$\therefore {S}_{\triangle AOC}=\frac{1}{2}OA\times \left|{y}_{C}\right|=\frac{1}{2}\times 2\times 2=\boxed{2}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\frac{1}{2}x+1$ intersects the x-axis and y-axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed inside the second quadrant with $AB$ as its side. What is the area of square $ABCD$?

Solution:
\textbf{Solution:} For the line $y=\frac{1}{2}x+1$, let $x=0$, then we obtain $y=1$; let $y=0$, then we obtain $x=-2$,\\
$\therefore A(-2, 0)$, $B(0, 1)$,\\
In the right triangle $Rt\triangle AOB$, $OA=2$, $OB=1$,\\
According to the Pythagorean theorem: $AB=\sqrt{2^2+1^2}=\sqrt{5}$;\\
Therefore, the area of square $ABCD$ is $\boxed{5}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the graph of the linear function $y=\frac{1}{2}x+1$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed in the second quadrant with $AB$ as its side. What are the coordinates of points $C$ and $D$?

Solution:
\textbf{Solution:} Construct $CE \perp y$-axis and $DF \perp x$-axis, we can get $\angle CEB = \angle AFD = \angle AOB = 90^\circ$,\\
since square $ABCD$,\\
thus $BC = AB = AD$, $\angle DAB = \angle ABC = 90^\circ$,\\
thus $\angle DAF + \angle BAO = 90^\circ$, $\angle ABO + \angle CBE = 90^\circ$,\\
since $\angle DAF + \angle ADF = 90^\circ$, $\angle BAO + \angle ABO = 90^\circ$,\\
thus $\angle BAO = \angle ADF = \angle CBE$,\\
thus $\triangle BCE \cong \triangle DAF \cong \triangle ABO$,\\
thus $BE = DF = OA = 2$, $CE = AF = OB = 1$,\\
thus $OE = OB + BE = 2 + 1 = 3$, $OF = OA + AF = 2 + 1 = 3$,\\
thus $\boxed{C(-1, 3)}$, $\boxed{D(-3, 2)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, it's known that the graph of the linear function $y=\frac{1}{2}x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. A square $ABCD$ is constructed inside the second quadrant with $AB$ as its side. Does there exist a point $M$ on the $x$-axis such that the perimeter of $\triangle MDB$ is minimized? Determine the coordinates of point $M$.

Solution:
\textbf{Solution:} To solve this, find the point $B'$ which is symmetrical to $B$ with respect to the $x$-axis, and connect $BD$, intersecting the $x$-axis at point $M$. At this time, the perimeter of $\triangle BMD$ is minimized,\\
$\because B(0, 1)$,\\
$\therefore B'(0, -1)$\\
Let the equation of line $B'D$ be $y=kx+b$,\\
Substituting the coordinates of $B'$ and $D$ yields: $\begin{cases}b=-1\\ -3k+b=2\end{cases}$,\\
Solving this, we get: $\begin{cases}k=-1\\ b=-1\end{cases}$, that is, the equation of line $B'D$ is $y=-x-1$,\\
Setting $y=0$, we find $x=-1$, i.e., \boxed{M(-1, 0)}.


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

Solution:
\textbf{Solution}: According to the problem, we have:\\
\[
\begin{cases}
y=\frac{3}{2}x-2 \\
y=-x+3
\end{cases}
\], solving this we get: \[
\begin{cases}
x=2 \\
y=1
\end{cases}
\], \\
$\therefore \boxed{P\left(2, 1\right)}$.


Question: As shown in the figure, the line $y=\frac{3}{2}x-2$ intersects with $y=-x+3$ at point P, and the line $y=\frac{3}{2}x-2$ intersects the x-axis and y-axis at points C and D respectively; the line $y=-x+3$ intersects the x-axis and y-axis at points A and B, respectively. Connect BC. What is the area of $\triangle BCD$?

Solution:
\textbf{Solution:} Substitute $x=0$ into the line $y=\frac{3}{2}x-2$ to get: $y=-2$,\\
thus $D(0, -2)$,\\
Substitute $x=0$ into the line $y=-x+3$ to get: $y=3$,\\
thus $B(0, 3)$,\\
Substitute $y=0$ into the line $y=\frac{3}{2}x-2$ to get: $x=\frac{4}{3}$,\\
thus $C\left(\frac{4}{3}, 0\right)$,\\
$\therefore BD=5, OC=\frac{4}{3}$,\\
$\therefore {S}_{\triangle BCD}=\frac{1}{2}BD\cdot OC=\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system, line $l_1$: $y=2x-1$ intersects the x-axis and y-axis at points $A$ and $B$, respectively. Line $l_2$: $y=-\frac{1}{2}x+1$ intersects the x-axis and y-axis at points $P$ and $C$, respectively. Connect $AC$, and lines $l_1$ and $l_2$ intersect at point $D$. Find the coordinates of point $D$, and directly write the solution set of the inequality $2x-1>-\frac{1}{2}x+1$.

Solution:
\textbf{Solution:} Solve the system of equations
\[
\begin{cases}
y=2x-1\\
y=-\frac{1}{2}x+1
\end{cases}
\]
to find
\[
\begin{cases}
x=\frac{4}{5}\\
y=\frac{3}{5}
\end{cases}
\]
$\therefore D\left(\frac{4}{5}, \frac{3}{5}\right)$

From the graph of the functions, it is known that the solution set for $2x-1>-\frac{1}{2}x+1$ is $\boxed{x>\frac{4}{5}}$;


Question: As shown in the diagram, in the Cartesian coordinate system, the line $l_1: y=2x-1$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and the line $l_2: y=-\frac{1}{2}x+1$ intersects the $x$-axis and the $y$-axis at points $P$ and $C$, respectively. Connecting $AC$, the lines $l_1$ and $l_2$ intersect at point $D$. What is the area of $\triangle ACD$?

Solution:
\textbf{Solution:} Given $y=2x-1$, let $y=0$, we find $x=\frac{1}{2}$, thus $A\left(\frac{1}{2}, 0\right)$,\\
let $x=0$, we find $y=-1$, thus $B\left(0, -1\right)$,\\
Given $y=-\frac{1}{2}x+1$, let $x=0$, we find $y=1$, thus $C\left(0, 1\right)$,\\
let $y=0$, we find $x=2$, thus $P\left(2, 0\right)$,\\
$\therefore BC=2, AO=\frac{1}{2}$,\\
$\therefore Area_{\triangle ACD}=Area_{\triangle BCD}-Area_{\triangle ABC}=\frac{1}{2}BC\times \left|x_{D}\right|-\frac{1}{2}BC\times OA=\frac{1}{2}\times 2\times \left(\frac{4}{5}-\frac{1}{2}\right)=\boxed{\frac{3}{10}}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=2x-1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and the line $l_{2}: y=-\frac{1}{2}x+1$ intersects the $x$-axis and $y$-axis at points $P$ and $C$, respectively. Connect $AC$, and lines $l_{1}$ and $l_{2}$ intersect at point $D$. If point $E$ is on line $l_{1}$ and $F$ is any point in the coordinate plane, investigate whether there exists a quadrilateral with $B$, $C$, $E$, $F$ as vertices which is a rectangle? Write down the coordinates of point $F$.

Solution:
\textbf{Solution:} Existence of coordinates for $F$ are either $\left(1, -1\right)$ or $\left(-\frac{4}{5}, -\frac{3}{5}\right)$,

(1) When $BC$ is the side of the rectangle, the quadrilateral $BCEF$ is a rectangle, \\
then $CE \perp BC$ \\
$\therefore CE \parallel x$-axis, $BF \parallel x$-axis \\
$\because C\left(0, 1\right)$, $B\left(0, -1\right)$ \\
Let $E\left(m, 1\right)$, $\because E$ is on line $l_1\colon y = 2x - 1$, \\
$\therefore 2m - 1 = 1$ \\
Solving, we get $m = 1$ \\
$\therefore x_F = 1$ \\
$\therefore F \left(1, -1\right)$; \\
(2) When $BC$ is the diagonal of the rectangle, \\
$\because OB = OC = 1$, $O$ is the intersection point of the diagonal, $C\left(0, 1\right), D\left(\frac{4}{5}, \frac{3}{5}\right), B\left(0, -1\right)$, \\
$\therefore BC=2, CD=\sqrt{\left(\frac{4}{5}\right)^2+\left(1-\frac{3}{5}\right)^2}=\frac{2\sqrt{5}}{5}, BD=\sqrt{\left(\frac{4}{5}\right)^2+\left(1+\frac{3}{5}\right)^2}=\frac{4\sqrt{5}}{5}$, \\
$\therefore CD^2+BD^2=\frac{4}{5}+\frac{16}{5}=4=BC^2$, \\
$\therefore \triangle BCD$ is a right triangle, with $\angle ADC = 90^\circ$ \\
$\therefore$ When $BC$ is the diagonal of rectangle $BECF$, points $D, E$ coincide, \\
$\therefore O$ is also the midpoint of $DF$, \\
Through central symmetry, we get $F\left(-\frac{4}{5}, -\frac{3}{5}\right)$ \\
In conclusion, the coordinates of $F$ are either $\boxed{\left(1, -1\right)}$ or $\boxed{\left(-\frac{4}{5}, -\frac{3}{5}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $l_{1}$ is $y=kx+b$ ($k\neq 0$), and the equation of line $l_{2}$ is $y=x$. Line $l_{1}$ passes through points $A(6, 0)$ and $B(0, 3)$, and line $l_{1}$ intersects with line $l_{2}$ at point C. What is the equation of line $l_{1}$?

Solution:
\textbf{Solution}: Because point $A(6, 0)$ and $B(0, 3)$ are on the line $y=kx+b$,\\
therefore $\begin{cases}6k+b=0\\ b=3\end{cases}$,\\
solving we get: $\begin{cases}k=-\frac{1}{2}\\ b=3\end{cases}$,\\
therefore the equation of line $l_1$ is: \boxed{y=-\frac{1}{2}x+3}.


Question: As shown in the figure, in the Cartesian coordinate system, the equation of line ${l}_{1}$ is $y=kx+b(k \ne 0)$, and the equation of line ${l}_{2}$ is $y=x$. Line ${l}_{1}$ passes through points $A(6, 0)$ and $B(0, 3)$, and line ${l}_{1}$ intersects line ${l}_{2}$ at point C. What is the area of $\triangle COB$?

Solution:
\textbf{Solution:} As shown in Figure 1, draw $CD\perp OB$ at point D through point C,\\
Solve $y=-\frac{1}{2}x+3$ and $y=x$ simultaneously,\\
$\begin{cases}y=-\frac{1}{2}x+3\\ y=x\end{cases}$,\\
to get $\begin{cases}x=2\\ y=2\end{cases}$.\\
Thus, the coordinates of the intersection point C of line $l_1$ and line $l_2$ are: (2, 2)\\
$\therefore CD=x_C=2$\\
Obviously, $OB=y_B=3$\\
$\therefore S_{\triangle COB}=\frac{1}{2}OB\cdot CD=\frac{1}{2}\times 3\times 2=\boxed{3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the equation of line $l_{1}$ is $y=kx+b \left(k \ne 0\right)$, and the equation of line $l_{2}$ is $y=x$. Line $l_{1}$ passes through point $A(6, 0)$ and $B(0, 3)$, and line $l_{1}$ intersects with line $l_{2}$ at point C. Is there a point P on the x-axis such that $\triangle POC$ is an isosceles triangle? Find the coordinates of point P.

Solution:
\textbf{Solution:} There exists a point $P$ on the $x$-axis such that $\triangle POC$ is an isosceles triangle. The reasoning is as follows:\\
Based on the problem statement, there are three scenarios:\\
(1) When point $O$ is the vertex angle, $OP=OC$. The two intersection points $P_1$ and $P_2$ of the circle centered at $O$ with radius $OC$ and the $x$-axis are the desired points $P$. It follows that $OC=2\sqrt{2}$.\\
$\therefore$ The coordinates of point $P_1$ are: $\left(-2\sqrt{2}, 0\right)$, and the coordinates of point $P_2$ are: $\left(2\sqrt{2}, 0\right)$.\\
(2) When point $C$ is the vertex angle, $CP=CO$. The intersection point $P_3$ of the circle centered at $C$ with radius $CO$ and the $x$-axis is the desired point $P$. It follows that the coordinates of point $P_3$ are: $\left(4, 0\right)$.\\
(3) When point $P$ is the vertex angle, the intersection point $P_4$ of the perpendicular bisector of $OC$ and the $x$-axis is the desired point $P$. It follows that the coordinates of point $P_4$ are: $\left(2, 0\right)$.\\
Combining (1), (2), and (3), it can be concluded that there are four points $P$ that form an isosceles triangle $\triangle POC$, which are: $\boxed{P_1\left(-2\sqrt{2}, 0\right)}$, $\boxed{P_2\left(2\sqrt{2}, 0\right)}$, $\boxed{P_3\left(4, 0\right)}$, $\boxed{P_4\left(2, 0\right)}$.


Question: As shown in the diagram, two vehicles depart simultaneously from locations A and B, respectively, along the same road, traveling at constant speeds. It is known that the sedan travels $20km$ per hour faster than the truck. After meeting, the two vehicles take a break for a while and then continue traveling at the same time. The relationship between the distance $y(km)$ between the two vehicles and the truck's travel time $x(h)$ is represented by the piecewise linear graph $AB-BC-CD-DE$. What are the respective speeds of the two vehicles in $km/h$?

Solution:
\textbf{Solution:} From the graph of the function, we learn that the distance between location A and B is 180 km. Let the speed of the truck be $x\,km/h$, and the speed of the sedan is $(x+20)\,km/h$. According to the problem, we have: $x + (x + 20) = 180$, solving this gives $x = \boxed{80}$,\\
thus, the speed of the truck is $80\,km/h$, and the speed of the sedan is $\boxed{100}\,km/h$.


Question: As shown in the figure, a truck is traveling from Place A to Place B, and a sedan is traveling from Place B to Place A. Both vehicles set off from their respective locations along the same road at the same time, moving at constant speeds. It is known that the sedan travels $20\,km$ per hour faster than the truck. After the vehicles meet, they rest for a while and then continue to travel at the same time. The relationship between the distance $y(km)$ separating the two vehicles and the travel time $x(h)$ of the truck is represented by the line segments $AB-BC-CD-DE$ in the figure. Based on the figure, answer the following questions: What is the function equation for segment $CD$? After how many hours from the truck's departure, is the distance between the truck and the sedan $20\,km$?

Solution:
\textbf{Solution:} Let the $x$-coordinate of point $D$ be $x$. Then we have:
\[80(x-1.5)+100(x-1.5)=144\]
Solving this gives $x=2.3$. Hence, the coordinates of point D are $(2.3, 144)$. Let the function of line segment $CD$ be $y=kx+b (k\ne 0)$. Then:
\[
\left\{
\begin{array}{l}
1.5k+b=0\\
2.3k+b=144
\end{array}
\right.
\]
Solving these equations gives:
\[
\left\{
\begin{array}{l}
k=180\\
b=-270
\end{array}
\right.
\]
Therefore, $y=180x-270$. When $180x-270=20$, solving for $x$ gives $x=\frac{29}{18}$. Let the equation for $AB$ be $y=mx+n (m\ne 0)$. Then:
\[
\left\{
\begin{array}{l}
n=180\\
m+n=0
\end{array}
\right.
\]
Solving this yields:
\[
\left\{
\begin{array}{l}
m=-180\\
n=180
\end{array}
\right.
\]
Therefore, the equation for line segment $AB$ is: $y=-180x+180$. When $-180x+180=20$, solving for $x$ gives $x=\frac{8}{9}$. Hence, the truck departs after $\boxed{\frac{8}{9}}$ hours or $\boxed{\frac{29}{18}}$ hours and will be $20km$ away from the car.


Question: As shown in the figure, the line $y_1 = x + 2$ intersects the x and y axes at points A and B, respectively. The line $y_2 = kx + 4$ intersects the x and y axes at points C and D, respectively. The two lines intersect at point $E\left(\frac{2}{3}, n\right)$. What is the value of $k$?

Solution:
\textbf{Solution:} Because the line $y_{1}=x+2$ intersects with the line $y_{2}=kx+4$ at point $E\left(\frac{2}{3}, n\right)$,\\
therefore, when $x= \frac{2}{3}$, $n= \frac{2}{3} +2= \frac{8}{3}$,\\
thus, the coordinates of point E are $\left( \frac{2}{3} , \frac{8}{3}\right)$. Substituting point E$\left( \frac{2}{3} , \frac{8}{3}\right)$ into $y_{2}=kx+4$, we get $ \frac{8}{3} = \frac{2}{3}k+4$. Solving this, we find $k=\boxed{-2}$,\\
therefore, $y_{2}=-2x+4$, which means the value of $k$ is $-2$.


Question: As shown in the figure, the line $y_1=x+2$ intersects the x-axis and y-axis at points A and B, respectively, and the line $y_2=kx+4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point $E\left(\frac{2}{3}, n\right)$. What is the area of $\triangle ACE$?

Solution: \textbf{Solution:} Given the line $y_1=x+2$, when $y=0$, we have $0=x+2$. Solving this, we get $x=-2$, hence the coordinates of point A are $(-2,0)$. For the line $y_2=-2x+4$, when $y=0$, we have $0=-2x+4$. Solving this, we get $x=2$, hence the coordinates of point C are $(2,0)$. Therefore, the length of $AC=2-(-2)=4$. As shown in the figure, draw $EH\perp AC$ at point H. Since the coordinates of point E are $E\left(\frac{2}{3}, \frac{8}{3}\right)$, it follows that $EH=\frac{8}{3}$. Therefore, the area of $\triangle ACE$ is $\frac{1}{2} AC\times EH=\frac{1}{2}\times 4\times \frac{8}{3}=\boxed{\frac{16}{3}}$.


Question: Place a rectangular sheet $OABC$ in the Cartesian coordinate system, where $O$ is the origin, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with $OA=9$ and $OC=15$. As shown in Figure 1, take a point $E$ on $OA$ and fold $\triangle EOC$ along $EC$, such that point $O$ falls on point $D$ on side $AB$. Find the equation of the line $EC$.

Solution: \textbf{Solution:} In rectangle $ABCO$, we have $AB=OC$, $OA=BC$, $\angle B=\angle BAO=\angle BCO=\angle EOC=90^\circ$, since $OA=9$, $OC=15$, therefore $AB=15$, $BC=9$. The coordinates of point $C$ are $(15,0)$. Based on the property of folding, we have: $DC=OC$, $EO=ED$, therefore $DC=15$. In $\triangle DBC$, $DB=\sqrt{DC^{2}-BC^{2}}=\sqrt{15^{2}-9^{2}}=12$. Therefore, $AD=AB-DB=15-12=3$. Since $OA=9$, $EO=ED$, thus $AE=OA-EO=9-ED$. In $\triangle AED$, we have $AE^{2}+AD^{2}=ED^{2}$, with $AD=3$, we get $(9-ED)^{2}+3^{2}=ED^{2}$. Solving this gives $DE=5$, therefore $EO=ED=5$. Thus, the coordinates of point $E$ are $(0,5)$. Let the equation of line $EC$ be $y=kx+b$. Substituting the coordinates of point $C$ as $(15,0)$ and point $E$ as $(0,5)$, we obtain $\begin{cases} 15k+b=0 \\ b=5 \end{cases}$. Solving this, we get $\begin{cases} k=-\frac{1}{3} \\ b=5 \end{cases}$, hence the equation of line $EC$ is \boxed{y=-\frac{1}{3}x+5}.


Question: As shown in the diagram, a rectangular paper $OABC$ is placed in the Cartesian coordinate system, where $O$ is the origin, point $C$ is on the $x$-axis, and point $A$ is on the $y$-axis, with $OA=9$ and $OC=15$. Under the given conditions, if point $T$ has the coordinates $(6,\frac{5}{2})$, and point $P$ is on the line $MN$, is there a point $Q$ on the coordinate axes such that the quadrilateral with vertices $M$, $D'$, $Q$, $P$ is a parallelogram? If such a point exists, please directly write the coordinates of point $Q$.

Solution:
\textbf{Solution:} There exists such a point Q, $\because$ in rectangle ABCO, $D'G \perp OC$, $\therefore$ quadrilateral $AD'GO$ is also a rectangle, $\therefore GD'=AO=9$, $AD'=OG$, $\because T(6, \frac{5}{2})$, $\therefore AD'=OG=6$, $TG=\frac{5}{2}$, $D'(6, 9)$, $\therefore TD'=GD'-TG=9-\frac{5}{2}=\frac{13}{2}$, $\therefore$ the side length of the rhombus $MD'TO$ is $\frac{13}{2}$, that is $OM=MD'=D'T=TO=\frac{13}{2}$, $\therefore$ the coordinates of point M are: $(0, \frac{13}{2})$.

Suppose the equation of the line MN is $y=kx+b$, with point T on line MN, substitute $M(0, \frac{13}{2})$, $T(6, \frac{5}{2})$ into it, yielding $\left\{\begin{array}{l}6k+b=\frac{5}{2}\\ b=\frac{13}{2}\end{array}\right.$, solve to get $\left\{\begin{array}{l}k=-\frac{2}{3}\\ b=\frac{13}{2}\end{array}\right.$, thus the equation of MN is $y=-\frac{2}{3}x+\frac{13}{2}$.

$\because$ Point P is on the line MN, and point Q is on the coordinate axis, $\therefore$ assume the coordinates of point P are $(x, -\frac{2}{3}x+\frac{13}{2})$,

When point Q is on the x-axis, let the coordinates of Q be $(a, 0)$, i.e.: $P(x, -\frac{2}{3}x+\frac{13}{2})$, $Q(a, 0)$, $M(0, \frac{13}{2})$, $D'(6, 9)$, these four points can form a parallelogram, knowing that the diagonals of a parallelogram bisect each other, it is solved that $x=6$, $a=0$, $\therefore$ at this time the coordinates of Q are $Q(0, 0)$; solved to get $x=\frac{27}{2}$, $a=\frac{39}{2}$, $\therefore$ at this time the coordinates of Q are $Q(\frac{39}{2}, 0)$; solved to get $x=-\frac{27}{2}$, $a=\frac{39}{2}$, $\therefore$ at this time the coordinates of Q are $Q(\frac{39}{2}, 0)$.

When point Q is on the y-axis, let the coordinates of Q be $(0, a)$, knowing that the diagonals of a parallelogram bisect each other, it is solved that $x=6$, $a=0$, $\therefore$ at this time the coordinates of Q are $Q(0, 0)$; solved to get $x=-6$, $a=13$, $\therefore$ at this time the coordinates of Q are $Q(0, 13)$; solved to get $x=6$, $a=13$, $\therefore$ at this time the coordinates of Q are $Q(0, 13)$.

In conclusion: The coordinates of the point Q that satisfy the conditions are: \boxed{Q(0, 0)}, \boxed{Q\left(\frac{39}{2}, 0\right)}, \boxed{Q(0, 13)}.


Question: Given: As shown in the figure, the linear function $y_{1}=kx-2$ intersects with the x-axis at point $B(-2, 0)$, $y_{2}=x+b$ intersects with the x-axis at point $C(4, 0)$, and the graphs of these two functions intersect at point $A$. Find the values of $k$ and $b$ and the coordinates of point $A$.

Solution:
\textbf{Solution:} Substitute $B(-2, 0)$ into $y_1=kx-2$ to get, $0=-2k-2$,\\
thus, $k=-1$;\\
Substitute $C(4, 0)$ into $y_2=x+b$ to get, $0=4+b$,\\
thus, $b=-4$;\\
Solving the system of equations yields, $\left\{\begin{array}{l}y=-x-2 \\ y=x-4\end{array}\right.$,\\
which gives, $\left\{\begin{array}{l}x=1 \\ y=-3\end{array}\right.$,\\
The coordinates of point A are: $A(1, -3)$.\\
Therefore, $k=\boxed{-1}$, $b=\boxed{-4}$, and the coordinates of point $A$ are $\boxed{(1, -3)}$.


Question: Given: As shown in the figure, the linear function $y_{1}=kx-2$ intersects the $x$-axis at point $B(-2, 0)$, and $y_{2}=x+b$ intersects the $x$-axis at point $C(4, 0)$. The graphs of these two functions intersect at point $A$. Connect $OA$. Is there a point $P$ on the line $y_{2}=x+b$ such that $S_{\triangle OCP}=\frac{1}{3}S_{\triangle OAC}$? If so, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Given that OC=4, and A(1, −3),

\[S_{\triangle OAC}=\frac{1}{2}\times 3\times 4=6,\]

\[S_{\triangle OCP}=\frac{1}{3}S_{\triangle OAC}=2,\]

Let the coordinates of point P be (x, y),

\[S_{\triangle OCP}=\frac{1}{2}OC\times |y|,\]

\[2=\frac{1}{2}\times 4\times |y|,\]

\[|y|=1,\]

When y=1, $1=x+b$,

x=5, the coordinates of point P are $\boxed{\left(5, 1\right)}$;

When y=−1, $−1=x+b$,

x=3, the coordinates of point P are $\boxed{\left(3, −1\right)}$.


Question: As shown in the figure, the linear function $y_1=kx-2$ intersects the x-axis at point $B(-2, 0)$, and $y_2=x+b$ intersects the x-axis at point $C(4, 0)$. The graphs of these two functions intersect at point $A$. Based on the graph, directly write the range of values for $x$ when $y_1\ge y_2$.

Solution:
\textbf{Solution:} From the graph, it is known that at point A or to the left of point A, ${y}_{1}\ge {y}_{2}$. \\
Hence, when $\boxed{x\leq 1, {y}_{1}\ge {y}_{2}}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the $x$ axis at point $A$, with $OA=4$, and intersects the graph of the direct proportion function $y=3x$ at point $B$, where the $x$-coordinate of point $B$ is 1. Find the equation of the linear function $y=kx+b$.

Solution:
\textbf{Solution:} Given that $OA=4$, we have $A(4,0)$. Substituting $x=1$ into $y=3x$, we get $y=3$, thus $B(1,3)$. Substituting $A(4,0)$ and $B(1,3)$ into $y=kx+b$, we have the system of equations
\[\left\{\begin{array}{l}
4k + b = 0 \\
k + b = 3
\end{array}\right.\]
Thus, we find that
\[\left\{\begin{array}{l}
k = -1 \\
b = 4
\end{array}\right.\]
Therefore, the linear equation is $\boxed{y = -x + 4}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the $x$ axis at point $A$, where $OA=4$, and intersects the graph of the direct proportion function $y=3x$ at point $B$, with the $x$-coordinate of point $B$ being 1. If point $C$ is on the $y$ axis, and $S_{\triangle BOC} = \frac{1}{2}S_{\triangle AOB}$, what are the coordinates of point $C$?

Solution:
\textbf{Solution:} Given $A(4, 0)$, $B(1, 3)$, \\
thus the area of $\triangle AOB = \frac{1}{2}OA\cdot y_B = \frac{1}{2}\times 4\times 3=6$, \\
thus the area of $\triangle BOC = \frac{1}{2}$ area of $\triangle AOB = $\frac{1}{2}\times 6=3$, \\
Since point $C$ is on the y-axis, let $C(0, y)$, \\
thus the area of $\triangle BOC = \frac{1}{2}|y|\cdot x_B = 3$, that is $\frac{1}{2}|y|\times 1=3$, solving this gives $y=\pm 6$, \\
thus the coordinates of point $C$ are $\boxed{(0, 6)}$ or $\boxed{(0, -6)}$;


Question: As shown in the diagram, the graph of the linear function $y=kx+b$ intersects the $x$-axis at point $A$, where $OA=4$, and intersects the graph of the direct proportion function $y=3x$ at point $B$, whose abscissa is 1. Please directly specify the range of $x$ for which $kx+b>3x$.

Solution: \textbf{Solution:} From the result of (1), we know the inequality is: $-x+4>3x$. Given $B(1, 3)$, when the graph of the linear function $y=-x+4$ is above the graph of the direct proportionality function $y=3x$, the range of the variable $x$ is: $\boxed{x<1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. Find the expressions for functions $y_1$ and $y_2$.

Solution:
\textbf{Solution:} Since the line $y_{1}=k_{1}x+b$ intersects the hyperbola $y_{2}=\frac{k_{2}}{x}$ at points $A(4, 2)$ and $B(m, -4)$,\\
we have $2=\frac{k_{2}}{4}$, solving which gives $k_{2}=8$,\\
therefore, the equation of the hyperbola $y_{2}$ is $y_{2}=\frac{8}{x}$,\\
subsequently, substituting $B(m, -4)$ into $y_{2}=\frac{8}{x}$, we get $-4=\frac{8}{m}$, solving which gives $m=-2$,\\
thus $B(-2, -4)$, substituting $A(4, 2)$ and $B(-2, -4)$ into $y_{1}=k_{1}x+b$ yields the system $\left\{\begin{array}{l}4k_{1}+b=2 \\ -2k_{1}+b=-4\end{array}\right.$, from which we solve $\left\{\begin{array}{l}k_{1}=1 \\ b=-2\end{array}\right.$,\\
therefore, the equation of the line $y_{1}$ is: \boxed{y_{1}=x-2}; and the equation of the hyperbola $y_{2}$ is: \boxed{y_{2}=\frac{8}{x}}.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. A line is drawn from point B perpendicular to the x-axis intersecting it at point P. What is the area of $\triangle ABP$?

Solution: \textbf{Solution:} Given $\because A(4, 2)$, $B(-2, -4)$, $\therefore BP=4$, $\therefore {S}_{\triangle ABP}=\frac{1}{2}\times 4\times (4+2)=\boxed{12}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}$ at points $A(4, 2)$ and $B(m, -4)$. Based on the graph of the function, directly write down the solution set of the inequality $k_1x+b>\frac{k_2}{x}$ in terms of $x$.

Solution: \textbf{Solution:} Observing the graph, the solution set of the inequality ${k}_{1}x+b>\frac{{k}_{2}}{x}$ with respect to $x$ is \boxed{-2<x<0} or \boxed{x>4}.


Question: As shown in the figure, the line $l_{1} \colon y_{1} = -2x + 6$ intersects the x-axis and the y-axis at points A and B, respectively. The line $l_{2}$ passes through the point C $(-5, 0)$ and intersects the line $l_{1}$ at the point D $(a, 8)$, and intersects the y-axis at the point E. Find the analytical expression of the line $l_{2}$?

Solution:
\textbf{Solution:} Given that line $l_{1}$ passes through point $D(-1,8)$,

\[
\therefore 8 = -2a + 6,
\]
\[
\therefore a = -1,
\]
\[
\therefore D(-1,8),
\]
Given that line $l_{2}$ passes through points $C(-5,0)$ and $D(-1,8)$,

\[
\therefore \begin{cases}
-5k + b = 0 \\
-k + b = 8
\end{cases},
\]
solving this, we get
\[
\begin{cases}
k = 2 \\
b = 10
\end{cases},
\]
\[
\therefore \text{the equation of line}\ l_{2}\ \text{is}\ \boxed{y = 2x + 10};
\]


Question: As shown in the figure, the line $l_{1}\colon y_{1}=-2x+6$ intersects the x-axis at point A and the y-axis at point B. The line $l_{2}$ passes through point C ($-5$, $0$) and intersects line $l_{1}$ at point D ($a$, $8$) and the y-axis at point E. What is the area of $\triangle BDE$?

Solution:
Solution: In $y=-2x+6$, let $x=0$, then $y=6$,\\
$\therefore B(0,6)$,\\
In $y=2x+10$, let $x=0$, then $y=10$,\\
$\therefore E(0,10)$,\\
$\therefore BE=10-6=4$,\\
$\therefore$ The area of $\triangle BDE$ is $\frac{1}{2} \times 4 \times 1 = \boxed{2}$.


Question: As shown in the figure, the line $y_{1}=x+1$ intersects the x-axis and y-axis at points A and B, respectively, and the line $y_{2}=-2x+4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point E. What are the coordinates of point E?

Solution:
\textbf{Solution:} Since $\begin{cases}y_1=x+1\\ y_2=-2x+4\end{cases}$,\\
we have $\begin{cases}x=1\\ y=2\end{cases}$,\\
thus, the coordinates of point E are $\boxed{(1,2)}$.


Question: As shown in the figure, the line $y_1 = x + 1$ intersects the x-axis and y-axis at points A and B, respectively. The line $y_2 = -2x + 4$ intersects the x-axis and y-axis at points C and D, respectively. The two lines intersect at point E. What is the area of $\triangle ACE$?

Solution: \textbf{Solution:} Solve when $y_1=x+1=0$, we get: $x=-1$, $\therefore A(-1,0)$. When $y_2=-2x+4=0$, we get: $x=2$, $\therefore C(2,0)$, $\therefore AC=2-(-1)=3$, $S_{\triangle ACE}=\frac{1}{2}\times AC\times y_E =\frac{1}{2}\times 3\times 2 =\boxed{3}.$


Question: As shown in the figure, the line $y_1 = x+1$ intersects the x-axis and the y-axis at points A and B, respectively, and the line $y_2 = -2x+4$ intersects the x-axis and the y-axis at points C and D, respectively. The two lines intersect at point E. Based on the figure, answer directly: For what values of $x$ is $y_1 < y_2$?

Solution: \textbf{Solution:} Given $E(1, 2),$ when $y_1<y_2$, the graph of $y_1=x+1$ is below the graph of $y_2=-2x+4$. By analyzing the graphs, we can conclude: $\boxed{x<1}.$


Question: As shown in the diagram, the linear function $y=x+2$ intersects the x-axis and the y-axis at points A and B, respectively. Point $M\left(1, m\right)$ is a point on line AB, and line MC intersects the x-axis at point $C\left(\frac{5}{2}, 0\right)$. Find the analytical expression of line MC.

Solution: \textbf{Solution:} Substituting the point $M\left(1, m\right)$ into $y=x+2$ yields: $m=1+2=3$, $\therefore$ the point $M(1, 3)$. Let the equation of line $MC$ be $y=kx+b\left(k\ne 0\right)$. Substituting the points $M(1, 3)$ and $C\left(\frac{5}{2}, 0\right)$ into it gives: $\begin{cases}\frac{5}{2}k+b=0 \\ k+b=3\end{cases}$, solving this system yields: $\begin{cases}k=-2 \\ b=5\end{cases}$, $\therefore$ the equation of the line $MC$ is $\boxed{y=-2x+5}$.


Question: As shown in the figure, a linear function $y=x+2$ intersects the x-axis and y-axis at points A and B, respectively. Point $M(1, m)$ is a point on line segment AB, and line segment MC intersects the x-axis at point $C\left(\frac{5}{2}, 0\right)$. If point P is a moving point on line segment AC, and BP, MP are connected, and if the area of $\triangle ABP$ is twice the area of $\triangle MPC$, find the coordinates of point P.

Solution: \textbf{Solution:} For $y=x+2$, when $x=0$, then $y=2$; when $y=0$, then $x=-2$. Therefore, points $A(-2, 0)$ and $B(0, 2)$ exist, and therefore $OA=OB=2$. Let point $P(a, 0)$, then $AP=a+2$, $PC=\frac{5}{2}-a$. Since the area of $\triangle ABP$ is twice the area of $\triangle MPC$, it follows that $\frac{1}{2}(a+2)\times 2=2\times \frac{1}{2}\left(\frac{5}{2}-a\right)\times 3$. Solving this, we find $a=\frac{11}{8}$. Therefore, the coordinates of point $P$ are $\boxed{\left(\frac{11}{8}, 0\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=-\frac{3}{4}x+m$ intersects the x-axis and y-axis at points B and A, respectively, with the coordinates of point B being $(12,0)$. The line $y=\frac{3}{8}x$ intersects with line AB at point C. Find the coordinates of point A.

Solution:
\textbf{Solution:} Given the line $y=-\frac{3}{4}x+m$, intersects the y-axis at point B$(12,0)$, \\
$\therefore$ $0=-\frac{3}{4}\times 12+m$, \\
$\therefore$ $m=9$, \\
$\therefore$ The equation of line AB is: $y=-\frac{3}{4}x+9$, \\
When $x=0$, $y=9$, \\
$\therefore$ The coordinates of point A are $\boxed{(0,9)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\frac{3}{4}x+m$ intersects with the x-axis and y-axis at points B and A, respectively, where the coordinate of point B is $(12, 0)$. The line $y=\frac{3}{8}x$ intersects with line AB at point C. What is the area of $\triangle BOC$?

Solution:
\textbf{Solution:} According to the problem, we have:
\[
\left\{
\begin{array}{l}
y=\frac{3}{8}x \\
y=-\frac{3}{4}x+9
\end{array}
\right.
\]
Solving this, we get:
\[
\left\{
\begin{array}{l}
x=8 \\
y=3
\end{array}
\right.
\]
Therefore, point C is at (8, 3), and hence the area of $\triangle BOC$ is $\frac{1}{2} \times 12 \times 3 = \boxed{18}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\frac{3}{4}x+m$ intersects the x-axis at point B and the y-axis at point A, where the coordinate of point B is (12, 0). The line $y=\frac{3}{8}x$ intersects line AB at point C. Point D is a moving point on line AB, and a perpendicular line is drawn from point D to the x-axis, intersecting line OC at point E. Let the x-coordinate of point D be $t$, and the length of segment DE be $d$. Find the functional formula of $d$ in terms of $t$ (including the range of the independent variable).

Solution:
\textbf{Solution:}

(1) As shown in the figure,\\
since the x-coordinate of point D is $t$, hence point D$(t, -\frac{3}{4}t+9)$, point E$(t, \frac{3}{8}t)$,\\
when $t<8$, $d= -\frac{3}{4}t+9-\frac{3}{8}t=-\frac{9}{8}t+9$,\\
when $t>8$, $d= \frac{3}{8}t+\frac{3}{4}t-9=\frac{9}{8}t-9$;\\
(2) since the line segment with endpoints H$(\frac{1}{2}, t)$ and G$(1, t)$ intersects with the edge of the square DEPQ at only one point,\\
hence $\frac{1}{2}\leq t \leq 1$ or $\left\{\begin{array}{l}
-\frac{9}{8}t+9\le t-\frac{1}{2} \\
-\frac{9}{8}t+9\ge t-1
\end{array}\right.$,\\
hence $\frac{1}{2}\leq t \leq 1$ or \boxed{\frac{76}{17}\le t \leq \frac{80}{17}}.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2, 6)$, intersects the $x$ axis at point $B$, and intersects the graph of the direct proportion function $y=3x$ at point $C$, where the $x$-coordinate of point $C$ is $1$. Find the values of $k$ and $b$?

Solution:
\textbf{Solution:} When $x=1$, $y=3x=3$. Therefore, the coordinates of point $C$ are $(1, 3)$. Since the line $y=kx+b$ passes through $(-2, 6)$ and $(1, 3)$, we have
\[
\begin{cases}
6=-2k+b \\
3=k+b
\end{cases}
\]
Solving this system of equations, we find
\[
\begin{cases}
k=\boxed{-1} \\
b=\boxed{4}
\end{cases}
\]
Therefore, the explicit formula for the linear function is $y=-x+4$, which means $k=-1$, $b=4$.


Question: As shown in the graph, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through point $A(-2,6)$, and intersects with the $x$-axis at point $B$, and intersects with the graph of the direct proportion function $y=3x$ at point $C$, where the abscissa of point $C$ is $1$. Please directly write down the solution set of the inequality $kx+b-3x>0$.

Solution: \textbf{Solution:} From (1), we know that the linear function $y=kx+b$ is $y=-x+4$, i.e., the inequality $kx+b-3x>0$ is transformed into: the inequality $-x+4>3x$. According to the graph of the function, the solution set of the inequality $-x+4>3x$ is $\boxed{x<1}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ passes through the point $A(-2,6)$ and intersects the $x$-axis at point $B$, and intersects the graph of the direct proportion function $y=3x$ at point $C$, with the $x$-coordinate of point $C$ being $1$. If point $D$ is on the $y$-axis, and the area of $\triangle BCD$ is twice the area of $\triangle BOC$, find the coordinates of point $D$.

Solution:
\textbf{Solution:} From (1), we know the linear function $y=kx+b$ is $y=-x+4$. When $y=0$, we have $0=-x+4$, solving this gives $x=4$, $\therefore B(4, 0)$, $\therefore S_{\triangle BCO}=\frac{1}{2}\times 4\times 3=6$, $\because S_{\triangle BCD}=2S_{\triangle BOC}$, $\therefore S_{\triangle BCD}=12$. Assume the coordinates of point D are (0, m), when point D is on the positive y-axis,\\
$S_{\triangle DBC}=S_{\triangle DOB}-S_{\triangle DCO}-S_{\triangle BOC}=\frac{1}{2}\times 4m-\frac{1}{2}\times 1\times m-6=\frac{3}{2}m-6$, $\because S_{\triangle BCD}=12$,\\
$\therefore \frac{3}{2}m-6=6\times 2$, $\therefore m=12$, $\therefore$ the coordinates of point D are $\boxed{(0,12)}$; when point D is on the negative y-axis,\\
$S_{\triangle DBC}=S_{\triangle DOB}-S_{\triangle DCO}+S_{\triangle BOC}=\frac{1}{2}\times 4\times (-m)+6-\frac{1}{2}\times 1\times (-m)=-\frac{3}{2}m+6$, $\because S_{\triangle BCD}=12$,\\
$\therefore -\frac{3}{2}m+6=6\times 2$, $\therefore m=-4$, $\therefore$ the coordinates of point D are $\boxed{(0,-4)}$,\\
In conclusion, the coordinates of point D are $\boxed{(0,12)}$ or $\boxed{(0,-4)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $AB: y=-\frac{1}{2}x+3$ and the line $CD: y=kx-2$ intersect at point $M(4, a)$, and intersect the coordinate axes at points A, B, C, and D, respectively. What are the values of $a$ and $k$?

Solution:
\textbf{Solution:} Let's substitute the coordinates of point M into $y=-\frac{1}{2}x+3$ and solve to get: $a=1$,\\
$\therefore$ for point $M(4, 1)$, substituting $M(4, 1)$ into $y=kx-2$, we get $4k-2=1$, solving this yields: $k=\frac{3}{4}$,\\
$\therefore a=\boxed{1}$, $k=\boxed{\frac{3}{4}}$;


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the line \(AB: y=-\frac{1}{2}x+3\) and the line \(CD: y=kx-2\) intersect at point \(M(4, a)\), intersecting the coordinate axes at points A, B, C, and D, respectively. As depicted, point \(P\) is a moving point on line \(CD\). When the area of triangle \(PBM\) is \(20\), what are the coordinates of point \(P\)?

Solution: \textbf{Solution:} Let $P\left(m, \frac{3}{4}m-2\right)$. The equation of line CD is: $y=\frac{3}{4}x-2$. Setting $x=0$, we find $y=-2$, thus point $D\left(0, -2\right)$. For line AB: $y=-\frac{1}{2}x+3$, setting $x=0$, we obtain $y=3$, thus point $B\left(0, 3\right)$, hence $BD=5$. Therefore, the area of $\triangle PBM$ equals $\frac{1}{2}\times BD\times \left|{x}_{M}-m\right|=\frac{1}{2}\times 5\times \left|4-m\right|=20$, therefore $\left|4-m\right|=8$. Solving this results in: $m=-4$ or $m=12$. In conclusion, the points are $P\left(-4, -5\right)$ or $P\left(12, 7\right)$; therefore, the answer is $\boxed{P\left(-4, -5\right)}$ or $\boxed{P\left(12, 7\right)}$.


Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and intersects with $y=2x$ at point $A(a,2)$. Point $M$ is on line $OA$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Since point A$(1,2)$ lies on the line $y=2x$, it follows that $2=2 \cdot 1$, hence $a=1$. Therefore, A$(1,2)$. Since the line $y=kx+b$ passes through $C(0,3)$ and $A(1,2)$, it follows that $\begin{cases} 3=0\cdot k+b \\ 2=k+b \end{cases}$. \\
Solving these equations, we get $\begin{cases} k=-1 \\ b=3 \end{cases}$. Therefore, the equation of line $AB$ is \boxed{y=-x+3};


Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $B$ and the $y$-axis at point $C(0,3)$, and intersects $y=2x$ at point $A(a,2)$. Point $M$ is on the line $OA$. What is the area of $\triangle OAB$?

Solution: \textbf{Solution:} Let $y=0$, we have $-x+3=0$. Solving this, we obtain: $x=3$, $\therefore B(3,0)$, $\therefore OB=3$, $\therefore$ the area of $\triangle OAB$ is $=\frac{1}{2}\times 3\times 2=\boxed{3}$.