Question: As shown in the plane Cartesian coordinate system, line $L_1\colon y=-\frac{1}{2}x+8$ intersects the $x$-axis and the $y$-axis at points $B$ and $C$, respectively, and intersects line $L_2\colon y=x$ at point $A$. If $D$ is a point on line $L_2$, and the area of $\triangle BOD$ is $32$, find the function expression of line $CD$.

Solution:
\textbf{Solution:} Given the line $L_1: y=-\frac{1}{2}x+8$ intersects with the x-axis at point $B$ and with the y-axis at point $C$, \\
$\therefore B(16, 0)$, $C(0, 8)$, \\
$\therefore OB=16$, \\
then the height of $\triangle BOD$ is: $32\times 2\div16=4$, \\
$\because D$ is a point on the line $L_2$, \\
$\therefore$ Coordinates of point D are: $(4, 4)$ or $(-4, -4)$, \\
Let the equation of line $CD$ be: $y=kx+b$, \\
Then $\left\{\begin{array}{l} b=8 \\ 4=4k+b \end{array}\right.$ or $\left\{\begin{array}{l} b=8 \\ -4=-4k+b \end{array}\right.$, \\
Solving, we get: $\left\{\begin{array}{l} k=-1 \\ b=8 \end{array}\right.$ or $\left\{\begin{array}{l} k=3 \\ b=8 \end{array}\right.$, \\
$\therefore$ The function expression of line $CD$ is: \boxed{y=-x+8} or \boxed{y=3x+8}.


Question: As shown in Figure 1, the coordinates of point C are $(4\sqrt{2}, 4\sqrt{2})$. Perpendiculars are drawn from point C to the x-axis and y-axis, meeting at points B and D, respectively. Point E is on the line segment OD (not coinciding with points O or D). Connect BE, construct the symmetric point $O'$ of point O with respect to line BE, connect $CO'$, point P is the midpoint of $CO'$, connect BP, extend $CO'$ to intersect the extended line of BE at point F, connect DF. $\angle PBF=45^\circ$; As shown in Figure 2, connect BD, when point $O'$ just falls on the line segment BD, what is the equation of line BF?

Solution:
\textbf{Solution:}

Connect $EO'$, as shown in the figure: \\
In $\triangle BOD$, we have $OB=OD=4\sqrt{2}$, \\
$\therefore BD=\sqrt{OB^2+OD^2}=8$, \\
$\because$ point O is the symmetric point of line BE with respect to O', \\
$\therefore OE=O'E$, $O'B=OB=4\sqrt{2}$, $\angle EO'B=\angle EOB=90^\circ$, \\
$\therefore \angle DOE=90^\circ$, $DO'=BD-O'B=8-4\sqrt{2}$, \\
Let $OE=O'E=x$, then $DE=4\sqrt{2}-x$, \\
In $\triangle DOE$, we have $DO'^2+O'E^2=DE^2$, \\
$\therefore (8-4\sqrt{2})^2+x^2=(4\sqrt{2}-x)^2$, \\
Solving for $x$ gives $x=8-4\sqrt{2}$, \\
$\therefore E(0,8-4\sqrt{2})$, \\
Let the equation of line BF be $y=kx+b$. Substituting $B(4\sqrt{2},0)$ and $E(0,8-4\sqrt{2})$ into this gives: \\
$\begin{cases} 0=4\sqrt{2}k+b \\ 8-4\sqrt{2}=b \end{cases}$, solving yields $\begin{cases} k=1-\sqrt{2} \\ b=8-4\sqrt{2} \end{cases}$, \\
$\therefore$ The equation of line BF is $\boxed{y=(1-\sqrt{2})x+8-4\sqrt{2}}$.


Question: As shown in Figure 1, it is known that the coordinates of point C are $(4\sqrt{2}, 4\sqrt{2})$. Perpendiculars are drawn from point C to the x-axis and y-axis, with the feet of the perpendiculars being points B and D, respectively. Point E is a point on line segment OD (not coinciding with points O or D). Line segment BE is connected. The symmetric point of O with respect to line BE is constructed as \(O'\). Line segment \(CO'\) is drawn, and point P is the midpoint of \(CO'\). Line segment BP is connected. Line \(CO'\) is extended to intersect the extended line of BE at point F. Line segment DF is connected. Is there a point M in the plane such that the quadrilateral formed by vertices M, O, \(O'\), and F is a parallelogram? If it exists, please write down the coordinates of point M directly; if it does not exist, please explain the reason.

Solution: \textbf{Solution:} Yes, it exists. The coordinates of $M$ are \(\boxed{\left(6\sqrt{2}-8, 4+2\sqrt{2}\right)}\), \(\boxed{\left(-2\sqrt{2}, 2\sqrt{2}-4\right)}\), or \(\boxed{\left(2\sqrt{2}, 4-2\sqrt{2}\right)}\).


Question: As shown in the figure, the line \(l_{1}\) intersects the y-axis at point A in the Cartesian coordinate system, and point B \((-3, 3)\) is also on line \(l_{1}\). By moving point B 1 unit to the right and then 2 units downward, we obtain point C, which is also on line \(l_{1}\). Determine the coordinates of point C and the equation of line \(l_{1}\).

Solution:
\textbf{Solution:} By the translation rule, the coordinates of point C are $(-3+1, 3-2)$, i.e., $(-2, 1)$.\\
Let the equation of line $l_{1}$ be $y=kx+c$,\\
then $\begin{cases}3=-3k+c\\ 1=-2k+c\end{cases}$, solving this gives: $\begin{cases}k=-2\\ c=-3\end{cases}$,\\
$\therefore$ the equation of line $l_{1}$ is $y=-2x-3$.\\
Therefore, the coordinates of point C are $\boxed{(-2, 1)}$ and the equation of line $l_{1}$ is $\boxed{y=-2x-3}$.


Question: As shown in the figure, line $l_{1}$ intersects the y-axis at point A in the Cartesian coordinate system, and point B$(-3, 3)$ is also on line $l_{1}$. Point B is translated 1 unit to the right and then 2 units down to get point C, which is also on line $l_{1}$. It is known that line $l_{2}: y=x+b$ passes through point B and intersects the y-axis at point E. Find the area of $\triangle ABE$.

Solution:
\textbf{Solution:} Substitute the coordinates of point B into $y=x+b$ to get,\\
$3=-3+b$, solving for $b$ gives $b=6$,\\
$\therefore y=x+6$.\\
When $x=0$, $y=6$,\\
$\therefore$ the coordinates of point E are $(0,6)$.\\
When $x=0$, $y=-3$,\\
$\therefore$ the coordinates of point A are $(0,-3)$,\\
$\therefore AE=6+3=9$,\\
$\therefore$ the area of $\triangle ABE$ is $\frac{1}{2} \times 9 \times |-3| = \boxed{13.5}$.


Question: As shown in the diagram, in the Cartesian coordinate system, points $E$, $F$, $G$ are on the sides of rectangle $ABCO$. When $\triangle EFO$ is folded along $EF$, point $O$ coincides exactly with point $G$, $GH\perp x$-axis at point $H$, and point $M$ is the intersection of $GH$ and $EF$. If $CG=2$, and $B(6,4)$, find the coordinates of point $F$.

Solution:
\textbf{Solution:} Let $OF=x$, then $CF=4-x$, and $GF=x$,\\
In $\triangle CGF$, by the Pythagorean theorem, we have\\
$CF^2+CG^2=GF^2$,\\
$\therefore (4-x)^2+2^2=x^2$, solving this gives $x=\frac{5}{2}$,\\
$\therefore F\boxed{\left(0,\frac{5}{2}\right)}$.


Question: As shown in the diagram, within a Cartesian coordinate system, points $E$, $F$, and $G$ are located on the sides of rectangle $ABCO$. When triangle $EFO$ is folded along $EF$, point $O$ coincides exactly with point $G$. $GH$ is perpendicular to the x-axis at point $H$, and point $M$ is the intersection of $GH$ and $EF$. Given that $CG=2$ and $B(6,4)$, determine the analytical expression of line $EF$.

Solution:
\textbf{Solution:} Since $GH \perp x$-axis, and $OC \perp x$-axis,\\
therefore $GH \parallel OC$, thus $\angle OFE = \angle GMF$.\\
Since $\angle OFE = \angle GFE$, therefore $\angle GFM = \angle GMF$,\\
thus $GF = GM = \frac{5}{2}$, and $HM = 4 - \frac{5}{2} = \frac{3}{2}$.\\
Therefore $M(2, \frac{3}{2})$.\\
Let the equation of line $EF$ be $y = kx + b$.\\
Therefore $\left\{ \begin{array}{l} b = \frac{5}{2}, \\ 2k + b = \frac{3}{2}, \end{array} \right.$ thus $\left\{ \begin{array}{l} k = -\frac{1}{2}, \\ b = \frac{5}{2}. \end{array} \right.$\\
Thus, the equation of line $EF$ is $y = -\frac{1}{2}x + \frac{5}{2} = \boxed{-\frac{1}{2}x + \frac{5}{2}}$.


Question: As shown in the figure, point A is a point in the Cartesian coordinate system with coordinates (1, $\sqrt{3}$). Line AB passes through point A, and point B is the intersection of the line with the x-axis. A parallelogram AOBC is formed with OA and OB as adjacent sides. If OD bisects $\angle AOB$ and D is the midpoint of AC. What are the coordinates of points B and D?

Solution:
\textbf{Solution:} Given that quadrilateral AOBC is a parallelogram, and OD is the bisector of $\angle AOB$,\\
thus $\angle ADO = \angle BOD$, $\angle AOD = \angle BOD$,\\
thus $\angle ADO = \angle AOD$,\\
thus $OA = AD$.\\
Since the coordinates of point A are $(1, \sqrt{3})$, and D is the midpoint of AC,\\
thus $OA = \sqrt{1^2+(\sqrt{3})^2} = 2$,\\
thus $AD = 2$, $AC = 4$,\\
thus $OB = AC = 4$,\\
thus the coordinates of point B are $\boxed{(4, 0)}$, and the coordinates of point D are $\boxed{(3, \sqrt{3})}$.


Question: As illustrated, point $A$ in the Cartesian coordinate system has coordinates $(1, \sqrt{3})$. Line $AB$ passes through point $A$, and point $B$ is the intersection of the line with the $x$-axis. A parallelogram $AOBC$ is formed with $OA$ and $OB$ as adjacent sides. If $OD$ is the bisector of $\angle AOB$, and $D$ is the midpoint of $AC$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Let the equation of the line AB be $y=kx+b$. Given that $A(1, \sqrt{3})$ and $B(4,0)$, according to the problem, we get
\[
\left\{
\begin{array}{l}
\sqrt{3}=k+b \\
0=4k+b
\end{array}
\right.
\]
Solving this, we find
\[
\left\{
\begin{array}{l}
k=-\frac{\sqrt{3}}{3} \\
b=\frac{4\sqrt{3}}{3}
\end{array}
\right.
\]
$\therefore$ The equation of the line AB is $\boxed{y=-\frac{\sqrt{3}}{3}x+\frac{4\sqrt{3}}{3}}$.


Question: As shown in the figure, point $A$ is located at $(1, \sqrt{3})$ in the Cartesian coordinate system. $AB$ is a line passing through point $A$, and $B$ is the intersection of this line with the $x$-axis. A parallelogram $AOBC$ is formed using $OA$ and $OB$ as adjacent sides. If $OD$ is the bisector of $\angle AOB$ and $D$ is the midpoint of $AC$. If $P$ is a moving point on the line $AB$, and $S_{\triangle POD} = \frac{1}{2} S_{\text{parallelogram} AOBC}$, please directly write down the coordinates of the point $P$ that satisfies the condition.

Solution:
\textbf{Solution:}

(1) Let the intersection of AB and the y-axis be M, and the intersection with OD be E, connect MD and BD. Set $x=0$, then $y=-\frac{\sqrt{3}}{3}x+\frac{4\sqrt{3}}{3}=\frac{4\sqrt{3}}{3}$.\\
Therefore, the intersection M of line AB and the y-axis is $\left(0,\frac{4\sqrt{3}}{3}\right)$, therefore MO$=\frac{4\sqrt{3}}{3}$. Let the equation of OD be $y=mx$, substitute the coordinates of point D $\left(3,\sqrt{3}\right)$ to get $m=\frac{3}{\sqrt{3}}=\sqrt{3}$. Therefore, the equation of OD is $y=\sqrt{3}x$, by solving the system
\[
\begin{cases}
y=\sqrt{3}x\\
y=-\frac{\sqrt{3}}{3}x+\frac{4\sqrt{3}}{3}
\end{cases}
\]
we get
\[
\begin{cases}
x=2\\
y=\frac{2\sqrt{3}}{3}
\end{cases}
\]
Therefore, the coordinates of point E are $\left(2,\frac{2\sqrt{3}}{3}\right)$, therefore OE$=\sqrt{2^2+\left(\frac{2\sqrt{3}}{3}\right)^2}=\frac{4\sqrt{3}}{3}$, ME$=\sqrt{(2-0)^2+\left(\frac{2\sqrt{3}}{3}-\frac{4\sqrt{3}}{3}\right)^2}=\frac{4\sqrt{3}}{3}$, AM$=\sqrt{(1-0)^2+\left(\sqrt{3}-\frac{4\sqrt{3}}{3}\right)^2}=\frac{2\sqrt{3}}{3}$, AE$=\sqrt{(2-1)^2+\left(\frac{2\sqrt{3}}{3}-\sqrt{3}\right)^2}=\frac{2\sqrt{3}}{3}$.\\
Therefore, OM=OE=EM, AE=AM$=\frac{2\sqrt{3}}{3}$, therefore $S_{\triangle OAM}$=$S_{\triangle OAE}$, $S_{\triangle AMD}$=$S_{\triangle AED}$, therefore $S_{\triangle ODM}$=$S_{\triangle OAM}$+$S_{\triangle OAE}$+$S_{\triangle AMD}$+$S_{\triangle AED}$=2$S_{\triangle OAD}$=$\frac{1}{2}S_{\text{parallelogram }AOBC}$. Therefore P and M coincide, so the coordinates of point P are $\boxed{\left(0,\frac{4\sqrt{3}}{3}\right)}$;\\
(2) Because point D is the midpoint of AC, therefore $S_{\triangle OBD}$=$\frac{1}{2}S_{\text{parallelogram }AOBC}$, therefore P coincides with B, so the coordinates of point P are $\boxed{\left(4,0\right)}$.\\
Therefore, the coordinates of point P are $\boxed{\left(0,\frac{4\sqrt{3}}{3}\right)}$ or $\boxed{\left(4,0\right)}$.


Question: In the Cartesian coordinate system, the coordinates of point A are $(6, 4)$. Through point A, lines $AB \perp x$-axis at point B and $AC \perp y$-axis at point C are drawn, respectively. The graph of the linear function $y = kx + b$ passes through point $P(2, 3)$. Express $b$ in an algebraic formula involving $k$.

Solution:
\textbf{Solution}: Substitute point P$(2,3)$ into $y=kx+b$, we get $3=2k+b$, \\
$\therefore b=\boxed{-2k+3}$.


Question: In the Cartesian coordinate system, the coordinates of point A are $(6,4)$. Lines $AB$ and $AC$ are drawn from point A and are perpendicular to the x-axis at point B and to the y-axis at point C, respectively. The graph of the linear function $y=kx+b$ passes through point $P(2,3)$. When $k=2$, how long is the segment intercepted by the line $y=kx+b$ on the rectangle $OBAC$?

Solution: \textbf{Solution:} For the line $y=kx+b$, the length of the segment intercepted by the rectangle $OBAC$ is $\boxed{2\sqrt{5}}$.


Question: In the Cartesian coordinate system, the coordinates of point $A$ are $(6, 4)$. The lines $AB$ and $AC$ are drawn from point $A$ perpendicular to the x-axis and y-axis at points $B$ and $C$ respectively. The graph of the linear function $y=kx+b$ passes through the point $P(2, 3)$. When $1\le x\le 5$, the function value $y$ satisfies $-1\le y\le 4$. Determine the range of values for $k$.

Solution:
\textbf{Solution:} Given the problem statement, when $k>0$, $y$ increases as $x$ increases. When $x=1$, we have ${y}_{\min}=k-2k+3=3-k$. When $x=5$, we have ${y}_{\max}=5k-2k+3=3k+3$. From the given conditions, we have
\[
\left\{
\begin{array}{l}
3-k\geq-1, \\
3k+3\leq4.
\end{array}
\right.
\]
Solving this, we find $k\leq\frac{1}{3}$. Therefore, $0<k\leq\frac{1}{3}$.

When $k<0$, $y$ decreases as $x$ increases. When $x=1$, we have ${y}_{\max}=k-2k+3=3-k$. When $x=5$, we have ${y}_{\min}=5k-2k+3=3k+3$. From the given conditions, we have
\[
\left\{
\begin{array}{l}
3k+3\geq-1, \\
3-k\leq4.
\end{array}
\right.
\]
Solving this, we find $k\geq-1$. Therefore, $-1\leq k<0$.

Therefore, combining the two cases, the range of values for $k$ is: $-1\leq k<0$, $0<k\leq\frac{1}{3}$. Hence, the solution is the range of values for $k$: \boxed{-1\leq k\leq\frac{1}{3}}.


Question: In the Cartesian coordinate system, the coordinate of point $A$ is $(6, 4)$. Line segments $AB$ and $AC$ are drawn, perpendicular to the x-axis at point $B$ and to the y-axis at point $C$, respectively. The graph of the linear function $y=kx+b$ passes through point $P(2, 3)$. When the line $y=kx+b$ ($k<0$) divides the rectangle $OBAC$ into two parts with an area ratio of $3:5$, directly write down the value of $k$.

Solution:
\textbf{Solution:} Solve: $k=-4, k=-\frac{4}{7}$.

Therefore, the value of $k$ is $k=\boxed{-4}$ or $k=\boxed{-\frac{4}{7}}$.


Question: As shown in the figure, the line $y=kx+b$ intersects the x-axis at point $A(4,0)$ and the y-axis at point $B(0,8)$. What is the equation of line $AB$?

Solution:
\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting A(4,0) and B(0,8) into it gives $\begin{cases}4k+b=0\\ b=8\end{cases}$, solving this system yields $\begin{cases}k=-2\\ b=8\end{cases}$, \\
$\therefore$ the equation of line AB is $\boxed{y=-2x+8}$.


Question: As shown in the figure, the line $y=kx+b$ intersects the $x$-axis at point $A(4,0)$ and intersects the $y$-axis at point $B(0,8)$. If point $P(2,m)$ and point $Q(n,2)$ are two points on line $AB$, what is the length of the segment $\overline{PQ}$?

Solution:
\textbf{Solution:} Given the problem statement,\\
$\because$ the point P(2, m) lies on the line $y=-2x+8$,\\
$\therefore m=-2\times 2+8=4$,\\
$\therefore$ the coordinates of point P are (2, 4);\\
$\because$ the point Q(n, 2) lies on the line $y=-2x+8$,\\
$\therefore -2n+8=2$, solving for $n$ yields $n=3$,\\
$\therefore$ the coordinates of point Q are (3, 2),\\
$\therefore PQ=\sqrt{{(2-3)}^2+{(4-2)}^2}=\boxed{\sqrt{5}}$.


Question: As shown in the figure, place the rectangular paper $OABC$ into the Cartesian coordinate system such that $OA$ and $OC$ lie on the positive x-axis and y-axis, respectively. Connect $AC$, and let $AC=4\sqrt{5}$ and $OA=2CO$. What is the equation of the line on which $AC$ lies?

Solution:
\textbf{Solution:} Let $\because$ OA=2CO, assume OC=x, then OA=2x. In $\triangle AOC$, by Pythagorean theorem we have $OC^2+OA^2=AC^2$,
$\therefore x^2+(2x)^2=(4\sqrt{5})^2$. Solving this, we get $x=4$ (discarding $x=-4$).
$\therefore$ OC=4, OA=8.
$\therefore$ A(8,0), C(0,4).
Let the equation of line AC be $y=kx+b$,
$\therefore \left\{\begin{array}{l}8k+b=0\\ b=4\end{array}\right.$, solving this we get $\left\{\begin{array}{l}k=-\frac{1}{2}\\ b=4\end{array}\right.$,
$\therefore$ the equation of line AC is $\boxed{y=-\frac{1}{2}x+4}$.


Question: As shown in the figure, a rectangular sheet $OABC$ is placed in the Cartesian coordinate system, with $OA$ and $OC$ lying on the positive halves of the x-axis and y-axis, respectively. Connect $AC$, and let $AC=4\sqrt{5}$ and $OA=2CO$. Fold the sheet $OABC$ so that point A coincides with point C (the fold line is $EF$). Find the area of the overlapping part of the sheet after folding.

Solution:
\textbf{Solution:} Given that $AE=CE$ from the fold, let $AE=CE=y$, then $OE=8-y$. In $\triangle OCE$, by the Pythagorean theorem, we have $OE^2+OC^2=CE^2$, \\
$\therefore$ $(8-y)^2+4^2=y^2$, solving this yields $y=5$\\
$\therefore$ $AE=CE=5$\\
In the rectangle $OABC$,\\
$\because$ $BC \parallel OA$,\\
$\therefore$ $\angle CFE=\angle AEF$, and from the fold $\angle AEF=\angle CEF$,\\
$\therefore$ $\angle CFE=\angle CEF$\\
$\therefore$ $CF=CE=5$\\
$\therefore$ $Area_{\triangle CEF}=\frac{1}{2}CF \cdot OC=\frac{1}{2} \times 5 \times 4=\boxed{10}$\\
Therefore, the area of the overlapping section is \boxed{10}.


Question: As shown in the figure, a rectangular piece of paper $OABC$ is placed in the Cartesian coordinate system, with $OA$ and $OC$ lying on the positive halves of the x-axis and y-axis, respectively. The line $AC$ is drawn, and it is given that $AC=4\sqrt{5}$ and $OA=2CO$. If any line passing through a fixed point $M$ always divides the area of rectangle $OABC$ into two equal parts, what are the coordinates of point $M$?

Solution: \textbf{Solution:} The coordinates of point M are $\boxed{(4,2)}$.


Question: As shown in the figure, the line $y=-2x+7$ intersects the x-axis at point C and the y-axis at point B, and intersects the line $y=\frac{3}{2}x$ at point A. What are the coordinates of point A?

Solution:
\textbf{Solution:} Solve the system of equations: $\begin{cases}y=-2x+7\\ y=\frac{3}{2}x\end{cases}$ to get: $\begin{cases}x=2\\ y=3\end{cases}$,\\
$\therefore$ The coordinates of point $A$ are $\boxed{(2, 3)}$;


Question: As shown in the figure, the line $y=-2x+7$ intersects the x-axis and y-axis at points C and B, respectively, and intersects the line $y=\frac{3}{2}x$ at point A. If there exists a point P on the y-axis such that $\triangle OAP$ is an isosceles triangle with $OA$ as its base, what are the coordinates of point P?

Solution:
\textbf{Solution:} Let the coordinates of point P be $(0, y)$,\\
since $\triangle OAP$ is an isosceles triangle with $OA$ as its base,\\
therefore $OP=PA$,\\
thus $2^2+(3-y)^2=y^2$,\\
solving this yields $y=\frac{13}{6}$,\\
therefore, the coordinates of point P are $\left(0, \boxed{\frac{13}{6}}\right)$;


Question: As shown in the figure, the straight line $y=-2x+7$ intersects the x-axis and y-axis at points C and B, respectively, and intersects the straight line $y=\frac{3}{2}x$ at point A. In the Cartesian coordinate system xOy, is there a point M such that the quadrilateral with vertices O, A, M, and C is a parallelogram? Write down the coordinates of all the points M that meet the conditions; if not, please explain the reason;

Solution:
\textbf{Solution:} The solution exists. Let $y=0$ and substitute it into $y=-2x+7$, we get $0=-2x+7$. Solving this we find $x=\frac{7}{2}$, thus $C\left(\frac{7}{2},0\right)$. Let $M(x,y)$ be as shown in the figure:
\begin{enumerate}
    \item When $AC$ is the diagonal, $x=2+\frac{7}{2}-0=\frac{7}{2}$, $y=3$. Therefore, the coordinates of point $M$ are $\left(\frac{11}{2},3\right)$;
    \item When $AO$ is the diagonal, $x=2+0-\frac{7}{2}=-\frac{3}{2}$, $y=3$. Therefore, the coordinates of point $M$ are $\left(-\frac{3}{2},3\right)$;
    \item When $CO$ is the diagonal, $x=0+\frac{7}{2}-2=\frac{3}{2}$, $y=-3$. Therefore, the coordinates of point $M$ are $\left(\frac{3}{2},-3\right)$,
\end{enumerate}
In summary: the coordinates of point $M$ are $\boxed{\left(\frac{11}{2},3\right)}$, $\boxed{\left(-\frac{3}{2},3\right)}$, $\boxed{\left(\frac{3}{2},-3\right)}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects the $x$ axis and $y$ axis at points $A$ and $B$ respectively, and intersects the line $y=x$ at point $C$. A line passing through point $C$ that bisects the area of $\triangle AOC$ intersects the $x$ axis at point $D$. Find the length of the line segment $CD$.

Solution:
\textbf{Solution:} From the problem statement, we have
\[
\left\{y = -\frac{1}{2}x+2 \\
y = x\right.
\]
Solving this, we get
\[
\left\{x = 2 \\
y = 2\right.
\]
Therefore, point C(2,2). Since the line that bisects the area of $\triangle AOC$ intersects the x-axis at point D, it follows that $OD=AD$. When x=0, $y=-\frac{1}{2}x+3=3$, therefore, point B(0,3). When y=0, $0=-\frac{1}{2}x+3$, therefore, x=6, thus point A(6,0), and therefore, point D(3,0). Construct $CH \perp OA$ at point H, so CH=2, and $DH=3-2=1$. In $\triangle CHD$, $CD=\sqrt{CH^2-HD^2}=\boxed{\sqrt{5}}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the line $y=x$ at point $C$. The line through point $C$ that bisects the area of $\triangle AOC$ intersects the $x$-axis at point $D$. Point $E$ is on the $y$-axis. When the perimeter of $\triangle DCE$ is minimized, find the coordinates of point $E$ (proof of minimum perimeter is not required).

Solution: \textbf{Solution:} Let point $D$ be symmetrically reflected across the $y$-axis to point $F$, and draw line $FC$ intersecting the $y$-axis at point $E$. Connecting $ED$, point $E$ is the required point. Due to symmetry, we have $F(-3, 0)$. Assume the equation of line $FC$ is $y=kx+b$. Substituting points $F(-3, 0)$ and $C(2, 2)$ into this equation, we get $\therefore \begin{cases}2k+b=2\\ -3k+b=0\end{cases}$, and solving these gives $\begin{cases}k=\frac{2}{5}\\ b=\frac{6}{5}\end{cases}$. $\therefore$ $y=\frac{2}{5}x+\frac{6}{5}$. Letting $x=0$, we find $y=\frac{6}{5}$, $\therefore$ the coordinates of point $E$ are $\boxed{\left(0, \frac{6}{5}\right)}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the line $y=x$ at point $C$. The line passing through point $C$ that bisects the area of $\triangle AOC$ intersects the $x$-axis at point $D$. Let $P$ be a moving point on line $AB$. Is there a point $Q$ in the plane such that the quadrilateral with vertices $A$, $D$, $P$, $Q$ is a parallelogram, and its area is equal to the area of $\triangle AOC$? Write down the coordinates of point $Q$.

Solution:
\textbf{Solution:} There exists a point $Q$, with coordinates $\left(-1, 2\right)$ or $\left(5, 2\right)$ or $\left(7, -2\right)$ or $\left(13, -2\right)$, which makes the quadrilateral formed by vertices $A$, $D$, $P$, $Q$ a parallelogram, and its area equals to the area of $\triangle AOC$.\\
Since $S_{\triangle AOC}=2S_{\triangle ADC}$ and $S_{\text{parallelogram}}=S_{\triangle AOC}$,\\
it follows that $S_{\text{parallelogram}}=2S_{\triangle PAD}$.\\
When $P$ coincides with point $C$ above the $x$-axis,\\
since $A\left(6, 0\right)$, $D\left(3, 0\right)$,\\
then $AD=3$,\\
since $P_1(2, 2)$,\\
by the property of translation, it's known that $Q_1\left(-1, 2\right)$ or $Q_2\left(5, 2\right)$;\\
When point $P$ lies on line $AB$ below the $x$-axis, and the distance to the $x$-axis is 2,\\
let $P_2(x, -\frac{1}{2}x+3)$, then $-\frac{1}{2}x+3=-2$,\\
solve to get $x=10$,\\
thus $P(10, -2)$,\\
by the property of translation, it's known that $Q_3\left(7, -2\right)$ or $Q_4\left(13, -2\right)$;\\
In conclusion, the coordinates of point $Q$ are $\boxed{\left(-1, 2\right)}$ or $\boxed{\left(5, 2\right)}$ or $\boxed{\left(7, -2\right)}$ or $\boxed{\left(13, -2\right)}$.


Question: As shown in the figure, it is known that the graph of the linear function $y_1=kx+b$ passes through the point $(-1, -5)$ and intersects the $y$-axis at point $B$, and it intersects the graph of the direct proportion function $y_2=\frac{1}{3}x$ at point $A(6, a)$. Find the equation of the line $y_1=kx+b$.

Solution:
\textbf{Solution:} Substitute $A(6, 2)$ into $y_2=\frac{1}{3}x$ to get $a=6\times \frac{1}{3}=2$,\\
$\therefore A(6, 2)$.\\
$\because$ Points $(-1, -5)$ and $A(6, 2)$ are on the graph of the linear function $y_1=kx+b$,\\
$\therefore \left\{\begin{array}{l}
6k+b=2 \\
-k+b=-5
\end{array}\right.$,\\
solving this, we get $\left\{\begin{array}{l}
k=1 \\
b=-4
\end{array}\right.$,\\
$\therefore \boxed{y_1=x-4}$.


Question: As shown in the figure, it is known that the graph of the linear function $y_1=kx+b$ passes through the point $(-1, -5)$ and intersects the $y$ axis at point $B$. It also intersects the graph of the direct proportion function $y_2=\frac{1}{3}x$ at point $A(6, a)$. What is the area of the triangle formed by these two lines and the $y$ axis?

Solution:
\textbf{Solution:} Since the linear function is given by $y_1=x-4$,\\
thus for $x=0$, $y_1=0-4=-4$,\\
hence the coordinates of point B are $(0,-4)$,\\
therefore, $OB=4$,\\
Since the direct proportion function is given by $y_2=\frac{1}{3}x$,\\
thus the direct proportion function intersects with the y-axis at $(0, 0)$,\\
Since point A is $(6,2)$,\\
therefore, in $\triangle AOB$, the height from the side OB is 6,\\
hence the area of $\triangle AOB$, $S_{\triangle AOB}=\frac{1}{2}\times 4\times 6=\boxed{12}$.


Question: As shown in the diagram, the line $y_1=kx+b$ (where $k$ is a constant, $k \neq 0$) intersects the hyperbola $y_2=\frac{2}{x}$ at points $A$ and $D$, and intersects the $x$-axis and $y$-axis at points $B$ and $C$, respectively. The coordinates of point $A$ are $(m, 2)$, and the coordinates of point $D$ are $(-2, n)$. Find the analytical expression of the line.

Solution:
\textbf{Solution:} Let point $A(1, 2)$ and point $D(-2, -1)$. Substitute into the hyperbola $y=\frac{2}{x}$, we get
\[
\begin{cases}
2=\frac{2}{m}\\
n=\frac{2}{-2}
\end{cases}
\]
Solving, we find
\[
\begin{cases}
m=1\\
n=-1
\end{cases}
\]
$\therefore$ The coordinates of point $A$ are $(1, 2)$, and the coordinates of point $D$ are $(-2, -1)$. Substituting the coordinates of point $A$ $(1, 2)$ and point $D$ $(-2, -1)$ into $y=kx+b$, we obtain
\[
\begin{cases}
k+b=2\\
-2k+b=-1
\end{cases}
\]
Solving, we find
\[
\begin{cases}
k=1\\
b=1
\end{cases}
\]
$\therefore$ The equation of the line is \boxed{y=x+1}.


Question: As shown in the figure, the line $y_1=kx+b$ (where $k$ is a constant and $k\ne 0$) intersects the hyperbola $y_2=\frac{2}{x}$ at points $A$ and $D$, and intersects the $x$-axis and $y$-axis at points $B$ and $C$ respectively. The coordinates of point $A$ are $(m, 2)$, and the coordinates of point $D$ are $(-2, n)$. Based on the graph, directly write the range of values for $x$ when $y_1 < y_2$.

Solution: \textbf{Solution:} From the graph, it is known that when $y_1 < y_2$, the range of $x$ is $\boxed{x < -2}$ or $\boxed{0 < x < 1}$.


Question: As shown in the figure, both quadrilateral $OABC$ and quadrilateral $ODEF$ are squares, with points F, O, A lying on a straight line, point D on the side $OC$. A Cartesian coordinate system $xOy$ is established with $FA$ as the x-axis and $OC$ as the y-axis. The line $y = \frac{1}{2}x + 3$ passes through points B and E. What are the side lengths of square $OABC$ and square $ODEF$?

Solution:
\textbf{Solution:} Let the side lengths of square $OABC$ and square $ODEF$ be: $a$, $b$,\\
$\therefore B(a,a)$, $A(-b,b)$,\\
$\because$ the line $y=\frac{1}{2}x+3$ passes through points B and E,\\
$\therefore \left\{\begin{array}{l}
\frac{1}{2}a+3=a \\
-\frac{1}{2}b+3=b
\end{array}\right.$, solving this we get:\\
$\left\{\begin{array}{l}
a=\boxed{6} \\
b=\boxed{2}
\end{array}\right.$,\\
$\therefore$ the side lengths of square $OABC$ and square $ODEF$ are: \boxed{6} and \boxed{2};


Question: As shown in the figure, in the diamond ABCD, AD is parallel to the x-axis, the coordinate of point A is (0, 3), and the coordinate of point B is (4, 0). The line where side CD is located, $y_{1}=mx+n$, intersects the x-axis at point C, and it intersects the hyperbola $y_{2}=\frac{k}{x}$ ($x<0$) at point D. Find the analytical expression of the line CD and the value of k.

Solution:
\textbf{Solution:} Given point A(0, 3), point B(4, 0), thus AO = 3, BO = 4. In $\triangle ABO$, by Pythagoras' theorem, we have $AB= \sqrt{3^2+4^2} = 5$. Since the quadrilateral ABCD is a rhombus, we have AD = BC = AB = 5, therefore OC = 1. Hence, the coordinates of point C are (-1, 0), and the coordinates of point D are (-5, 3). For the line $y_1=mx+n$, we have
\[
\begin{cases}
-m+n=0\\
-5m+n=3
\end{cases}
\]
Solving for this, we get:
\[
\begin{cases}
m=-\frac{3}{4}\\
n=-\frac{3}{4}
\end{cases}
\]
Therefore, $y_1=-\frac{3}{4}x-\frac{3}{4}$. For $y_2= \frac{k}{x}$ (x < 0), we have $3= \frac{k}{-5}$. Solving for this yields $k=\boxed{-15}$.


Question: As shown in the rhombus $ABCD$, where $AD\parallel$ the $x$-axis, the coordinates of point $A$ are $(0,3)$, and the coordinates of point $B$ are $(4,0)$. The line containing side $CD$, denoted by $y_1=mx+n$, intersects the $x$-axis at point $C$ and intersects the hyperbola $y_2= \frac{k}{x}$ ($x<0$) at point $D$. What is the range of values for the independent variable $x$ such that $y_1 - y_2 \leq 0$ when $x<0$?

Solution: \textbf{Solution:} When $x<0$, the range of the independent variable $x$ that makes $y_1-y_2\leq 0$ is $\boxed{-5\leq x<0}$.


Question: As shown in the figure, the line \(y = -\frac{4}{3}x + 8\) intersects the x-axis and y-axis at points A and B, respectively. Point M is a point on OB. If \(\triangle ABM\) is folded along AM, point B exactly falls on point \(B^{\prime}\) on the x-axis. Find the coordinates of point M.

Solution:
\textbf{Solution:} Solve: When $x=0$, $y=8$, that is, $B(0,8)$. When $y=0$, $0=-\frac{4}{3}x+8$, solving this, we get $x=6$, that is, $A(6,0)$;\\
$\therefore$ $OA=6$, $OB=8$,\\
$\because$ $\angle AOB=90^\circ$,\\
$\therefore$ $AB= \sqrt{OA^{2}+OB^{2}} =10$,\\
By the property of folding, we get: $AB=AB^{\prime}=10$,\\
$\therefore$ $OB^{\prime}=AB^{\prime}-OA=10-6=4$,\\
Let $MO=x$, then $MB=MB^{\prime}=8-x$,\\
In $\triangle OMB^{\prime}$, $OM^{2}+OB^{\prime 2}=B^{\prime}M^{2}$,\\
that is, $x^{2}+4^{2}=(8-x)^{2}$,\\
Solving for $x$ yields: $x=3$,\\
$\therefore$ The coordinates of point M are $\boxed{(0,3)}$.


Question: As shown in the figure, the line $y = -\frac{4}{3}x + 8$ intersects the x-axis and y-axis at points A and B, respectively. Point M is on OB. If $\triangle ABM$ is folded along AM, point B falls exactly on point B$^{\prime}$ on the x-axis. Find the equation of line AM.

Solution:
Solution: Let the equation of line AM be \(y=kx+b\). Substituting \((0,3)\) and \((6,0)\) into it, we get
\[
\left\{
\begin{array}{l}
b=3\\
6k+b=0
\end{array}
\right.
\]
Solving this, we find
\[
\left\{
\begin{array}{l}
b=3\\
k=-\frac{1}{2}
\end{array}
\right.
\]
Therefore, the equation of line AM is \(\boxed{y=-\frac{1}{2}x+3}\).


Question: As shown in the figure, the equation of line $AB$ is $y=x+2$, and the equation of line $AC$ is $y=-x+4$. The two lines intersect at point $A$ and intersect the $x$-axis at points $B$ and $C$, respectively. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} Draw $AE \perp x$ axis at point $E$ through point A. By solving the system of equations given by the problem statement
\[
\left\{
\begin{array}{l}
y=x+2 \\
y=-x+4
\end{array}
\right.
\]
we find:
\[
\left\{
\begin{array}{l}
x=1 \\
y=3
\end{array}
\right.
\]
Therefore, $A(1, 3)$ and thus $AE=3$. When $y=0$, we have $x+2=0$, therefore $x=-2$, leading to $B(-2, 0)$. When $y=0$, $-x+4=0$, therefore $x=4$, leading to $C(4, 0)$. Therefore, $BC=4-(-2)=6$, therefore the area of $\triangle ABC = \frac{1}{2}BC\cdot AE = \frac{1}{2} \times 6 \times 3 = \boxed{9}$.


Question: As shown in the figure, the equation of line $AB$ is $y=x+2$, and the equation of line $AC$ is $y=-x+4$. The two lines intersect at point $A$, and they intersect the $x$-axis at points $B$ and $C$, respectively. Point $D$ is a point on line segment $AC$, and line $BD$ is drawn. If $BD=\sqrt{26}$, what are the coordinates of point $D$?

Solution: \textbf{Solution:} Draw $DF\perp x$-axis at point $F$ through point $D$. Suppose the x-coordinate of point $D$ is $m$. Since point $D$ is on line $AC$, we have $y=-m+4$. Thus, $D(m, -m+4)$. Therefore, $DF=-m+4$ and $BF=m+2$. In right triangle $\triangle DBF$, with $\angle DFB=90^\circ$, according to the Pythagorean theorem, we get $BD^{2}=DF^{2}+BF^{2}$. Hence, $\left(\sqrt{26}\right)^{2}=\left(-m+4\right)^{2}+\left(m+2\right)^{2}$. After simplification, we obtain $m^{2}-2m-3=0$. Solving this yields $m_{1}=3$, $m_{2}=-1$ (this solution is rejected as it is not consistent with the conditions of the problem). Therefore, $D(3, 1)$. Thus, the coordinates of point $D$ are $\boxed{D(3, 1)}$.


Question: As shown in the figure, the graph of the linear function $y=mx+2$ passes through point $A(2, 4)$ and $B(n, -1)$. Find the values of $m$ and $n$.

Solution:
\textbf{Solution:} Since the graph of the linear function $y=mx+2$ passes through point $A(2, 4)$,\\
thus $2m+2=4$, we find $m=\boxed{1}$.\\
Therefore, the expression for the linear function is $y=x+2$.\\
Since the graph of the linear function $y=x+2$ passes through point $B(n, -1)$,\\
thus $n+2=-1$, we find $n=\boxed{-3}$.


Question: As shown in the figure, the graph of the linear function $y=mx+2$ passes through points $A(2, 4)$ and $B(n, -1)$. Connect $OA$ and $OB$. Find the area of $\triangle OAB$.

Solution:
\textbf{Solution:} As shown in the figure, let the intersection of line $AB$ with the x-axis be $C$. Let $y=0$, then $x=-2$, $\therefore$ the coordinates of point $C$ are $(-2, 0)$, $\therefore$ $OC=2$, $\therefore$ the area of $\triangle OAB$ equals the area of $\triangle AOC$ plus the area of $\triangle BOC$\\
$=\frac{1}{2}\times OC\times y_A+\frac{1}{2}\times OC\times |y_B|$\\
$=\frac{1}{2}\times 2\times 4+\frac{1}{2}\times 2\times 1=\boxed{5}$.


Question: As shown in the figure, the graph of the function $y=mx+\frac{4}{3}$ is a straight line $l_{1}$, and the graph of the function $y=kx+b$ is a straight line $l_{2}$. The lines $l_{1}$ and  $l_{2}$ intersect the x-axis at point $B$ and point $C$ ($3$, $0$) respectively, and intersect the y-axis at point $D$ and $E$ respectively. Lines $l_{1}$ and $l_{2}$ intersect at point $A$ ($2$, $2$). Directly write down the solution set of the inequality $mx+\frac{4}{3}<kx+b$.

Solution:
\textbf{Solution:} Since lines $l_1$ and $l_2$ intersect at point $A(2, 2)$,\\
it is known from the graph that when $x<2$, $mx+\frac{4}{3}<kx+b$,\\
therefore, the solution set of the inequality $mx+\frac{4}{3}<kx+b$ is $x<2$;\\
Hence, the answer is: $\boxed{x<2}$;


Question: As illustrated, the graph of function $y=mx+\frac{4}{3}$ is the straight line $l_{1}$, and the graph of function $y=kx+b$ is the straight line $l_{2}$. Lines $l_{1}$ and $l_{2}$ intersect the $x$-axis at points $B$ and $C$ ($3$, $0$) respectively, and intersect the $y$-axis at points $D$ and $E$ respectively. Lines $l_{1}$ and $l_{2}$ intersect at point $A$ ($2$, $2$). Calculate the area of $\triangle ADE$.

Solution:
\textbf{Solution:} Since line $l_2$ passes through points $A$ and $C$, \\
we have $\begin{cases}2k+b=2\\ 3k+b=0\end{cases}$, solving this we get $k=-2$, $b=6$, \\
thus the coordinates of point $D$ are $(0, \frac{4}{3})$, \\
hence the area of $\triangle ADE$ is $S=\frac{1}{2}\times |b-{y}_{D}|\times |{x}_{A}|=\frac{1}{2}\times |6-\frac{4}{3}|\times 2=\boxed{\frac{14}{3}}$.


Question: As shown in the figure, it is known that the line $l_1: y=x+n-2$ and the line $l_2: y=mx+n$ intersect at the point $P(1, 2)$. Find the values of $m$ and $n$.

Solution:
\textbf{Solution:} Since point $P$ is the intersection of two lines, \\
$\therefore$ by substituting point $P\left(1, 2\right)$ into $y=x+n-2$ and $y=mx+n$ respectively, we get $2=1+n-2$, $2=m+n$, \\
solving these, we find $m=\boxed{-1}$, $n=\boxed{3}$.


Question: As shown in the figure, it is known that the lines $y=2x+3$ and $y=-2x-1$ intersect. Find the coordinates of points $A$ and $B$, which are the respective points of intersection of each line with the y-axis.

Solution:
\textbf{Solution:} In $y=2x+3$, when $x=0$, we have $y=3$, that is point \boxed{A(0, 3)}; \\
In $y=-2x-1$, when $x=0$, we have $y=-1$, that is point \boxed{B(0, -1)};


Question: As shown in the figure, it is known that the graph of the line $y = 2x + 3$ and the line $y = -2x - 1$ are given. Find the coordinates of the point of intersection $C$ of the two lines.

Solution:
\textbf{Solution:} According to the problem, we have
\[
\left\{
\begin{array}{l}
y=2x+3 \\
y=-2x-1
\end{array}
\right.
\]
Solving this system of equations, we get
\[
\left\{
\begin{array}{l}
x=-1 \\
y=1
\end{array}
\right.
\]
$\therefore$ The coordinates of point C are $\boxed{(-1, 1)}$.


Question: As shown in the figure, it is known that the graph of the line $y=2x+3$ and the graph of the line $y=-2x-1$ are as shown in the figure. Find the area of $\triangle ABC$.

Solution:
\textbf{Solution:} Draw $CD\perp AB$ through point C, intersecting the y-axis at point D;\\
$\therefore CD=1$;\\
$\because AB=3-(-1)=4$;\\
$\therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot CD=\frac{1}{2}\times 4\times 1=\boxed{2}$.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the coordinates of point A are $(2,1)$. $OA \perp OB$, $OA = OB$, $BC \parallel x$-axis, and intersects the y-axis at point D, $\angle BCA = 135^\circ$. What are the coordinates of point B?

Solution:
\textbf{Solution:} As shown in the diagram, draw $AE \perp x$ axis at point E,

$\therefore \angle AEO = 90^\circ$

$\because BC \parallel x$ axis

$\therefore \angle BDO = \angle AEO = 90^\circ$

$\because OA \perp OB$

$\therefore \angle BOA = 90^\circ$

$\therefore \angle BOD + \angle DOA = \angle AOE + \angle DOA$

$\therefore \angle BOD = \angle AOE$

$\because OA = OB$

$\therefore \triangle BOD \cong \triangle AOE$

$\therefore BD = AE = 1, OD = OE = 2$

$\therefore$ The coordinates of $B$ are $\boxed{(-1, 2)}$


Question: As shown in Figure 1, in the Cartesian coordinate system xOy, the coordinate of point A is $(2,1)$. It is given that $OA \perp OB$, $OA = OB$, and $BC \parallel$ to the x-axis, intersecting the y-axis at point D. The angle $\angle BCA = 135^\circ$. What is the length of the segment BC?

Solution:
As shown in the figure, draw $AF\perp BC$ at point F, \\
$\therefore \angle AFC=90^\circ$\\
$\because \angle BCA=135^\circ$\\
$\therefore \angle ACF=45^\circ$\\
$\therefore \triangle ACF$ is an isosceles right triangle\\
$\therefore CF=AF=2-1=1$\\
$\therefore CD=2-1=1$\\
$\therefore BC=BD+CD=\boxed{2}$;


Question: As shown in Figure 1, in the Cartesian coordinate system xOy, the coordinates of point A are (2, 1). OA is perpendicular to OB, and the length of OA is equal to OB. BC is parallel to the x-axis and intersects the y-axis at point D. The angle $\angle BCA$ equals $135^\circ$. Point P(0, m) is on segment OD. A line parallel to the axis is drawn through point P (as shown in Figure 2), and this line intersects the sides of quadrilateral AOBC, resulting in a segment of length $d$. Find the function expression of $d$ in terms of $m$, and directly state the range of the independent variable $m$.

Solution:
\textbf{Solution:} From the problem, we have $d=\begin{cases}
\frac{5}{2}m, & 0\le m<1, \\
-\frac{m}{2}+3, & 1\le m\le 2
\end{cases}$.

Therefore, the range of the independent variable $m$ is $\boxed{0\le m\le 2}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=k_{1}x+b$ ($k_{1}\neq 0$) passes through points $A(4,0)$ and $B(0,2)$, and intersects with the line $l_2: y=k_{2}x$ ($k_{2}\neq 0$) at point $P(a,1)$. What are the equations of lines $l_{1}$ and $l_{2}$?

Solution:
\textbf{Solution:} Let $\because$ the line ${l}_{1}\colon y={k}_{1}x+b\left({k}_{1}\ne 0\right)$ passes through point A (4, 0) and B (0, 2),\\
$\therefore$ $\begin{cases}b=2\\ 4{k}_{1}+b=0\end{cases}$,\\
$\therefore$ $\begin{cases}{k}_{1}=-\frac{1}{2}\\ b=2\end{cases}$,\\
$\therefore$ the equation of line ${l}_{1}$ is $y=-\frac{1}{2}x+2$,\\
When $y=1$, then $-\frac{1}{2}x+2=1$,\\
$\therefore x=2$,\\
$\therefore$ point P (2, 1),\\
$\therefore$ $1=2{k}_{2}$,\\
$\therefore$ ${k}_{2}=\frac{1}{2}$,\\
$\therefore$ the equation of line ${l}_{2}$ is $\boxed{y=\frac{1}{2}x}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}: y=k_{1}x+b$ ($k_{1}\neq 0$) passes through points $A(4,0)$ and $B(0,2)$, and intersects with line $l_2: y=k_{2}x$ ($k_{2}\neq 0$) at point $P(a,1)$. Point $C$ is on line $l_{1}$. A line $m$ is drawn through $C$ and is perpendicular to the x-axis at point $E$, and line $m$ intersects $l_2$ at point $D$. When $CD=3ED$, find the coordinates of point $C$.

Solution:
\textbf{Solution:} Let the point $C(t, -\frac{1}{2}t+2)$, then the point $D(t, \frac{1}{2}t)$, and the point $E(t, 0)$, \\
$\therefore CD=\left|-\frac{1}{2}t+2-\frac{1}{2}t\right|=\left|-t+2\right|$, $DE=\left|\frac{1}{2}t\right|$, \\
$\because CD=3DE$, \\
$\therefore \left|-t+2\right|=3\left|\frac{1}{2}t\right|$, \\
$\therefore -t+2=\frac{3}{2}t$ or $-t+2-\frac{3}{2}t=0$, \\
$\therefore t=\boxed{\frac{4}{5}}$ or $t=-4$, \\
$\therefore$ the point $C\boxed{\left(\frac{4}{5}, \frac{8}{5}\right)}$ or $\boxed{\left(-4, 4\right)}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through two points $A$ and $B$, where $A(-2,-1)$ and $B(1,3)$, and it intersects the $y$-axis at point $D$. What are the equation of the linear function and the coordinates of point $D$?

Solution:
\textbf{Solution:} Substitute the coordinates $(-2, -1)$ and $(1, 3)$ into $y=kx+b$, \\
we have $\begin{cases}
-2k+b=-1 \\
k+b=3
\end{cases}$\\
Solving this, we find $\begin{cases}
k=\frac{4}{3} \\
b=\frac{5}{3}
\end{cases}$\\
Therefore, the equation of the line is $\boxed{y=\frac{4}{3}x+\frac{5}{3}}$.\\
Furthermore, since the line intersects the y-axis at point $D$\\
Thus, by substituting $x=0$ into $y=\frac{4}{3}x+\frac{5}{3}$\\
we get $y=\frac{5}{3}$,\\
Therefore, the coordinates of point $D$ are $\boxed{(0, \frac{5}{3})}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through two points $A$ and $B$, where $A(-2, -1)$ and $B(1, 3)$, and it intersects the $y$-axis at point $D$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Since the coordinates of point D are $ (0, \frac{5}{3})$.

Hence, $S_{\triangle AOB}=S_{\triangle AOD}+S_{\triangle BOD}$

$=\frac{1}{2}\times \frac{5}{3}\times 2+\frac{1}{2}\times \frac{5}{3}\times 1$

$=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, the line $l_1$ intersects the x-axis and y-axis at points A and B, respectively, with the coordinates of A and B being $(2, 0)$ and $(0, 3)$, respectively. The line $l_2$ passing through point B is given by $y=\frac{1}{2}x+3$ and intersects the x-axis at point C. Point $D(n, 6)$ lies on the line $l_1$, and line $CD$ is drawn. What is the analytical expression for $l_1$?

Solution:
\textbf{Solution:} Let the equation of line $l_{1}$ be $y=kx+b$, \\
Substituting points A(2, 0) and B(0,3) into the equation gives\\
$\begin{cases}2k+b=0\\ b=3\end{cases}$,\\
Solving this system of equations yields $\begin{cases}k=-\frac{3}{2}\\ b=3\end{cases}$,\\
$\therefore$ the equation of line $l_{1}$ is $\boxed{y=-\frac{3}{2}x+3}$.


Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with coordinates $(2, 0)$ and $(0, 3)$. The line $l_{2}: y=\frac{1}{2}x+3$ passing through point B intersects the x-axis at point C. Point $D(n, 6)$ is a point on the line $l_{1}$. Connect $CD$. What are the coordinates of C and D?

Solution:
\textbf{Solution}: When $y=0$, $\frac{1}{2}x+3=0$, solving this gives $x=-6$,\\
hence the coordinates of point C are $\boxed{(-6,0)}$,\\
substituting point D$(n,6)$ into $y=-\frac{3}{2}x+3$ we get $-\frac{3}{2}n+3=6$, solving for $n$ gives $n=-2$,\\
therefore, the coordinates of point D are $\boxed{(-2,6)}$;


Question: As shown in the figure, the line $l_{1}$ intersects the x-axis and y-axis at points A and B, respectively, with the coordinates of A and B being $(2, 0)$ and $(0, 3)$, respectively. The line $l_{2}\colon y=\frac{1}{2}x+3$ passes through point B and intersects the x-axis at point C. Point $D(n, 6)$ lies on the line $l_{1}$. Connect $CD$. What is the area of $\triangle BCD$?

Solution:
\textbf{Solution:} By the problem statement, we have
\[
S_{\triangle BCD} = S_{\triangle DAC} - S_{\triangle BAC} = \frac{1}{2} \times (2+6) \times 6 - \frac{1}{2} \times (2+6) \times 3 = \boxed{12}.
\]


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line $l_1$ intersects the $x$-axis at point $A\left(m, 0\right)$ and intersects the $y$-axis at point $B\left(0, mn\right)$ ($m<0$, $n>0$). If $m=-4$ and $n=\frac{1}{2}$, what is the equation of the line $l_1$?

Solution:
\textbf{Solution:} Given the problem, we know:\\
Point A ($-4$, $0$), Point B ($0$, $-2$),\\
Let the equation of line $l_{1}$ be: $y=kx+b$,\\
$\begin{cases} b=-2 \\ -4k+b=0 \end{cases}$, solving this gives: $\begin{cases} b=-2 \\ k=-\frac{1}{2} \end{cases}$,\\
$\therefore y=-\frac{1}{2}x-2$;\\
$\therefore$ The equation of line $l_{1}$ is $\boxed{y=-\frac{1}{2}x-2}$.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line $l_1$ intersects the $x$-axis at point $A(m, 0)$ and the $y$-axis at point $B(0, mn)$ ($m<0$, $n>0$). As shown in Figure 2, under the condition of (1), the line $l_2: y=\frac{1}{2}x$ intersects the line $l_1$ at point $C$, and let's consider point $D(0, 2)$. Does there exist a point $G$ on the line $l_2$ such that $\frac{S_{\triangle ACD}}{S_{\triangle CDG}}=\frac{4}{3}$? If it exists, find the coordinates of point $G$.

Solution:
\textbf{Solution:} Solve the system $\begin{cases}y=-\frac{1}{2}x-2\\ y=\frac{1}{2}x\end{cases}$. We find $\begin{cases}x=-2\\ y=-1\end{cases}$, thus $C(-2,-1)$.

Let the equation of line CD be: $y=mx+n$.

Hence $\begin{cases}n=2\\ -2m+n=-1\end{cases}$. Solving this, we find $\begin{cases}n=2\\ m=\frac{3}{2}\end{cases}$, thus $y=\frac{3}{2}x+2$.

Setting $y=0$, $0=\frac{3}{2}x+2$ leads to $x=-\frac{4}{3}$, thus $E\left(-\frac{4}{3},0\right)$.

Therefore, $AE=4-\frac{4}{3}=\frac{8}{3}$.

Therefore, the area of $\triangle ACD$, $S_{\triangle ACD}=\frac{1}{2}AE\cdot DF=\frac{1}{2}\times\frac{8}{3}\times3=4$.

Since $\frac{S_{\triangle ACD}}{S_{\triangle CDG}}=\frac{4}{3}$,

Thus, $S_{\triangle CDG}=3$.

Let $G(x, \frac{1}{2}x)$.

Thus, $\frac{1}{2}OD\cdot |x+2|=3$,

or $\frac{1}{2}\times2\cdot |x+2|=3$,

Thus, $x_{1}=1, x_{2}=-5$,

Therefore, \boxed{G(1, \frac{1}{2})} or \boxed{G\left(-5, -\frac{5}{2}\right)}.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the line ${l}_{1}$ intersects the $x$-axis at point $A(m, 0)$ and the $y$-axis at point $B(0, mn)$ ($m<0$, $n>0$). There is a point $P$ below the line ${l}_{1}$, whose abscissa is $m+n$. Connect $PB$. If $\angle PBA=2\angle BAO$, find the range of values for $\frac{n}{OA}$.

Solution:
Solution: As shown in Figure 2,
(1) When $m+n<0$, that is $-\frac{n}{m}<1$,\\
extend line $AO$ and take $OC=OA$,\\
since $OB\perp AC$,\\
it follows that $AB=BC$,\\
therefore $\angle BCO=\angle BAO$,\\
hence $\angle APB=\angle BAO+\angle BCO=2\angle BAO$,\\
hence point $P$ lies on the extension of line $CB$,\\
thus, there exists a point $P$ below $l_1$, satisfying $\angle PBA=2\angle BAO$,\\
as shown in Figure 3,\\
(2) Extend line $AO$ and take $OC=OA$,\\
when $m+n>0$, that is $-\frac{n}{m}>1$,\\
from (1) it is known that: $\angle ABE=2\angle BAO$,\\
thus $\angle PBA=\angle ABE+\angle PBE$,\\
therefore $\angle PBA>\angle ABE$,\\
therefore $\angle PBA\neq 2\angle BAO$,\\
In summary: $\boxed{\frac{n}{OA}<1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_{1}\colon y=3x$ intersects with the line $l_{2}\colon y=kx+b$ at point $A(a, 3)$, and point $B(2, 4)$ lies on the line $l_{2}$. Find the value of $a$ and the explicit equation of the line $l_{2}$.

Solution:
\textbf{Solution}: The line $l_{1}\colon y=3x$ intersects with the line $l_{2}\colon y=kx+b$ at the point $A(a,3)$, which yields $3a=3$. \\
Solving this gives $a=\boxed{1}$. \\
$\therefore$ For the point $A(1,3)$, \\
the line $l_{2}\colon y=kx+b$ passes through the point $A(1,3)$, and point $B(2,4)$, \\
hence $\begin{cases}k+b=3 \\ 2k+b=4 \end{cases}$, \\
solving this leads to $\begin{cases}k=1 \\ b=2 \end{cases}$, \\
thus, the equation of the line $l_{2}$ is $\boxed{y=x+2}$.


Question: As shown in the Cartesian coordinate system, the line $l_{1}\colon y=3x$ intersects with the line $l_{2}\colon y=kx+b$ at point $A(a, 3)$, and point $B(2, 4)$ lies on the line $l_{2}$. Write directly the solution set of the inequality $kx+b<3x$ in terms of $x$.

Solution: \textbf{Solution:} The solution to the inequality $kx + b < 3x$ in terms of $x$ is $x > \boxed{1}$.


Question: As shown in the diagram, within the same coordinate system, the line $l_{1}: y=-x+2$ intersects the x-axis at point P, and the line $l_{2}: y=ax-4$ passes through point P. What is the value of $a$?

Solution:
\textbf{Solution:} Since $l_{1}\colon y=-x+2$ intersects the x-axis at point P, \\
it follows that $P(2,0)$, \\
Also, since $l_{2}\colon y=ax-4$ passes through point P, \\
thus $0=2a-4$ yields $a=\boxed{2}$.


Question: As shown in the coordinate system, the line $l_{1}\colon y=-x+2$ intersects the $x$-axis at point $P$, and the line $l_{2}\colon y=ax-4$ passes through point $P$. Points $M$ and $N$ lie on lines $l_{1}$ and $l_{2}$, respectively, and are symmetric about the origin. What is the area of $\triangle PMN$?

Solution: \textbf{Solution:} Given $a=2$, we have $l_2\colon y=2x-4$. Let the x-coordinate of point $M$ be $x$, hence $M\left(x, -x+2\right)$, $N\left(-x, x-2\right)$. Furthermore, since $N\left(-x, x-2\right)$ lies on the line $l_2\colon y=2x-4$, $\therefore$ $x-2=-2x-4$ yields $x= -\frac{2}{3}$. Thus, $M\left(-\frac{2}{3}, \frac{8}{3}\right)$, $N\left(\frac{2}{3}, -\frac{8}{3}\right)$. $\therefore S_{\triangle PMN}=\frac{1}{2} \cdot OP\cdot |My|+ \frac{1}{2} \cdot OP\cdot |Ny| = \frac{1}{2} \times 2\times \frac{8}{3} + \frac{1}{2} \times 2\times \frac{8}{3} = \boxed{\frac{16}{3}}$.


Question: Given the figure, it is known that the line \(l_1\) passing through the point \(B(1, 0)\) intersects with the line \(l_2: y=2x+4\) at the point \(P(-1, a)\). Find the equation of the line \(l_1\).

Solution: \textbf{Solution:} Since point P$(-1, a)$ lies on the line $y=2x+4$, we have $2 \times (-1) + 4 = a$, which means $a=2$. Therefore, the coordinates of P are $(-1, 2)$. Let the equation of line $l_1$ be $y=kx+b$ (where $k \neq 0$). Substituting points B$(1, 0)$ and P$(-1, 2)$ into this equation, we get $\begin{cases} k + b = 0 \\ -k + b = 2 \end{cases}$. Solving this system of equations, we find $\begin{cases} k = -1 \\ b = 1 \end{cases}$. Hence, the equation of line $l_1$ is \boxed{y = -x + 1}.


Question: As shown in the figure, it is known that the line $l_1$ passing through point $B(1,0)$ intersects with line $l_2\colon y=2x+4$ at point $P(-1,a)$. Find the area of the quadrilateral $PAOC$.

Solution:
\textbf{Solution:}

Since the line $l_1\colon y=-x+1$ intersects with the y-axis at point C,

the coordinates of C are (0, 1),

And since the line $l_2\colon y=2x+4$ intersects with the x-axis at point A,

the coordinates of A are (-2, 0),

Because $S_{\text{quadrilateral} PAOC}=S_{\triangle PAB}-S_{\triangle BOC}$,

therefore, $S_{\text{quadrilateral} PAOC}=\frac{1}{2}\times3\times2-\frac{1}{2}\times1\times1=\boxed{\frac{5}{2}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ are such that vertex $O$ coincides with the origin, and vertices $A$ and $C$ are respectively on the coordinate axes, with the coordinates of vertex $B$ being $\left(6, 4\right)$. $E$ is the midpoint of $AB$, and the line passing through points $D\left(8, 0\right)$ and $E$ intersects $BC$ and the y-axis at points $F$ and $G$, respectively. What is the equation of the line $DE$?

Solution:
\textbf{Solution:} Let the equation of $DE$ be: $y=kx+b\ (k\ne 0)$,\\
since $B(6, 4)$ and $E$ is the midpoint of $AB$,\\
thus $E(6, 2)$,\\
substituting $E(6, 2)$ and $D(8, 0)$ into the equation gives $\left\{\begin{array}{l}6k+b=2 \\ 8k+b=0 \end{array}\right.$,\\
solving this yields $\left\{\begin{array}{l}k=-1 \\ b=8 \end{array}\right.$,\\
therefore $\boxed{y=-x+8}$.


Question: As shown in the diagram, within the Cartesian coordinate system, the rectangle $OABC$ has its vertex O coincide with the coordinate origin, and vertices A and C respectively lie on the coordinate axes, with vertex B at the coordinates $\left(6, 4\right)$. E is the midpoint of $AB$, and the line passing through points $D\left(8, 0\right)$ and E intersects $BC$ and the y-axis at points F and G, respectively. The graph of the function $y=mx-2$ goes through point F and intersects the x-axis at point H. Determine the coordinates of point F and the value of $m$?

Solution:
\textbf{Solution:} When $y=4$, $-x+8=4$, hence $x=4$,\\
therefore, the coordinates of point F are $\boxed{\left(4, 4\right)}$,\\
since the graph of the function $y=mx-2$ passes through point F,\\
therefore $4m-2=4$, yielding $m=\boxed{\frac{3}{2}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices of the rectangle $OABC$ with vertex O coinciding with the origin, and vertices A and C located on the coordinate axes respectively, whereas the coordinates of vertex B are $\left(6, 4\right)$. E is the midpoint of $AB$, and the line passing through points $D\left(8, 0\right)$ and E intersects with $BC$ and the y-axis at points F and G, respectively. Calculate the area of the quadrilateral $OHFG$.

Solution:
\textbf{Solution:}

The equation of line $FH$ is: $y=\frac{3}{2}x-2$,

When $y=0$, we have $\frac{3}{2}x-2=0$, thus $x=\frac{4}{3}$,

Therefore, point $H\left(\frac{4}{3}, 0\right)$,

Since $G$ is the intersection of line $DE$ with the y-axis, thus point $G\left(0, 8\right)$,

Therefore, $OH=\frac{4}{3}$, $CF=4$, $OC=4$, $CG=OG-OC=4$,

Therefore, the area of quadrilateral $OHFG$ is the sum of the area of trapezoid $OHFC$ and the area of triangle $CFG$: ${S}_{\text{quadrilateral} OHFG}={S}_{\text{trapezoid} OHFC}+{S}_{\triangle CFG}=\frac{1}{2}\times \left(\frac{4}{3}+4\right)\times 4+\frac{1}{2}\times 4\times 4=\boxed{18\frac{2}{3}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through points $A(-2, -2)$ and $B(1, 4)$, and intersects the x-axis at point C and the y-axis at point D. What is the analytical expression of this linear function?

Solution:
\textbf{Solution}: Substituting $A(-2, -2)$ and $B(1, 4)$ into $y=kx+b$, we get
\[
\begin{cases}
-2k+b=-2 \\
k+b=4
\end{cases},
\]
solving this yields
\[
\begin{cases}
k=2 \\
b=2
\end{cases},
\]
\[
\therefore
\] the equation of the linear function is $\boxed{y=2x+2}$;


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ passes through points $A(-2, -2)$, $B(1, 4)$, and intersects the x-axis at point C, and the y-axis at point D. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} When $x=0$, $y=2x+2=2$,\\
$\therefore$ the coordinates of point D are $\left(0, 2\right)$;\\
$\therefore S_{\triangle AOB}=\frac{1}{2}\times 2\times 1+\frac{1}{2}\times 2\times 2=\boxed{3}$\\
$\therefore$ The area of $\triangle AOB$ is \boxed{3}.


Question: As shown in the figure, the line \(l_1\) passes through points \(A(0, 2)\) and \(C(6, -2)\). The coordinates of point B are \( (4, 2) \). Point P is a moving point on segment AB (point P does not coincide with point A). The line \(l_2: y = kx + 2k\) passes through point P and intersects \(l_1\) at point M. Through point P, draw \(PN \perp l_2\), intersecting \(l_1\) at point N. Find the function expression of \(l_1\)?

Solution:
\textbf{Solution:} Let the equation of ${l}_{1}$ be: $y={k}_{1}x+b$,\\
Substituting points $A(0, 2)$ and $C(6, -2)$ into the equation, we get: $\begin{cases}b=2 \\ 6{k}_{1}+b=-2\end{cases}$,\\
Solving these equations, we find: $\begin{cases}{k}_{1}=-\frac{2}{3} \\ b=2\end{cases}$,\\
Therefore, the equation of ${l}_{1}$ is: \boxed{y=-\frac{2}{3}x+2}.


Question: As shown in the figure, in the Cartesian coordinate system, the line \(l\) passes through points \(A(0, 1)\) and \(B(-2, 0)\). Find the equation of the line \(l\).

Solution:
\textbf{Solution:} Set the expression for line $l$ as: $y=kx+b$, passing through $A(0,1)$, $B(-2,0)$,\\
$\therefore \left\{\begin{array}{l}b=1 \\ -2k+b=0 \end{array}\right.$,\\
$\therefore \left\{\begin{array}{l}k=\frac{1}{2} \\ b=1 \end{array}\right.$,\\
$\therefore$ the expression for line $l$ is: $\boxed{y=\frac{1}{2}x+1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line \(l\) passes through points \(A(0, 1)\) and \(B(-2, 0)\). If point \(M(3, m)\) is on line \(l\), find the value of \(m\).

Solution:
\textbf{Solution:} Since $M(3,m)$ lies on line $l$,\\
therefore $m=\frac{1}{2}\times3+1=\boxed{\frac{5}{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l$ passes through points $A(0,1)$ and $B(-2,0)$. If $y=-x+b$ passes through point $A$ and intersects the $x$-axis at point $C$, what is the area of $\triangle ABC$?

Solution:
Solution: The line $y = -x + b$ passes through point $A(0,1)$, thus $b = 1$. Therefore, the line $y = -x + 1$ intersects the $x$-axis at point $C(1,0)$.\\
Therefore, the area of $\triangle ABC$ is $= \frac{1}{2} \times (2+1) \times 1 = \boxed{\frac{3}{2}}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects the coordinate axes at points A and B, respectively, and intersects the line $y=x$ at point C. A point Q on the line segment OA moves from point O towards point A at a constant speed of 1 unit per second. Let the movement time be $t(s)$. Connect C and Q. Find the coordinates of point C.

Solution:
\textbf{Solution:} Since the line $y=-\frac{1}{2}x+3$ intersects the line $y=x$ at point C,\\
it follows that $\begin{cases} y=-\frac{1}{2}x+3 \\ y=x \end{cases}$,\\
solving these equations yields $\begin{cases} x=2 \\ y=2 \end{cases}$,\\
therefore, the coordinates of point C are $\boxed{(2,2)}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects the coordinate axes at points A and B, and intersects the line $y=x$ at point C. A point Q on the line segment OA moves from point O to point A at a uniform speed of 1 unit per second. Let the motion time be $t(s)$. Connect C and Q. If $\triangle OQC$ is an isosceles right triangle, find the value of $t$.

Solution:
\textbf{Solution:}\\
(1) As shown in Figure 1, when $\angle CQO=90^\circ$, CQ=OQ,\\
$\because C(2,2)$,\\
$\therefore OQ=CQ=2$,\\
$\therefore t=2$;\\
(2) As shown in Figure 2, when $\angle OCQ=90^\circ$, OC=CQ, draw CM$\perp$OA at M through point C,\\
$\because C(2,2)$,\\
$\therefore CM=OM=2$,\\
$\therefore QM=OM=2$,\\
$\therefore t=2+2=4$,\\
thus the value of t is 2 or 4;\\
Therefore, the value of $t$ is $\boxed{2}$ or $\boxed{4}$.


Question: As shown in the figure, the line $y=-\frac{1}{2}x+3$ intersects with the coordinate axes at points A and B, respectively, and intersects with the line $y=x$ at point C. A point Q on line segment OA moves from point O to point A at a uniform speed of 1 unit per second, with the time of motion denoted as $t(s)$. If line segment CQ bisects the area of $\triangle OAC$, what is the functional expression of line CQ?

Solution:
\textbf{Solution:} Let $x=6$,\\
hence A$(6,0)$,\\
$\because$ CQ bisects the area of $\triangle OAC$,\\
$\therefore$ Q$(3,0)$,\\
Assume the equation of line CQ is $y=kx+b$,\\
Substituting C$(2,2)$ and Q$(3,0)$ into it, we get:\\
$\begin{cases} 3k+b=0 \\ 2k+b=2 \end{cases}$,\\
Solving the system, we find:\\
$\begin{cases} k=-2 \\ b=6 \end{cases}$,\\
$\therefore$ the function corresponding to line CQ is: \boxed{y=-2x+6}.


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

Solution:
\[ \textbf{Solution:} \]
Since point C is on the line \( l_{2} \) with an x-coordinate of 1,\\
thus substituting \( x=1 \) into \( y=2x \) gives \( y=2\times 1=2 \),\\
hence, the coordinates of point C are \( (1, 2) \).\\
Assume the equation of line \( l_{1} \) is \( y=kx+b \). Substituting the coordinates of points B and C, we get
\[
\left\{ \begin{array}{l}
3=k\times 0+b, \\
2=k\times 1+b,
\end{array} \right.
\]
from which we derive
\[
\left\{ \begin{array}{l}
k=-1, \\
b=3.
\end{array} \right.
\]
Therefore, the equation of line \( l_{1} \) is \(\boxed{y=-x+3}\).


Question: As shown in the Cartesian coordinate system, line $l_1$ intersects the x-axis at point A and the y-axis at point B, with the coordinates of point B being $(0, 3)$. Line $l_2$: $y=2x$ intersects line $l_1$ at point C, with the x-coordinate of point C being 1. If point D is a point on the y-axis, and the area of $\triangle OCD$ is $\frac{2}{3}$ of the area of $\triangle AOC$, find the coordinates of point D.

Solution:
\textbf{Solution:} Given that point A is the intersection of line $l_{1}$ and the $x$-axis,\\
therefore, substituting $y=0$ into $y=-x+3$, we find $x=3$,\\
therefore, the coordinates of point A are $(3, 0)$.\\
As shown in the figure, draw $CM\perp x$-axis through point C, with M being the foot of the perpendicular; draw $CN\perp y$-axis through point C, with N being the foot of the perpendicular.\\
Given the coordinates of point C as $(1, 2)$, we obtain $CM=2$, $CN=1$,\\
Let the coordinates of point D be $(0, y)$.\\
According to the problem, we have $\frac{1}{2}\cdot OD\cdot CN=\frac{2}{3}\times \frac{1}{2}\cdot OA\cdot CM$,\\
which is $\frac{1}{2}\times |y|\times 1=\frac{2}{3}\times \frac{1}{2}\times 3\times 2$.\\
Solving this, we get $|y|=4$, that is $y=\pm 4$.\\
Therefore, the coordinates of point D are $\boxed{(0, 4)}$ or $\boxed{(0, -4)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1$ intersects the x-axis at point A and the y-axis at point B, with the coordinates of point B being $\left(0, 3\right)$. The line $l_2$: $y=2x$ intersects the line $l_1$ at point C, where the x-coordinate of point C is 1. Is there a point E in the plane such that the quadrilateral with vertices O, A, C, and E is a parallelogram? If such a point exists, directly write down the coordinates of the point E that meet the conditions.

Solution: \textbf{Solution:} There exists a point E with coordinates $\boxed{(4,2)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y =\frac{3}{4}x +\frac{3}{2}$ intersects the x-axis at point A and passes through point B $(2, a)$. There is a moving point P on the y-axis, and a moving point M on line BC, where point C is $(3, 0)$. What is the value of $a$?

Solution: \textbf{Solution:} From the given problem, we have \(a=\boxed{3}\).


Question: As shown in the figure, in the Cartesian coordinate system, the line $y =\frac{3}{4}x +\frac{3}{2}$ intersects the x-axis at point A, and passes through point B (2, a). There is a moving point P on the y-axis, and a moving point M on the line BC, with C (3, 0). If $\triangle APM$ is an isosceles right triangle with segment AM as its hypotenuse, what are the coordinates of point M?

Solution: \textbf{Solution:} The coordinates of point M are $\boxed{\left(\frac{7}{2}, -\frac{3}{2}\right)}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle BAD=121^\circ$, $\angle B=\angle D=90^\circ$, points $M$ and $N$ are respectively on $BC$ and $CD$. When $\angle MAN=\angle C$, what is the degree measure of $\angle AMN+\angle ANM$?

Solution:
\textbf{Solution:} Given that, $\triangle MAN$ and $\triangle C$ are similar,\\
$\because$ in any triangle, the sum of the interior angles equals $180^\circ$,\\
$\therefore \angle AMN + \angle ANM = 180^\circ - \angle MAN = 180^\circ - \angle C = \boxed{180^\circ - \angle C}$.


Question: As shown in the figure, in the quadrilateral $ABCD$, $\angle BAD=121^\circ$, $\angle B=\angle D=90^\circ$, and points $M,N$ are on $BC,CD$ respectively. When the perimeter of $\triangle AMN$ is minimized, what is the sum of $\angle AMN+\angle ANM$ in degrees?

Solution:
\textbf{Solution:} It is known from the problem statement that when the perimeter of $\triangle AMN$ is minimized, $\angle AMN + \angle ANM = 180^\circ - \angle MNA$.\\
Since $\angle MNA$ is the external angle of $\triangle AMN$,\\
$\therefore \angle AMN + \angle ANM = 180^\circ - \angle MNA = \boxed{180^\circ}$.


Question: As shown in the figure, in $\triangle ABC$ and $\triangle DCE$, we have $AC = DE$, $\angle B = \angle DCE = 90^\circ$, and the points A, C, D are collinear in that order, with $AB \parallel DE$. Therefore, $\triangle ABC \cong \triangle DCE$. Connecting $AE$, when $BC = 5$ and $AB = 12$, find the length of $AD$.

Solution:
\textbf{Solution:} Since $\triangle ABC \cong \triangle DCE$,\\
it follows that $BC=EC$, $AC=DE$, and $AB=CD$.\\
In the right triangle $\triangle ABC$, with $AB=12$ and $BC=5$,\\
by the Pythagorean theorem, we have $AC=\sqrt{AB^2+BC^2}=\sqrt{12^2+5^2}=13$,\\
thus, $AD=AC+CD=13+12=\boxed{25}$.


Question: Given: As shown in the diagram, the graph of the direct proportion function $y=2x$ intersects with the graph of the linear function $y=-\frac{2}{3}x+4$ at the point $A\left(\frac{3}{2}, 3\right)$, and the graph of the linear function $y=-\frac{2}{3}x+4$ intersects with the $x$-axis at point $B$. Find the coordinates of point $B$.

Solution:
\textbf{Solution}: Given that the equation of the linear function is $y=-\frac{2}{3}x+4$,\\
setting $y=0$ yields: $0=-\frac{2}{3}x+4$,\\
solving for $x$, we get: $x=6$,\\
$\therefore$ the coordinates of point $B$ are $\boxed{(6, 0)}$.


Question: Given: As shown in the diagram, the graph of the directly proportional function $y=2x$ and the graph of the linear function $y=-\frac{2}{3}x+4$ intersect at point $A\left(\frac{3}{2}, 3\right)$, and the graph of the linear function $y=-\frac{2}{3}x+4$ intersects the $x$-axis at point $B$. Find the area of $\triangle AOB$.

Solution:
\textbf{Solution:} As shown in Figure 1 \\
$\because A\left(\frac{3}{2}, 3\right)$, $B(6, 0)$ \\
$\therefore OB = 6$, the height on side OB is 3, \\
$\therefore S_{\triangle AOB} = \frac{1}{2} \times 6 \times 3 = \boxed{9}$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $D$ is the midpoint of side $BC$, $\angle B=25^\circ$. Find the measure of $\angle DAC$?

Solution:
\textbf{Solution:} Since $AB=AC$,\\
it follows that $\angle C=\angle B=25^\circ$,\\
Since $AD\perp BC$,\\
it implies $\angle ADB=90^\circ$,\\
thus $\angle DAC=90^\circ-25^\circ=\boxed{65^\circ};$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, $D$ is the midpoint of side $BC$, $\angle B=25^\circ$. If $AB=13$ and $AD=5$, what is the length of $BC$?

Solution:
\textbf{Solution:} Given that $\because AB=13$ and $AD=5$,\\
$\therefore BD= \sqrt{AB^{2}-AD^{2}}=\sqrt{13^{2}-5^{2}}=12$,\\
$\because AD\perp BD$,\\
$\therefore BC=2BD=2\times 12=\boxed{24}.$


Question: As shown in the figure, in $\triangle ABC$, $AB=AC$, the perpendicular bisector of $AB$, $DE$, intersects $AC$ at point $D$ and $AB$ at point $E$, $\angle C=75^\circ$. What is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Given that $AB=AC$ and $\angle C=75^\circ$, \\
thus $\angle ABC=\angle C=75^\circ$, \\
therefore $\angle A=180^\circ-75^\circ-75^\circ=30^\circ$, \\
thus the measure of $\angle A$ is $\boxed{30^\circ}$.


Question: Could you please provide the text you need to be translated into English and formatted in LaTeX?

Solution:
Solution: Since $DE$ is the perpendicular bisector of $AB$, \\
$\therefore DA=DB$, \\
$\therefore \angle DBA=\angle A=30^\circ$, \\
$\therefore \angle CBD=\angle ABC-\angle ABD=75^\circ-30^\circ=\boxed{45^\circ}$.


Question: As shown in the diagram, within $ \triangle ABC$, $ AD$ and $ AF$ are the median and the altitude of $ \triangle ABC$, respectively, and $ BE$ is the angle bisector of $ \triangle ABD$. Given that the area of $ \triangle ABC$ is 40 and $ BD=5$, find the length of $ AF$.

Solution:
\textbf{Solution:} Since $AD$ is the median of $\triangle ABC$ and $BD=5$,\\
it follows that $BC=2BD=2\times 5=10$.\\
Given $AF\perp BC$ and the area of $\triangle ABC$ is $40$,\\
we have $\frac{1}{2}BC\cdot AF=\frac{1}{2}\times 10\cdot AF=40$,\\
which leads to $AF=\boxed{8}$.


Question: As shown in the figure, in $\triangle ABC$, $AD$ and $AF$ are respectively the median and height of $\triangle ABC$, $BE$ is the angle bisector of $\triangle ABD$. If $\angle BED=40^\circ$ and $\angle BAD=25^\circ$, find the measure of $\angle BAF$.

Solution:
\textbf{Solution:} In $\triangle ABE$, $\angle BED$ is an exterior angle, and $\angle BED=40^\circ$, $\angle BAD=25^\circ$,\\
$\therefore \angle ABE=\angle BED-\angle BAD=40^\circ-25^\circ=15^\circ$.\\
Since $BE$ is the angle bisector of $\triangle ABD$,\\
$\therefore \angle ABC=2\angle ABE=2\times 15^\circ=30^\circ$.\\
Since $AF\perp BC$,\\
$\therefore \angle AFB=90^\circ$,\\
In $Rt\triangle ABF$, $\angle BAF=90^\circ-\angle ABC=90^\circ-30^\circ=\boxed{60^\circ}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C = 90^\circ$, $AB = 4$, and $AC = \sqrt{7}$. What is the length of $BC$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle C=90^\circ$, $AB=4$, $AC=\sqrt{7}$.\\
$BC=\sqrt{AB^2-AC^2}=\sqrt{4^2-(\sqrt{7})^2}=\boxed{3}$.


Question: Given that in $\triangle ABC$, $\angle C=90^\circ$, $AB=4$, $AC=\sqrt{7}$. Find $\sin A$.

Solution: \textbf{Solution:} We have $\sin A= \frac{BC}{AB}=\boxed{\frac{3}{4}}$.


Question: As shown in the figure, a mathematics interest group measures the height of an ancient tree $BH$ and a teaching building $CG$. First, at point $A$, they use an altimeter with a height of $1.5$ meters to measure the angle of elevation $\angle HDE=45^\circ$ of the top of the ancient tree $H$. At this moment, the top of the teaching building $G$ is exactly in the line of sight $DH$. Then, they move forward $7$ meters to point $B$ and measure the angle of elevation $\angle GEF=60^\circ$ of the top of the teaching building $G$. Points $A$, $B$, $C$ are on the same horizontal line. Calculate the height of the ancient tree $BH$ in meters.

Solution:
\textbf{Solution:} Solution: According to the problem statement: Quadrilateral ABED is a rectangle, so we get $DE=AB=7$ meters.\\
In $\triangle DEH$, $\because \angle EDH=45^\circ$,\\
$\therefore HE=DE=7$ meters,\\
$\therefore BH=EH+BE=\boxed{8.5}$ meters.


Question: As shown in the figure, in order to measure the heights of an ancient tree BH and a teaching building CG, a math interest group first takes a measurement at point A with a theodolite that is 1.5 meters tall. They measure the angle of elevation to the top of the ancient tree H to be $\angle HDE=45^\circ$. At this moment, the top of the teaching building G is exactly in line with sight DH. Then, moving forward 7 meters to point B, they measure the angle of elevation to the top of the teaching building G to be $\angle GEF=60^\circ$. Points A, B, and C are on the same horizontal line. Calculate the height of the teaching building CG. (Reference data: $\sqrt{2} \approx 1.4$, $\sqrt{3} \approx 1.7$)

Solution:
\textbf{Solution:} Let $HJ \perp CG$ at $G$. Then $\triangle HJG$ is an isosceles triangle, and quadrilateral $BCJH$ is a rectangle. Let $HJ=GJ=BC=x$. In $\triangle EFG$, $\tan 60^\circ = \frac{GF}{EF}$,

\[
\therefore \sqrt{3} = \frac{7+x}{x},
\]

\[
\therefore x = \frac{7}{2}(\sqrt{3}+1),
\]

\[
\therefore GF = \sqrt{3}x \approx 16.45,
\]

\[
\therefore CG = CF + FG = 1.5 + 16.45 \approx \boxed{18.0} \text{ meters.}
\]


Question: As shown in the figure, in $ \triangle PMN$, $\angle MPN=90^\circ$, $PN=4$, $\sin\angle PMN=\frac{4}{5}$. What are the lengths of $BC$ and $FG$?

Solution:
\textbf{Solution:} Let the height from the hypotenuse of $\triangle PMN$ be $h$,\\
since the orthographic projection of the right prism is given,\\
thus $BC=MN$, $FG=h$, $AB=EF$,\\
$\sin\angle PMN=\frac{PN}{MN}=\frac{4}{5}$,\\
Solving gives: $MN=BC=\boxed{5}$,\\
thus $PM=3$,\\
therefore the area of $\triangle PMN=\frac{1}{2}PM\cdot PN=\frac{1}{2}MN\cdot h$,\\
thus $\frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\cdot h$,\\
Solving gives: $h=FG=\boxed{\frac{12}{5}}$.


Question: The three views of a geometric body are interrelated. Given the three views of a right triangular prism as shown in the figure, in $\triangle PMN$, $\angle MPN=90^\circ$, $PN=4$, and $\sin\angle PMN=\frac{4}{5}$. If the rectangles in the front view and the left view are similar, what is the length of $AB$?

Solution:
\textbf{Solution:} Since rectangle ABCD is similar to rectangle EFGH, and AB=EF, \\
it follows that $\frac{AB}{FG} = \frac{BC}{EF}$, which means $\frac{AB}{\frac{12}{5}} = \frac{5}{EF}$. Solving this gives: $AB = \boxed{2\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\sin\angle BAC=\frac{5}{13}$, $AB=13$, $AC=7.2$, $BD\perp AC$ with the foot of the perpendicular at point D, point E is the midpoint of BD, and AE intersects BC at point F. Find the tangent of $\angle CBD$?

Solution:
\textbf{Solution:} Since $BD\perp AC$,\\
it follows that $\angle D=90^\circ$,\\
therefore $\sin\angle BAD=\frac{BD}{AB}=\frac{5}{13}$,\\
hence $BD=\frac{5}{13}AB=5$,\\
thus $AD=\sqrt{AB^{2}-BD^{2}}=12$,\\
therefore $CD=AD-AC=4.8$,\\
hence $\tan\angle CBD=\frac{CD}{BD}=\frac{4.8}{5}=\boxed{\frac{24}{25}}$.


Question: As shown in the figure, in $\triangle ABC$, $\sin\angle BAC=\frac{5}{13}$, $AB=13$, $AC=7.2$, $BD\perp AC$ with the foot of the perpendicular at point $D$, point $E$ is the midpoint of $BD$, and $AE$ intersects $BC$ at point $F$. Find the value of $\frac{BF}{CF}$.

Solution:
\textbf{Solution:} As shown in the diagram, draw $FH\parallel BD$ through point F and intersect AD at H, \\
$\therefore \angle FHC=\angle EDC=90^\circ$, $\angle HFC=\angle DBC$, \\
$\therefore \triangle CHF\sim \triangle CDE$, $\triangle AHF\sim \triangle ADE$, $\tan\angle HFC=\tan\angle CBD=\frac{CH}{FH}=\frac{24}{25}$, \\
$\therefore \frac{HF}{BD}=\frac{CF}{CB}$, $CH=\frac{24}{25}FH$, $\frac{AH}{AD}=\frac{HF}{DE}$, \\
Let $FH=x$, then $CH=\frac{24}{25}x$, $AH=AC+CH=7.2+\frac{24}{25}x$, \\
Since E is the midpoint of BD, \\
$\therefore DE=\frac{1}{2}BD=\frac{5}{2}$, \\
$\therefore \frac{7.2+\frac{24}{25}x}{12}=\frac{x}{\frac{5}{2}}$, \\
$\therefore$ solving yields $x=\frac{15}{8}$, \\
$\therefore FH=\frac{15}{8}$, \\
$\therefore \frac{\frac{15}{8}}{5}=\frac{CF}{CB}$, that is $\frac{CF}{CB}=\frac{3}{8}$, \\
$\therefore CF=\frac{3}{8}CB$, \\
$\therefore BF=\frac{5}{8}CB$, \\
$\therefore \frac{BF}{CF}=\boxed{\frac{5}{3}}$.


Question: As shown in the diagram, AB is a diameter of circle $O$, point P is on circle $O$, and $PA=PB$. Point M is outside circle $O$, and $MB$ is tangent to circle $O$ at point B. Connect $OM$, and draw $AC\parallel OM$ through point A intersecting circle $O$ at point C, then connect $BC$ and intersect $OM$ at point D. $MC$ is a tangent to circle $O$; if $OB=\frac{15}{2}$ and $BC=12$, connect $PC$. Find the length of $PC$.

Solution:
\textbf{Solution:} Given that AB is the diameter of circle O,\\
it follows that $\angle ACB=\angle APB=90^\circ$,\\
given that $OB=\frac{15}{2}$,\\
it follows that $AB=15$,\\
thus $PA=PB=\frac{15\sqrt{2}}{2}$,\\
given that $BC=12$,\\
it follows that $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{15^{2}-12^{2}}=9$,\\
draw line AH perpendicular to PC at point H,\\
thus $\angle AHC=\angle AHP=90^\circ$,\\
given that $\angle ACH=\angle ABP=45^\circ$,\\
it follows that $AH=CH=AC\times \sin 45^\circ=9\times \frac{\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}$,\\
$PH=\sqrt{PA^{2}-AH^{2}}=\sqrt{(\frac{15\sqrt{2}}{2})^{2}-(\frac{9\sqrt{2}}{2})^{2}}=6\sqrt{2}$,\\
thus $PC=PH+CH=\boxed{\frac{21\sqrt{2}}{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, $OB=5$, $\sin\angle AOB= \frac{3}{5}$, and the coordinate of point $A$ is $(10,0)$. What are the coordinates of point $B$?

Solution:
\textbf{Solution:} Draw $BC\perp OA$ at $C$ through point $B$. In $\triangle BOC$, $\sin\angle AOB= \frac{BC}{OB}=\frac{3}{5}$,\\
therefore $BC=3$. According to the problem, $OB^2=BC^2+OC^2$,\\
therefore $OC=\sqrt{OB^2-BC^2}=\sqrt{5^2-3^2}=4$,\\
therefore the coordinates of point $B$ are $\boxed{(4,3)}$.


Question: As shown in the figure, in the Cartesian coordinate system, $OB=5$, $\sin\angle AOB= \frac{3}{5}$, and the coordinates of point $A$ are $(10,0)$. What is the value of $\sin\angle OAB$?

Solution:
\textbf{Solution:} Given the information in the problem, the coordinates of point A are (10, 0),\\
$\therefore OA=10$,\\
$\because OC=4$,\\
$\therefore AC=6$,\\
In $\triangle ABC$, by Pythagoras' theorem we have\\
$AB=\sqrt{AC^2+BC^2}=\sqrt{6^2+3^2}=3\sqrt{5}$,\\
$\therefore \sin\angle OAB=\frac{BC}{AB}=\frac{3}{3\sqrt{5}}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, the line $y=-\frac{1}{3}x+1$ intersects the x-axis and y-axis at points A and B, respectively. Find the coordinates of points A and B.

Solution:
\textbf{Solution:} For $y = -\frac{1}{3}x + 1$, let $x = 0$, then $y = 1$, let $y = 0$, that is $-\frac{1}{3}x + 1 = 0$, solving gives $x = 3$,\\
hence the coordinates of points A and B are \boxed{(3,0)} and \boxed{(0,1)}, respectively.


Question: The straight line $y=-\frac{1}{3}x+1$ intersects the x-axis and y-axis at points A and B, respectively. Given that the coordinates of point G are $(2,7)$, the straight line BG is drawn through points G and B, and AG is connected. What is the value of $\tan(\angle AGB)$?

Solution:
\textbf{Solution:} Given the coordinates of A, B, and G, $BG^{2}=2^{2}+(7-1)^{2}=40$, similarly $AB^{2}=10$, $AG^{2}=50$, thus $AG^{2}=AB^{2}+BG^{2}$,\\
$\therefore \triangle ABG$ is a right triangle,\\
$\therefore \tan\angle AGB=\frac{AB}{BG}=\frac{\sqrt{10}}{\sqrt{40}}=\boxed{\frac{1}{2}}$.


Question: The line $y=-\frac{1}{3}x+1$ intersects the x-axis and the y-axis at points A and B, respectively. Does there exist a point Q on the line BG such that the triangle formed by points A, B, and Q is similar to $\triangle AOB$? If such a point exists, find the coordinates of point Q; if it does not exist, explain why.

Solution:
\textbf{Solution:} Let the equation of line BG be $y=kx+b$, then
\[
\left\{
\begin{array}{l}
7=2k+b \\
b=1
\end{array}
\right.
\]
Solving gives
\[
\left\{
\begin{array}{l}
k=3 \\
b=1
\end{array}
\right.
\]
Thus, the equation of line BG is $y=3x+1$. Let point Q be $(m,3m+1)$,

(1) When $\triangle ABQ \sim \triangle AOB$,
then $\frac{AB}{AO}=\frac{BQ}{OB}$, i.e., $\frac{\sqrt{10}}{3}=\frac{\sqrt{m^2+(3m+1-1)^2}}{1}$, solving gives $m=\pm \frac{1}{3}$,
$\therefore$ $Q_1\left(\frac{1}{3}, 2\right)$, $Q_2\left(-\frac{1}{3}, 0\right)$

(2) When $\triangle ABQ \sim \triangle BOA$,
$\frac{AB}{OB}=\frac{BQ}{AO}$, i.e., $\frac{\sqrt{10}}{1}=\frac{\sqrt{m^2+(3m+1-1)^2}}{3}$, solving gives $m=\pm3$,
$\therefore$ $Q_3\left(3, 10\right)$, $Q_4\left(-3, -8\right)$

Hence, the coordinates of point Q are $\boxed{\left(\frac{1}{3}, 2\right)}$ or $\boxed{\left(-\frac{1}{3}, 0\right)}$ or $\boxed{\left(3, 10\right)}$ or $\boxed{\left(-3, -8\right)}$.


Question: As shown in the figure, in $\triangle ABC$, CD is the median to side AB, $\angle B=45^\circ$, $\tan\angle ACB=3$, $AC= \sqrt{10}$. Find: the area of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the diagram, construct $AH \perp BC$ at $H$. In $\triangle ACH$, since $\tan \angle ACB = 3$ and $AC= \sqrt{10}$, let $CH=x$ and $AH=3x$. According to the Pythagorean theorem, we find $AC= \sqrt{10}x$, thus $CH=1$ and $AH=3$. In $\triangle ABH$, $\angle B=45^\circ$, so $BH=AH=3$,\\
therefore $S_{\triangle ABC}= \frac{1}{2} \times 4 \times 3 = \boxed{6}$;


Question: As shown in the figure, in $\triangle ABC$, $CD$ is the median to side $AB$, $\angle B=45^\circ$, $\tan \angle ACB=3$, and $AC= \sqrt{10}$. Find the value of $\sin \angle ACD$?

Solution:
\textbf{Solution:} Draw $DF \perp BC$ at $F$, and $DE$ perpendicular to $AC$ at $E$. \\
Since the area of $\triangle ACD= \frac{1}{2} \times \sqrt{10} \times DE=3$, \\
it follows that $DE= \frac{3}{5}\sqrt{10}$, \\
Because $AH \perp BC$, $DF \perp BC$, and $CD$ is the median to side $AB$, \\
it follows that $DF= \frac{1}{2}AH= \frac{3}{2}$, \\
Therefore, $BF=DF= \frac{3}{2}$, \\
In $\triangle CDF$, $CD= \sqrt{DF^{2}+FC^{2}}=\sqrt{\left(\frac{3}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}}=\frac{\sqrt{34}}{2}$, \\
Therefore, in $\triangle CDE$, $\sin\angle ACD= \frac{DE}{CD}=\boxed{\frac{6\sqrt{85}}{85}}$.


Question: As shown in the figure, in the right-angled triangle $\triangle ABO$ where $\angle O = 90^\circ$, a circle with center $O$ and radius $OB$ intersects $AB$ at point $C$, and intersects $OA$ at point $D$. If $C$ is the midpoint of $AB$, what is the measure of $\angle A$?

Solution:
\textbf{Solution:} As shown in the diagram, connect OC, where $\angle AOB=90^\circ$, and C is the midpoint of AB, \\
therefore $OC=AC=BC$, \\
since $OB=OC$, \\
hence $\triangle BOC$ is an equilateral triangle, \\
therefore $\angle B=60^\circ$, \\
thus $\angle A=90^\circ-60^\circ=\boxed{30^\circ}$.


Question: As shown in the figure, in the right-angled triangle $\mathrm{Rt}\triangle ABO$, $\angle O=90^\circ$. A circle with point $O$ as its center and $OB$ as its radius intersects $AB$ at point $C$ and intersects $OA$ at point $D$. If $BO=2$ and $AO=4$, find the length of $BC$.

Solution:
\textbf{Solution:} As shown in the figure, draw $OQ\perp AB$ at Q, and $\angle AOB=90^\circ$,\\
$\therefore \angle BOQ+\angle B=90^\circ=\angle B+\angle A=90^\circ$,\\
$\therefore \angle BOQ=\angle A$,\\
$\because OB=2$, $AO=4$,\\
$\therefore AB=\sqrt{2^2+4^2}=2\sqrt{5}$,\\
$\therefore \sin A=\frac{OB}{AB}=\frac{2}{2\sqrt{5}}=\frac{\sqrt{5}}{5}$,\\
$\therefore \sin\angle BOQ=\frac{BQ}{OB}=\frac{BQ}{2}=\frac{\sqrt{5}}{5}$,\\
$\therefore BQ=\frac{2\sqrt{5}}{5}$,\\
$\because OQ\perp AB$,\\
$\therefore BC=2BQ=\boxed{\frac{4\sqrt{5}}{5}}.$


Question: As shown in the diagram, rectangle $OABC$ has vertices $A$ and $C$ located on the positive semiaxes of the $x$-axis and $y$-axis, respectively, with the coordinate of point $B$ being $(2\sqrt{3}, 4)$. The graph of the linear function $y=-\frac{\sqrt{3}}{3}x+b$ intersects sides $OC$, $AB$, and the $x$-axis at points $D$, $E$, and $F$, respectively, with $\angle DFO=30^\circ$, and it satisfies $OD=BE$, where point $M$ is a moving point on line segment $DF$. What is the value of $b$?

Solution:
\textbf{Solution:} Given $y=-\frac{\sqrt{3}}{3}x+b$, let $x=0$, then $y=b$,\\
$\therefore$ the coordinates of point $D$ are $(0, b)$,\\
$\because OD=BE$ and $B(2\sqrt{3}, 4)$,\\
$\therefore E(2\sqrt{3}, 4-b)$,\\
Substituting $E(2\sqrt{3}, 4-b)$ into $y=-\frac{\sqrt{3}}{3}x+b$ we get:\\
$4-b=-\frac{\sqrt{3}}{3}\times 2\sqrt{3}+b$,\\
Solving for $b$ yields: $b=\boxed{3}$;


Question: As shown in the diagram, the vertices $A$ and $C$ of the rectangle $OABC$ are on the positive half-axes of $x$ and $y$, respectively, and the coordinates of point $B$ are $(2\sqrt{3}, 4)$. The graph of the linear function $y=-\frac{\sqrt{3}}{3}x+b$ intersects with sides $OC$, $AB$, and the $x$ axis at points $D$, $E$, and $F$, respectively, with $\angle DFO=30^\circ$, and satisfying $OD=BE$. Point $M$ is a moving point on the segment $DF$. Connect $OM$. If the ratio of the area of $\triangle ODM$ to the area of the quadrilateral $OAEM$ is $1:3$, what are the coordinates of point $M$?

Solution:
\textbf{Solution:} From (1), we obtain the linear function $y=-\frac{\sqrt{3}}{3}x+3$, with points $D(0, 3)$ and $E(2\sqrt{3}, 1)$,\\
$\therefore OD=3$, $AE=1$, $OA=2\sqrt{3}$,\\
$\therefore S_{\text{quadrilateral }OADE}=\frac{1}{2}(OD+AE)\cdot OA=\frac{1}{2}\times (3+1)\times 2\sqrt{3}=4\sqrt{3}$,\\
$\because$ the area ratio of $\triangle ODM$ to the quadrilateral $OAEM$ is $1:3$,\\
$\therefore$ the area ratio of $\triangle ODM$ to the quadrilateral $OADE$ is $1:4$,\\
$\therefore S_{\triangle ODM}=\frac{1}{4}S_{\text{quadrilateral }OADE}=\sqrt{3}$,\\
Let the x-coordinate of point $M$ be $a$, hence $\frac{1}{2}\times 3a=\sqrt{3}$,\\
Solving this yields $a=\frac{2\sqrt{3}}{3}$,\\
Substituting $x=\frac{2\sqrt{3}}{3}$ into $y=-\frac{\sqrt{3}}{3}x+3$ gives $y=\frac{7}{3}$,\\
$\therefore \boxed{M\left(\frac{2\sqrt{3}}{3}, \frac{7}{3}\right)}$.


Question: As shown in the diagram, the vertices $A$ and $C$ of the rectangle $OABC$ are located on the positive semi-axes of the $x$ and $y$ axes, respectively, and the coordinate of point $B$ is $(2\sqrt{3}, 4)$. The graph of the linear function $y=-\frac{\sqrt{3}}{3}x+b$ intersects the sides $OC$, $AB$, and the $x$ axis at points $D$, $E$, and $F$, respectively, with $\angle DFO=30^\circ$, and it satisfies $OD=BE$. Point $M$ is a movable point on segment $DF$. Find the minimum value of $OM+\frac{1}{2}MF$.

Solution:
\textbf{Solution:} As shown in the diagram, draw $MN\perp x$ axis through point $M$, intersecting at point $N$,\\
$\because \angle DFO=30^\circ$,\\
$\therefore MN=\frac{1}{2}MF$,\\
$\therefore OM+\frac{1}{2}MF=OM+MN$,\\
Construct the symmetric point ${O}^{\prime}$ of point $O$ with respect to the linear function, and let $OO’$ intersect line $DF$ at point $Q$. Draw a perpendicular to the $x$ axis from point ${O}^{\prime}$, intersecting the $x$ axis at point ${N}^{\prime}$,\\
$\therefore OM={O}^{\prime}M$,\\
$\therefore OM+\frac{1}{2}MF=OM+MN={O}^{\prime}M+MN$,\\
When ${O}^{\prime}$, $M$, and $N$ are on the same line, ${O}^{\prime}M+MN$ is minimized,\\
That is, the minimum value of $OM+\frac{1}{2}MF=OM+MN={O}^{\prime}M+MN$ is ${O}^{\prime}{N}^{\prime}$,\\
$\because \angle DFO=30^\circ$,\\
$\therefore \angle ODF=60^\circ$, $\angle DOQ=30^\circ$, $\angle {O}^{\prime}O{N}^{\prime}=60^\circ$,\\
In $Rt\triangle ODQ$, $OQ=OD\cdot \sin60^\circ=3\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$,\\
$\therefore OO^{\prime}=2OQ=3$,\\
In $Rt\triangle O{N}^{\prime}{O}^{\prime}$, ${O}^{\prime}{N}^{\prime}=OO^{\prime}\sin60^\circ=3\times \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$,\\
Therefore, the minimum value of $OM+\frac{1}{2}MF$ is $\boxed{\frac{3\sqrt{3}}{2}}$.


Question: As shown in Figure 1, in the Cartesian coordinate system xOy, point $A(-1, 0)$ and point $B(4, 0)$ are given. Point $C$ is located on the negative semi-axis of the y-axis. The lines $AC$ and $BC$ are connected and it is given that $\angle ACO = \angle CBO$. What is the equation of the line $BC$?

Solution:
\textbf{Solution:} Given $\because$ $\angle AOC = \angle BOC = 90^\circ$ and $\angle ACO = \angle CBO$,
\\
$\therefore$ $\triangle AFC \sim \triangle CFG$,
\\
$\therefore$ $\frac{OA}{OC} = \frac{OC}{OB}$,
\\
$\because$ OA = 1 and OB = 4,
\\
$\therefore$ OC = 2, and the coordinates of point C are (0, $-2$),
\\
Let the equation of line BC be $y = kx - 2$,
\\
Substituting point $B(4, 0)$ into the equation yields $0 = 4k - 2$,
\\
Solving for $k$ gives: $k = \frac{1}{2}$,
\\
$\therefore$ the equation of line BC is $\boxed{y = \frac{1}{2}x - 2}$.


Question: As shown in Figure 1, in the rhombus ABCD, AC is a diagonal, and AB = AC = 6. Points E and F are moving points on sides AB and BC, respectively, satisfying AE = BF. Lines AF and CE intersect at point G. Find the measure of $\angle CGF$.

Solution:
\textbf{Solution:} Given that quadrilateral ABCD is a rhombus, and AB=AC, \\
$\therefore$ AB=AC=BC=AD=CD, \\
$\therefore$ $\triangle ABC$, $\triangle ACD$ are equilateral triangles. \\
$\therefore$ $\angle ABC=\angle CAE$, in $\triangle ABF$ and $\triangle CAE$, \\
$\left\{\begin{array}{l}
AB=CA\\
\angle ABF=\angle CAE\\
BF=AE
\end{array}\right.$ \\
$\therefore$ $\triangle ABF\cong \triangle CAE$ (SAS), \\
$\therefore$ $\angle BAF=\angle ACE$, \\
$\therefore$ $\angle CGF=\angle GAC+\angle ACG=\angle GAC+\angle BAF=\angle BAC=60^\circ$; \\
Hence, the measure of $\angle CGF$ is $\boxed{60^\circ}$.


Question: As shown in Figure 1, in rhombus $ABCD$, $AC$ is the diagonal, and $AB=AC=6$. Points $E$ and $F$ are points on edges $AB$ and $BC$, respectively, satisfying $AE=BF$. Lines $AF$ and $CE$ intersect at point $G$. As shown in Figure 2, draw $DH\perp CE$ meeting $CE$ at point $H$. If $CF=4$ and $AF=2\sqrt{7}$, determine the value of $GH$.

Solution:
\textbf{Solution:} As shown in the diagram, extend GA to point K so that $AK = CG$. \\
$\because \angle BAF = \angle ACE$, \\
$\therefore \angle FAC = \angle GCF$, \\
$\because \angle GFC = \angle AFC$, \\
$\therefore \triangle AFC \sim \triangle CFG$, \\
$\therefore \frac{FG}{FC} = \frac{FC}{FA} = \frac{CG}{AC}$, \\
$\because CF = 4$, $AF = 2\sqrt{7}$, $AC = 6$, \\
$\therefore GF = \frac{8\sqrt{7}}{7}$, $AG = \frac{6\sqrt{7}}{7}$, $CG = \frac{12\sqrt{7}}{7}$, \\
$\because \angle FGC = 60^\circ$, $\angle ADC = 60^\circ$, \\
$\therefore \angle AGC + \angle ADC = 180^\circ$, \\
$\therefore \angle GAD + \angle GCD = 180^\circ$, \\
$\because \angle KAD + \angle GAD = 180^\circ$, \\
$\therefore \angle KAD = \angle GCD$, \\
Moreover, $\because DC = DA$, \\
$\therefore \triangle ADK \cong \triangle CDG$ (SAS), \\
$\therefore DK = DG$, $\angle KDA = \angle GDC$, \\
$\therefore \angle KDG = \angle ADC = 60^\circ$, \\
$\therefore \triangle DKG$ is an equilateral triangle, \\
$\therefore \angle AGD = \angle DGH = 60^\circ$, $DG = KG = AK + AG = AG + CG = \frac{18\sqrt{7}}{7}$, \\
$\because DH \perp CE$, $\angle DGH = 60^\circ$, \\
$\therefore GH = \frac{1}{2}DG = \boxed{\frac{9\sqrt{7}}{7}}$;


Question: As shown in the figure, in rhombus $ABCD$, $AC$ is a diagonal, and $AB=AC=6$. Points $E$ and $F$ are on sides $AB$ and $BC$ respectively, satisfying $AE=BF$. The segments $AF$ and $CE$ intersect at point $G$. Point $O$ is the midpoint of segment $CE$, and segment $EO$ is rotated clockwise by $60^\circ$ around point $E$ to obtain segment $EM$. When $\triangle MAC$ forms an isosceles triangle, please directly state the length of $AE$.

Solution:
Solution:

(1) If $AM=MC$, then $\triangle MAC$ is an isosceles triangle. In this case, let point $P$ be the midpoint of $AC$, and connect $OP$, $OM$, and $BM$,

$\because \angle MEO=60^\circ$, $EO=EM$,

$\therefore \triangle OEM$ is an equilateral triangle,

$\because \angle FGC=60^\circ$,

$\therefore \angle MEO=\angle FGC$,

$\therefore ME\parallel AF$,

$\because O$ is the midpoint of $CE$, $P$ is the midpoint of $AC$,

$\therefore OP$ is the midline of $\triangle AEC$, $OP\parallel AB$,

$\because \triangle ABC$ is an equilateral triangle, $\triangle MAC$ is an isosceles triangle, $P$ is the midpoint of $AC$,

$\therefore from$ "three lines coincide," points $B$, $M$, $P$ are collinear, and $BP\perp AC$, $AP=PC= \frac{1}{2}AC=3$, $\angle ABP= \frac{1}{2} \angle ABC=30^\circ$,

$\because \triangle OEM$ is an equilateral triangle,

$\therefore OE=OM$, $\angle OEM=\angle OME$,

$\because OE=OC$,

$\therefore OM=OC$, $\angle OMC=\angle OCM$,

$\therefore \angle OEM+\angle OCM=\angle OME+\angle OMC$,

that is: $\angle OEM+\angle OCM=\angle EMC$,

$\therefore \angle EMC=90^\circ$, $CM\perp EM$,

$\therefore in \triangle CEM$, $\angle ECM=90^\circ-60^\circ=30^\circ$,

At this point, as shown in the figure, rotate $\triangle AEC$ counterclockwise around point $C$ through $60^\circ$ until $\triangle BNC$, connect $MN$,

then $\angle ACE=\angle BCN$, $\angle NBC=60^\circ$,

$\because \angle ECM=30^\circ$,

$\therefore \angle ACE+\angle MCB=30^\circ$,

$\therefore \angle BCN+\angle MCB=\angle MCN=30^\circ$,

$\therefore \angle MCN=\angle MCE=30^\circ$,

$\because CE=CN$, $\angle MCN=\angle MCE$, $CM=CM$,

$\therefore \triangle MCE\cong \triangle MCN$,

$\therefore \angle CMN=\angle CME=90^\circ$,

$\therefore points $E, $M$, $N$ are collinear,

$\therefore \triangle ECN$ is an equilateral triangle,

$\because \angle NBC=\angle ACB=60^\circ$,

$\therefore BN\parallel AC$,

$\because \angle BPC=90^\circ$,

$\therefore \angle NBM=90^\circ$,

$\because \angle CMN=90^\circ$,

$\therefore \angle BMN+\angle CMP=90^\circ$,

$\because \angle BMN+\angle BNM=90^\circ$,

$\therefore \angle BNM=\angle CMP$,

$\therefore \triangle BMN\sim \triangle PCM$,

$\therefore \frac{BM}{PC}=\frac{NM}{MC}$,

$\because \frac{NM}{MC}=\tan\angle MCN=\tan30^\circ$,

$\therefore \frac{BM}{PC}=\tan30^\circ=\frac{\sqrt{3}}{3}$,

$\because PC=3$,

$\therefore BM= \sqrt{3}$,

In $\triangle ABP$, $AP=3$, $\angle ABP=30^\circ$,

$\therefore BP= 3\sqrt{3}$,

$\therefore PM=BP-BM= 2\sqrt{3}$,

$\because \angle MBC=30^\circ$, $\angle OMC=90^\circ-\angle OME=30^\circ$,

$\therefore \angle MBC+\angle MCB=\angle OMC+\angle OMP$,

$\therefore \boxed{AE=3}$.


Question: As shown in the figure, quadrilateral $ABCD$ is inscribed in circle $\odot O$, with diagonal $AC$ being the diameter of $\odot O$. Through point $C$, draw $CE \perp AC$, intersecting the extension of $AD$ at point $E$. Let $F$ be the midpoint of $CE$, and connect $DB$ and $DF$. What is the measure of $\angle CDE$?

Solution:
\textbf{Solution:} Since $AC$ is the diameter of $\odot O$,\\
\(\therefore \angle ADC=90^\circ\),\\
\(\therefore \angle CDE=\boxed{90^\circ}\);


Question: As shown in the figure, the quadrilateral $ABCD$ is inscribed in circle $\odot O$, with the diagonal $AC$ being the diameter of $\odot O$. Through point $C$, draw $CE \perp AC$ to intersect the extension of line $AD$ at point $E$. Let $F$ be the midpoint of $CE$, and connect $DB$, $DF$. $DF$ is the tangent to $\odot O$. If $\tan\angle ABD=3$, find the value of $\frac{AC}{DE}$.

Solution:
\textbf{Solution:} \\
Since $\angle ACD = \angle ABD$, \\
it follows that $\tan \angle ACD = \frac{AD}{CD} = 3$, \\
Let $AD=3$ and $CD=1$, \\
thus $AC = \sqrt{AD^2 + CD^2} = \sqrt{10}$, \\
Since $\angle ACD + \angle DCE = \angle DEC + \angle DCE$, \\
it follows that $\angle ACD = \angle DEC$, \\
thus $\tan \angle DEC = \frac{DC}{DE} = 3$, \\
therefore $DE = \frac{1}{3}$, \\
hence $\frac{AC}{DE} = \frac{\sqrt{10}}{\frac{1}{3}} = \boxed{3\sqrt{10}}$.


Question: As shown in the figure, $AB$ is the diameter of circle $O$, points $C$ and $D$ are on circle $O$, and $AC=CD$. Line segments $AD$ and $BC$ intersect at point $E$, point $F$ is on the extension of $BC$, and $AF=AE$. $AF$ is the tangent to circle $O$. If $EF=12$, $\sin \angle BAC = \frac{4}{5}$, what is the radius of circle $O$?

Solution: \textbf{Solution:} Given that $AE=AF$, $\angle ACB=90^\circ$, hence $CF=CE=\frac{1}{2}EF=6$, since $\angle CAB=\angle CEA$, then $\sin\angle CAB=\sin\angle CEA=\frac{4}{5}$. It follows that $\frac{AC}{AE}=\frac{4}{5}$, thus $AC=\frac{4}{5}AE$. Consequently, $\left(\frac{4}{5}AE\right)^2+6^2=AE^2$, from which we find $AE=10$, thereby $AC=8$. Given that $\sin\angle CAB=\frac{BC}{AB}=\frac{4}{5}$, it leads to $BC=\frac{4}{5}AB$. Therefore, $8^2+\left(\frac{4}{5}AB\right)^2=AB^2$, hence $AB=\frac{40}{3}$, which implies the radius of $\odot O$ is $\boxed{\frac{20}{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $O$ is a point on $AC$. Taking point O as the center and $OC$ as the radius, a circle is drawn tangent to $BC$ at point C. A line is drawn through point O such that $OE \perp AB$ and intersects $AB$ at point E. Through point A, $AD \perp BO$ is drawn, and it intersects the extension of $BO$ at point D, and $\angle AOD = \angle BAD$. Given $OE=OC$, if $BC=6$ and $\tan \angle ABC = \frac{4}{3}$, find the length of $AD$.

Solution:
\textbf{Solution:} Given that $BC=6$, $\tan\angle ABC =\frac{4}{3}$, and $\angle ACB=90^\circ$,\\
it follows that $AC=BC\cdot\tan\angle ABC=8$\\
Thus, $AB =\sqrt{6^2+8^2}=10$\\
Since $OE=OC$,\\
it implies $BE=BC=6$\\
Hence, $AE=AB-BE=4$\\
Let $OC=OE=x$, then in $\triangle AEO$, $(8-x)^2=4^2+x^2$ yields $x=3$,\\
Therefore, $OB= \sqrt{OC^2+BC^2}=3\sqrt{5}$\\
Given that $S_{\triangle BOA}= \frac{1}{2} AB \cdot OE= \frac{1}{2} BO \cdot AD$,\\
it follows that $AB \cdot OE= BO \cdot AD$\\
Therefore, $AD =\boxed{2\sqrt{5}}$


Question: Given that $AB$ is the diameter of circle $O$, and $\angle ACD$ is the angle subtended by the arc $\overset{\frown}{AD}$ on the circumference of the circle with $\angle ACD=30^\circ$. What is the measure of $\angle DAB$?

Solution:
\textbf{Solution:} As shown in the diagram, connect BD,\\
$\because \angle ACD=30^\circ$,\\
$\therefore \angle B=\angle ACD=30^\circ$,\\
$\because AB$ is the diameter of $\odot O$,\\
$\therefore \angle ADB=90^\circ$,\\
$\therefore \angle DAB=90^\circ-\angle B= \boxed{60^\circ};$


Question: Given that $AB$ is the diameter of $\odot O$ and $\angle ACD$ is the inscribed angle corresponding to the arc $\overset{\frown}{AD}$, where $\angle ACD=30^\circ$. Through point $D$, draw $DE\perp AB$ with $E$ as the foot of the perpendicular. The extension of $DE$ intersects $\odot O$ at point $F$. If $AB=4$, find the length of $DF$.

Solution:
\textbf{Solution:} Since $\angle ADB=90^\circ$ and $\angle B=30^\circ$, with $AB=4$,\\
it follows that $AD=\frac{1}{2} AB=2$.\\
Since $\angle DAB=60^\circ$, $DE\perp AB$, and $AB$ is the diameter,\\
it results in $EF=DE=AD\sin60^\circ=\sqrt{3}$,\\
hence $DF=2DE=2\sqrt{3}$.\\
Therefore, the length of $DF$ is $\boxed{2\sqrt{3}}$.


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, and intersects the y-axis at point C. Point D is on the graph of the parabola in the first quadrant, and \(\angle BAD=30^\circ\). If the x-coordinate of point D is 5, what is the equation of the parabola?

Solution: \textbf{Solution:} Let the parabola be $y=\frac{k}{8}(x-2)(x+4)$. Setting $y=0$, we obtain $(x-2)(x+4)=0$, which solves to $x=2$ or $x=-4$. Thus, $A(-4,0)$ and $B(2,0)$. Therefore, $AO=4$. Construct $DE\perp$ x-axis at point $E$, hence $OE=5$, $AE=AO+OE=4+5=9$. Since $\angle BAD=30^\circ$, it follows that $DE=AE\cdot\tan{30^\circ}=9\times \frac{\sqrt{3}}{3}=3\sqrt{3}$. Hence, point $D$ is at $(5,3\sqrt{3})$. Given that point $D$ lies on the parabola $y=\frac{k}{8}(5-2)(5+4)=3\sqrt{3}$, it follows that $k= \frac{8\sqrt{3}}{9}$. Thus, the expression for the parabola is: $y= \frac{\sqrt{3}}{9}(x-2)(x+4)$. In other words, $\boxed{y= \frac{\sqrt{3}}{9}x^{2}+ \frac{2\sqrt{3}}{9}x- \frac{8\sqrt{3}}{9}}$.