Solutions to this question by accurate drawing will not be accepted. The diagram below shows a triangle ABC, with $A(3,6)$ and $B(-11,0)$. Triangle $ABC$ has an area of $22.5\ \text{units}^2$. $E$ is a point on the line $AB$ passing through the $y$-axis.

1. Substitute the coordinates into the distance formula.

For $A(3,6)$ and $B(-11,0)$,
$AB=\sqrt{(3-(-11))^2+(6-0)^2}$

2. Evaluate and simplify the length.

$AB=\sqrt{14^2+6^2}$
$=\sqrt{196+36}$
$=\sqrt{232}=2\sqrt{58}$

1. Find the gradient of line $AB$.

Using $A(3,6)$ and $B(-11,0)$,
$m=\frac{6-0}{3-(-11)}=\frac{6}{14}=\frac{3}{7}.$

2. Substitute the gradient and one point into $y=mx+c$ to find the equation.

Using $y=mx+c$ and point $A(3,6)$,
$6=\frac{3}{7}(3)+c$
$6=\frac{9}{7}+c$
$c=\frac{33}{7}.$
So the equation is
$y=\frac{3}{7}x+\frac{33}{7}.$

1. Use the fact that a point on the $y$-axis has $x=0$ and substitute into the equation of $AB$.

From part (b),
$y=\frac{3}{7}x+\frac{33}{7}.$
At the $y$-axis, $x=0$, so
$y=\frac{3}{7}(0)+\frac{33}{7}=\frac{33}{7}.$
Hence $E=(0,\frac{33}{7})$.

1. Use the triangle area formula to find the length of $BC$.

Since $B$ and $C$ lie on the $x$-axis, $BC$ is horizontal.

The height from $A(3,6)$ to the $x$-axis is $6$ units.

$\frac{1}{2}\times BC\times 6=22.5$

$3BC=22.5$

$BC=7.5$

2. Find the coordinates of $C$ using its position on the $x$-axis.

Point $B$ is $(-11,0)$ and $C$ is on the $x$-axis between $B$ and $O$.

So $C$ is $7.5$ units to the right of $B$:

$x_C=-11+7.5=-3.5=-\frac{7}{2}$

Since $C$ lies on the $x$-axis, $y_C=0$.

1. Use equal opposite sides of the parallelogram to find the coordinates of $D$.

From part (d), $C\left(-\dfrac{7}{2},0\right)$.

Since $ABDC$ is a parallelogram,
$\overrightarrow{BD}=\overrightarrow{AC}.$

$\overrightarrow{AC}=\left(-\dfrac{7}{2}-3,\ 0-6\right)=\left(-\dfrac{13}{2},-6\right).$

So
$D=B+\overrightarrow{AC}$
$=\left(-11,0\right)+\left(-\dfrac{13}{2},-6\right)$
$=\left(-\dfrac{35}{2},-6\right).$

1. Write the equation of line $l$ in gradient-intercept form and compare with line $AB$.

For line $l$,
$7y-3x+15=0$
$7y=3x-15$
$y=\frac{3}{7}x-\frac{15}{7}$
From part (b), line $AB$ is
$y=\frac{3}{7}x+\frac{33}{7}.$

2. Use the gradients and intercepts to decide whether the lines intersect.

Both lines have gradient $\dfrac{3}{7}$, but their $y$-intercepts are different:
$-\frac{15}{7}\ne \frac{33}{7}.$
Therefore, the two lines are parallel and distinct, so they will never cut each other even when extended.