A, B, C and D are four points such that the coordinates of A and C are (4,2) and (10,−34) respectively. $\overrightarrow{AB}=\begin{pmatrix}-9\\-12\end{pmatrix}$ and $\overrightarrow{AD}=\begin{pmatrix}24\\-12\end{pmatrix}$.

1. Add the displacement vector $\overrightarrow{AB}$ to the coordinates of A to find B.

Since $A=(4,2)$ and $\overrightarrow{AB}=\begin{pmatrix}-9\\-12\end{pmatrix}$,
$B=(4+(-9),\ 2+(-12))=(-5,-10).$

1. Find the magnitude of the given vector $\overrightarrow{AB}$.

Given $\overrightarrow{AB}=\begin{pmatrix}-9\\-12\end{pmatrix}$,
$|\overrightarrow{AB}|=\sqrt{(-9)^2+(-12)^2}$
$=\sqrt{81+144}=\sqrt{225}=15$

1. Find the coordinates of D using the position of A and the vector $\overrightarrow{AD}$.

$A=(4,2)$ and $\overrightarrow{AD}=\begin{pmatrix}24\\-12\end{pmatrix}$.

So
$D=(4+24,\ 2-12)=(28,-10).$

2. Find the vector $\overrightarrow{BD}$ and hence $\overrightarrow{BE}$.

From part (i), $B=(-5,-10)$.

$\overrightarrow{BD}=\begin{pmatrix}28-(-5)\\-10-(-10)\end{pmatrix}=\begin{pmatrix}33\\0\end{pmatrix}.$

Given that
$\overrightarrow{BE}=\frac13\overrightarrow{BD},$
$\overrightarrow{BE}=\frac13\begin{pmatrix}33\\0\end{pmatrix}=\begin{pmatrix}11\\0\end{pmatrix}.$

3. Find the coordinates of E.

Since $B=(-5,-10)$ and $\overrightarrow{BE}=\begin{pmatrix}11\\0\end{pmatrix}$,

$E=(-5+11,\ -10+0)=(6,-10).$

4. Show that $\overrightarrow{AE}$ and $\overrightarrow{AC}$ are parallel by comparing them as scalar multiples.

$\overrightarrow{AE}=\begin{pmatrix}6-4\\-10-2\end{pmatrix}=\begin{pmatrix}2\\-12\end{pmatrix}.$

Also,
$\overrightarrow{AC}=\begin{pmatrix}10-4\\-34-2\end{pmatrix}=\begin{pmatrix}6\\-36\end{pmatrix}.$

Therefore,
$\overrightarrow{AE}=\frac13\overrightarrow{AC}.$

Since $\overrightarrow{AE}$ is a scalar multiple of $\overrightarrow{AC}$, A, C and E are collinear.

1. Find the gradient of $AB$, which is also the gradient of line $L$ since $L \parallel AB$.

For $A(2,6)$ and $B(-4,-2)$,

$m_{AB}=\frac{-2-6}{-4-2}=\frac{-8}{-6}=\frac{4}{3}.$

Since $L \parallel AB$, the gradient of $L$ is $\dfrac{4}{3}$.

2. Use point $C(6,-2)$ and the gradient to find the equation of line $L$.

Line $L$ has equation $y=\dfrac{4}{3}x+c$.

Substitute $C(6,-2)$:

$-2=\frac{4}{3}(6)+c$

$-2=8+c$

$c=-10.$

Therefore,

$y=\frac{4}{3}x-10.$

1. Use $AD \parallel BC$ to find the equation of the line through $A$ on which $D$ lies.

Since $B(-4,-2)$ and $C(6,-2)$ have the same $y$-coordinate, $BC$ is horizontal.

Since $AD \parallel BC$ and $A(2,6)$, the line $AD$ is also horizontal.

Hence, the line $AD$ has equation
$y=6.$

2. Find the intersection of line $AD$ with line $L$.

From part (i), line $L$ has equation
$y=\frac{4}{3}x-10.$

Since $D$ lies on both $L$ and $AD$, substitute $y=6$:
$6=\frac{4}{3}x-10$
$16=\frac{4}{3}x$
$x=12.$

Therefore,
$D=(12,6).$

1. State that ABCD is a parallelogram using the parallel sides.

Line $L$ passes through $C$ and $D$, and $L \parallel AB$, so
$AB \parallel CD.$
It is also given that
$AD \parallel BC.$
Hence $ABCD$ is a parallelogram.

2. Show that adjacent sides of the parallelogram are equal, then conclude it is a rhombus.

Using $A(2,6)$, $B(-4,-2)$ and $C(6,-2)$,
$AB=\sqrt{(-4-2)^2+(-2-6)^2}=\sqrt{36+64}=10,$
$BC=\sqrt{(6-(-4))^2+(-2-(-2))^2}=\sqrt{100}=10.$
So $AB=BC$.

Since $ABCD$ is a parallelogram with adjacent equal sides, $ABCD$ is a rhombus.