The diagram shows a trapezium in which $AD$ is parallel to $BC$ and angle $ADC =$ angle $BCD = 90^\circ$. The points $B$ and $C$ are $(12, -2)$ and $(2, -7)$ respectively. The line $AD$ meets the y-axis at $(0, 7)$. Given that $AB = 2BC$, find

1. Find the gradient of $BC$ and hence the equation of line $AD$.

Since $B=(12,-2)$ and $C=(2,-7)$,
$m_{BC}=\frac{-2-(-7)}{12-2}=\frac{5}{10}=\frac12.$
Since $AD\parallel BC$, $m_{AD}=\frac12$.

Line $AD$ passes through $(0,7)$, so
$y=\frac12x+7.$

2. Express the coordinates of $A$ using the equation of $AD$.

Let $A=(x,y)$.

Since $A$ lies on $AD$,
$y=\frac12x+7.$
So
$A=\left(x,\frac12x+7\right).$

3. Use $AB=2BC$ to form and solve an equation for $x$.

First find $BC$:
$BC=\sqrt{(12-2)^2+[-2-(-7)]^2}$
$=\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt5.$

Given $AB=2BC$,
$AB=10\sqrt5.$

Using $A=\left(x,\frac12x+7\right)$ and $B=(12,-2)$,
$AB^2=(x-12)^2+\left(\frac12x+7-(-2)\right)^2.$
So
$500=(x-12)^2+\left(\frac12x+9\right)^2.$
Expanding,
$500=x^2-24x+144+\frac14x^2+9x+81,$
$500=\frac54x^2-15x+225.$
Hence
$\frac54x^2-15x-275=0.$
Multiplying by $4$,
$5x^2-60x-1100=0,$
$x^2-12x-220=0.$
Factorising,
$(x-22)(x+10)=0.$
Thus
$x=22\quad \text{or}\quad x=-10.$

4. Choose the correct point for $A$ and find its $y$-coordinate.

If $x=22$,
$y=\frac12(22)+7=18,$
so $A=(22,18)$.

If $x=-10$,
$y=\frac12(-10)+7=2,$
so the other possible point is $(-10,2)$.

From the diagram, $A$ is the upper right vertex on line $AD$, so choose
$A=(22,18).$

1. Find the equation of the line $DC$.

Gradient of $BC$:
$m_{BC}=\frac{-2-(-7)}{12-2}=\frac{5}{10}=\frac12.$
Since $DC\perp BC$,
$m_{DC}=-2.$
Line $DC$ passes through $C(2,-7)$:
$y+7=-2(x-2)$
$y=-2x-3.$

2. Find the intersection of $DC$ with $AD$.

From part (i), line $AD$ has gradient $\frac12$ and passes through $(0,7)$, so
$y=\frac12x+7.$
At $D$, the two lines meet:
$\frac12x+7=-2x-3$
$2.5x=-10$
$x=-4.$
Then
$y=\frac12(-4)+7=5.$