(a) Find the coordinates of A.

$$

\begin{array}{l}

2 y-3 x=6 \\

3 y+x=20 \\

\hline 2 y-3 x=6 \\

\dfrac{9 y+3 x=60}{11 y=60} \\

y=6 \\

x=20-18 \\

=2

\end{array}

$$

Coordinates of $\mathrm{A}$ are $(2,6)$

(b) A third line $L_{3}$ is perpendicular to $L_{2}$ at point

A. Find the equation of $L_{3}$ in the form $y=m x+$

c, Where $m$ and $c$ are constants. (3 marks)

(b) $\mathrm{L}_{2}: 3 y=-x+20$

$$

y=-\frac{1}{3} x+20

$$

Gradient of perpendicular $=3$

$$

\begin{array}{l}

\dfrac{y-6}{x-2}=3 \\

y=3 x-6+6 \\

y=3 x

\end{array}

$$

\begin{array}{l}

\text { (c) Gradient of } L 4=\text { gradient of } L \text { , }\\

=\dfrac{3}{2}\\

\dfrac{y-3}{x+1}=\frac{3}{2}\\

2 y-6=3 s+3\\

2 y-3 x=9\\

\text { When } x=0 \quad y-4.5\\

\text { When } \mathrm{y}-0 \mathrm{x}=-3

\end{array}