Question 2
a) The line with equation $y=x+12$ intersects the line with equation $y=2-4 x$ at point $P$.
Find the coordinates of point P. You must show your working.
b) Point $\mathrm{A}$ lies at $(2 p,-q)$ and point $\mathrm{B}$ lies at $(5 p, 8 q)$. A line which is perpendicular to $A B$ has gradient $-\frac{2}{3}$.
Express $p$ in terms of $q$.

Step 1 ：Set the two equations equal to each other: \(x + 12 = 2 - 4x\).

Step 2 ：Solve for x: \(5x = -10\) so \(x = -2\).

Step 3 ：Substitute \(x = -2\) into either of the original equations to find y. Using the first equation: \(y = -2 + 12 = 10\).

Step 4 ：So, the coordinates of point P are \(\boxed{(-2, 10)}\).

Step 5 ：The gradient of line AB is \(\frac{8q-(-q)}{5p-2p} = \frac{9q}{3p} = \frac{3q}{p}\).

Step 6 ：Since the line perpendicular to AB has gradient \(-\frac{2}{3}\), and the product of the gradients of two perpendicular lines is -1, we have \(-\frac{2}{3} = -\frac{1}{\frac{3q}{p}}\).

Step 7 ：Solving for p, we get \(p = \frac{3q}{2}\).