In the diagram, the points P, Q and R have coordinates (9, -3), (6,1) and (18, -3) respectively.

1. Find the distance between P and Q using their coordinates.

$P(9,-3)$ and $Q(6,1)$

$PQ=\sqrt{(9-6)^2+(-3-1)^2}$
$=\sqrt{3^2+(-4)^2}$
$=\sqrt{9+16}=\sqrt{25}=5$

1. Identify the adjacent side and hypotenuse for $\angle QPR$, then apply the cosine ratio.

From P to Q, the horizontal change is
$|9-6|=3.$
Also, from part (a),
$PQ=5.$
For $\angle QPR$,
$\cos \angle QPR=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{3}{5}.$

1. Identify the base and perpendicular height, then apply the triangle area formula.

Since $P(9,-3)$ and $R(18,-3)$ lie on the horizontal line $y=-3$,

$PR=18-9=9$ units.

The perpendicular height from $Q(6,1)$ to $PR$ is

$1-(-3)=4$ units.

$\text{Area of }\triangle PQR=\frac12 \times 9 \times 4=18$

1. Find the gradient of the line $PQ$.

Using $P(9,-3)$ and $Q(6,1)$,
$m=\frac{1-(-3)}{6-9}=\frac{4}{-3}=-\frac{4}{3}.$

2. Use the gradient and one point to find the equation of $PQ$.

Substitute $m=-\dfrac{4}{3}$ and $Q(6,1)$ into $y=mx+c$:
$1=-\frac{4}{3}(6)+c$
$1=-8+c$
$c=9.$
Hence,
$y=-\frac{4}{3}x+9.$

1. Write the equation of line $PQ$ in a form suitable for simultaneous equations.

From part (d),
$y=-\dfrac{4}{3}x+9$.

Multiply by $3$:
$3y=-4x+27$
$4x+3y=27$

2. Solve the two linear equations simultaneously to find $y$.

Solve
$4x+3y=27 \quad \text{and} \quad 4x+21y=9.$

Subtract the first equation from the second:
$(4x+21y)-(4x+3y)=9-27$
$18y=-18$
$y=-1$

3. Substitute $y$ into one equation to find $x$, then state the intersection point.

Substitute $y=-1$ into $4x+3y=27$:
$4x+3(-1)=27$
$4x-3=27$
$4x=30$
$x=\dfrac{15}{2}.$

So the point of intersection is
$\left(\dfrac{15}{2}, -1\right).$