The diagram (not drawn to scale) shows a right-angled triangle PQR in which point P is (5, 4) and point R is (9, 2). Point S is (-1.5, 11) and PR is parallel to SQ.

1. Find the gradient of the line PR

Gradient of $PR = \frac{2 - 4}{9 - 5} = \frac{-2}{4} = -\frac{1}{2}$

2. Find the gradient of PQ using the perpendicularity condition

Since $\angle QPR = 90^\circ$, $PQ \perp PR$.
Gradient of $PQ = -\frac{1}{-\frac{1}{2}} = 2$

3. Find the equation of the line PQ

Using point $P(5, 4)$ and gradient $m = 2$:
$y - 4 = 2(x - 5)$
$y = 2x - 10 + 4$
$y = 2x - 6$

1. Find the equation of line SQ

From part (a), the gradient of $PR$ is $-\frac{1}{2}$.
Since $SQ$ is parallel to $PR$, the gradient of $SQ$ is also $-\frac{1}{2}$.
Using point $S(-1.5, 11)$:
$y - 11 = -\frac{1}{2}(x - (-1.5))$
$y = -0.5x - 0.75 + 11$
$y = -0.5x + 10.25$

2. Set up simultaneous equations to find Q

From part (a), the equation of $PQ$ is $y = 2x - 6$.
Equating the two equations for $y$:
$2x - 6 = -0.5x + 10.25$

3. Solve for the coordinates of Q

$2.5x = 16.25$
$x = 6.5$
Substitute $x = 6.5$ into $y = 2x - 6$:
$y = 2(6.5) - 6 = 13 - 6 = 7$
The coordinates of $Q$ are $(6.5, 7)$.

1. Find the coordinates of Q from earlier parts

Gradient of $PR = \frac{2 - 4}{9 - 5} = -\frac{1}{2}$.
Since $PQ \perp PR$, gradient of $PQ = 2$.
Equation of $PQ$: $y - 4 = 2(x - 5) \implies y = 2x - 6$.
Since $SQ \parallel PR$, gradient of $SQ = -\frac{1}{2}$.
Equation of $SQ$: $y - 11 = -\frac{1}{2}(x - (-1.5)) \implies y = -\frac{1}{2}x + 10.25$.
Solving simultaneously:
$2x - 6 = -\frac{1}{2}x + 10.25 \implies 2.5x = 16.25 \implies x = 6.5$.
$y = 2(6.5) - 6 = 7$.
So $Q = (6.5, 7)$.

2. Calculate the length of QR

$QR = \sqrt{(9 - 6.5)^2 + (2 - 7)^2} = \sqrt{2.5^2 + (-5)^2} = \sqrt{6.25 + 25} = \sqrt{31.25}$
$\sqrt{31.25} = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2} = 2.5\sqrt{5}$ units.

3. Use the ratio of lengths to find the coordinates of T

Since T is on QR extended, Q, R, and T are collinear in that order.
The ratio of lengths is $\frac{QT}{QR} = \frac{10\sqrt{5}}{2.5\sqrt{5}} = 4$.
This means the horizontal and vertical displacements from Q to T are 4 times those from Q to R.
Displacement from Q to R: $\Delta x = 9 - 6.5 = 2.5$, $\Delta y = 2 - 7 = -5$.
Displacement from Q to T: $\Delta x_T = 4 \times 2.5 = 10$, $\Delta y_T = 4 \times (-5) = -20$.
Coordinates of T: $(6.5 + 10, 7 - 20) = (16.5, -13)$.