The points Q(1, 3) and R(7, 0) lie on the line l1, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 11 - 2008 - Paper 1

To determine the equation of line l2, which is perpendicular to l1 and passes through point Q(1, 3), we start by finding the gradient of line QR:

$\text{Gradient of } QR = \frac{0 - 3}{7 - 1} = -\frac{3}{6} = -\frac{1}{2}$.

The gradient of line l2 is the negative reciprocal of the gradient of QR:

$\text{Gradient of } l2 = 2$.

Using point-slope form of the linear equation:

$y - y_1 = m(x - x_1)$,

we plug in the values to get:

$y - 3 = 2(x - 1)$ $y - 3 = 2x - 2$ $y = 2x + 1$.

Therefore, the equation for l2 is:

$y = 2x + 1$.

Point P is where line l2 crosses the y-axis. At the y-axis, the value of x is 0. Plugging x = 0 into the equation of l2:

$y = 2(0) + 1 = 1$.

Thus, the coordinates of point P are:

$P(0, 1)$.

To find the area of triangle PQR, we can use the formula for the area of a triangle given its vertices:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$,

where P(0, 1), Q(1, 3), and R(7, 0).

Using the coordinates:

$\text{Area} = \frac{1}{2} \left| 0(3 - 0) + 1(0 - 1) + 7(1 - 3) \right|$ $= \frac{1}{2} \left| 0 - 1 - 14 \right|$ $= \frac{1}{2} \left| -15 \right| = \frac{15}{2} = 7.5$.

Thus, the area of triangle ΔPQR is:

$\text{Area} = 7.5$.