The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

To determine the gradient of line AC, we first compute the gradient using the coordinates of A(1, 7) and C(15, -3):

$m_{AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 7}{15 - 1} = \frac{-10}{14} = -\frac{5}{7}$

The gradient of line l, which is perpendicular to AC, can be found using the negative reciprocal:

$m_l = -\frac{1}{m_{AC}} = -\frac{7}{5}$

Now using point-slope form to find the equation of line l, which passes through point D(8, 2):

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

Substituting in the values gives:

$y - 2 = -\frac{7}{5}(x - 8)$

To convert to standard form, rearranging yields:

$7x + 5y - 46 = 0$

To find the x-coordinate at point E where line l intersects AB, we first note that line AB is horizontal (since both A and B have the same y-coordinate of 7). Thus, the equation of line AB is simply:

$y = 7$

Substituting this into the equation of line l:

$7x + 5(7) - 46 = 0$

This simplifies to:

$7x + 35 - 46 = 0$ $7x - 11 = 0$ $7x = 11$ $x = \frac{11}{7}$

Thus, the exact x-coordinate of E is:

$\frac{11}{7}$