The points P and Q have coordinates (–1, 6) and (9, 0) respectively - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 1

The gradient (slope) m of line PQ is calculated using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Thus,

$m = \frac{0 - 6}{9 - (-1)} = \frac{-6}{10} = -\frac{3}{5}$

The line l is perpendicular to PQ, so its gradient is the negative reciprocal of m:

$m_l = -\frac{1}{m} = -\frac{1}{-\frac{3}{5}} = \frac{5}{3}$

Using the point-slope form of a line's equation, which is:

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

Substituting M (4, 3) and the gradient of l (\frac{5}{3}):

$y - 3 = \frac{5}{3}(x - 4)$

Expanding this gives:

$y - 3 = \frac{5}{3}x - \frac{20}{3}$

Rearranging to standard form:

$3(y - 3) = 5x - 20$

This simplifies to:

$5x - 3y - 11 = 0$

The equation of line l in the required form is:

$5x - 3y - 11 = 0$

Here, a = 5, b = -3, and c = -11, meeting the requirement of integers.