The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a. - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 1

The slope of line PQ, which has the points P(3, 4) and Q(a, b), is given by:

$m_{PQ} = \frac{b - 4}{a - 3}$

For two lines to be perpendicular, the product of their slopes must equal -1. Therefore:

$m_{PQ} \cdot m_{perpendicular} = -1$

Substituting the slopes we find:

$\frac{b - 4}{a - 3} \cdot \left(-\frac{3}{2}\right) = -1$

To solve for b, we can multiply both sides of the equation by -2:

$-2 \cdot \left(\frac{b - 4}{a - 3}\right) \cdot \left(-\frac{3}{2}\right) = -2 \cdot (-1) \\ \Rightarrow 3(b - 4) = 2(a - 3)$

Expanding yields:

$3b - 12 = 2a - 6$

Rearranging gives:

$3b = 2a + 6 \\ b = \frac{2a + 6}{3}$

Thus, the expression for b in terms of a is:

$b = \frac{2a + 6}{3}$