P is the point (0, -1) and Q is the point (5, 9) - OCR - GCSE Maths - Question 18 - 2018 - Paper 4

For a line to be perpendicular to another, the product of their gradients must equal -1. Therefore, if the gradient of line PQ is 2, the gradient of the line perpendicular to it (m_perpendicular) is:

$m_{perpendicular} = -\frac{1}{m_{PQ}} = -\frac{1}{2}$

We can use the point-slope form of the line equation, which is written as:

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

In this case, we will use point P (0, -1) and the perpendicular gradient we calculated:

$y - (-1) = -\frac{1}{2}(x - 0)$

Simplifying this gives:

$y + 1 = -\frac{1}{2}x$ $y = -\frac{1}{2}x - 1$

Thus, the final equation of the line through P that is perpendicular to line PQ is:

$y = -\frac{1}{2}x - 1$