Sketching a quick diagram may help to see the connection between the coordinates and to make sure you are using the correct gradients in the correct places.

edexcel-tq-mis-coordinate-geo-vh-q2

The gradient of AC can be found in terms of p and q and set equal to $- \frac{6}{7}$ .

$m_{A C} = \frac{10 - q}{p - 8} = - \frac{6}{7}$

Expression for the gradient [1]
Forming the equation [1]

Angle ABC is a right angle, so AB and BC are perpendicular to each other.

The gradients of perpendicular lines are negative reciprocals so the gradients of AB and BC can be found in terms of p and q.

$m_{A B} = \frac{10 - - 5}{p - - 1} = \frac{15}{p + 1} m_{B C} = \frac{q - - 5}{8 - - 1} = \frac{q + 5}{9}$

[1]

Set the gradient of AB equal to the negative reciprocal of the gradient of BC.

$m_{A B} = \frac{1}{m_{B C}} \frac{15}{p + 1} = - \frac{1}{\frac{q + 5}{9}} = - \frac{9}{q + 5}$

[1]

Solve the two equations simultaneously to find the values of p and q.

Number the equations.

$\begin{array}{rcl} \frac{10 - q}{p - 8} & = & - \frac{6}{7} (1) \\ \frac{15}{p + 1} & = & - \frac{9}{q + 5} (2) \end{array}$

Rearrange equation (1) to get rid of the fractions.

$\begin{array}{rcl} 7 (10 - q) & = & - 6 (p - 8) (1) \end{array}$

Expand the brackets on both sides of equation (1).

$\begin{array}{rcl} 70 - 7 q & = & - 6 p + 48 (1) \end{array}$

Simplify.

$\begin{array}{rcl} 6 p - 7 q & = & - 22 (1) \end{array}$

Rearrange equation (2) to get rid of the fractions.

$\begin{array}{rcl} 15 (q + 5) & = & - 9 (p + 1) (2) \end{array}$

Expand the brackets on both sides of equation (1).

$\begin{array}{rcl} 15 q + 75 & = & - 9 p - 9 (2) \end{array}$

Simplify by bringing unknown terms to one side.

$\begin{array}{rcl} 9 p + 15 q & = & - 84 (2) \end{array}$

Multiply equation (1) by 3 and equation (2) by 2 and eliminate p.

$18 p - 21 q = - 66 3 (1) (-) 18 p + 30 q = - 168 2 (2) - 51 q = 102$

[1]

Solve to find q.

$\begin{array}{rcl} q & = & \frac{102}{- 51} = - 2 \end{array}$

Find p by substituting q = -2 into equation (1) and solving.

$\begin{array}{rcl} (1) 6 p - 7 (- 2) & = & - 22 \\ 6 p + 14 & = & - 22 \\ 6 p & = & - 36 \\ p & = & - 6 \end{array}$

$p = - 6, q = - 2$ [1]

Coordinate Geometry (OCR GCSE Maths)

Exam Questions