Although you do not know all of the information, sketching a graph with a rough idea of where the lines go will help you to visualise this question. It does not matter if your lines are not in the exact places but make sure that they intersect each other according to the question.

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We can see from the graph that to find the area of the triangle we will need to find the length of AB (the base of the triangle) and the length of CM (the height of the triangle).

First let's consider CM. We can find the coordinates of M using the midpoint formula, $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$\begin{array}{rcl} M & = & (\frac{5 + (- 6)}{2}, \frac{3 + 2}{2}) \\ = & (- \frac{1}{2}, \frac{5}{2}) \end{array}$

[1]

To find C, we can find the equation of CM and examine where it intersects the line $x = - 1$ .

To find the equation of CM, start by finding the gradient of AB, which is perpendicular to CM, using gradient $m = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$ .

$\begin{array}{rcl} m_{A B} & = & \frac{3 - 2}{5 - (- 6)} \\ = & \frac{1}{11} \end{array}$

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Perpendicular gradients are negative reciprocals so from this we can deduce the gradient of CM .

$\begin{array}{rcl} m_{C M} & = & - 1 \div \frac{1}{11} \\ = & - 11 \end{array}$

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Now find the equation of CM by substituting the point $M \begin{array}{rcl} (- \frac{1}{2}, \frac{5}{2}) \end{array}$ and the gradient $- 11$ into $y = m x + c$ , and solving for c.

$\begin{array}{rcl} y & = & m x + c \\ \frac{5}{2} & = & - 11 (- \frac{1}{2}) + c \\ \frac{5}{2} & = & \frac{11}{2} + c \\ \frac{5 - 11}{2} & = & c \\ c & = & \frac{- 6}{2} = - 3 \end{array}$

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Now we know that the equation of CM is:

$\begin{array}{rcl} y & = & - 11 x - 3 \end{array}$

Substitute $x = - 1$ into the equation of CM to find the y-coordinate of C.

$\begin{array}{rcl} y & = & - 11 (- 1) + 3 \\ = & 11 - 3 \\ = & 8 \end{array}$

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Now we can find the length of CM using the length of a line segment formula, $length = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$ .

$\begin{array}{rcl} C M & = & \sqrt{{(- \frac{1}{2} - (- 1))}^{2} + {(\frac{5}{2} - 8)}^{2}} \\ = & \sqrt{{(\frac{1}{2})}^{2} + {(\frac{11}{2})}^{2}} \\ = & \frac{\sqrt{122}}{2} \end{array}$

And finding the length of AB is straightforward as we already have the coordinates of A and B.

$\begin{array}{rcl} A B & = & \sqrt{{(3 - 2)}^{2} + {(5 - (- 6))}^{2}} \\ = & \sqrt{122} \end{array}$

length of CM and AB found [1]

Finally, we are ready to find the area of triangle ABC using area of a triangle = $\frac{1}{2}$ × base × height.

$Area = \frac{1}{2} \times \sqrt{122} \times \frac{\sqrt{122}}{2}$

[1]

$\begin{array}{rcl} = & \frac{122}{4} \\ = & \frac{61}{2} \end{array}$

$Area = 30.5 {units}^{2}$ [1]

$x$	$- 2$	$- 1$	$0$	$1$	$2$
$y$	$1$		$5$

Linear Graphs y = mx + c (AQA GCSE Maths)

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