Review question

# Given a right angle and the equations of two sides, can we find the third? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5661

## Solution

In the triangle $ABC$ the equations of two of the sides are $2y=x+2$ and $y=2x-2$. Given that $A$ is the point $(4,3)$ and that $A\hat{B}C$ is $90^\circ$, find the equation of the third side.

Our best starting plan is to draw a diagram.

The point $A$ must be on either $y = 2x-2$, or $2y = x + 2$, and we can see by substituting in that $A$ is on the second of these but not the first.

So the first line ($y=2x-2$) must give the side $BC$, so $B$ lies on this line. We are also told that the angle at $B$ is a right angle.

We can now sketch our diagram. (At this stage, we have not worked out whether $2y=x+2$ passes through $B$ or not.)

Now the gradient of $BC$ is $2$, so the gradient of $AB$ is $-\frac12$, since $AB$ is perpendicular to $BC$. Since the line $AB$ passes through $A(4,3)$, it follows that it has equation $y-3=-\frac12(x-4)$ (using the general formula $y-y_1=m(x-x_1)$), which rearranges to $y=-\frac12x+5$ or $x+2y=10$. This is the equation of the third side.

It also follows that $2y=x+2$ is the line $AC$ (since it has gradient $+\frac12$, so is not perpendicular to $BC$). This would allow us to work out the coordinates of $C$, though we are not asked to find these.