Review question

# Can we find $\alpha$ if one length is twice another? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5365

## Solution

Given the points $A(\alpha,3)$, $B(-2,1)$ and $C(3,2)$ find the possible values of $\alpha$ if the length of $AB$ is twice the length of $BC$.

Drawing a sketch would probably help us to see what is going on.

We can use Pythagoras’ theorem to find the distance between two points.

We have that $AB=\sqrt{(\alpha+2)^2+(3-1)^2}=\sqrt{(\alpha+2)^2+4},$ and $BC=\sqrt{(3+2)^2+(2-1)^2}=\sqrt{5^2+1^2}=\sqrt{26}.$ Then if $AB=2BC$, we must have $\sqrt{(\alpha+2)^2+4}=2\sqrt{26}.$ Squaring both sides, we find that $(\alpha+2)^2+4=104,$ which we rearrange to find $(\alpha+2)^2=100.$ Therefore $\alpha=-2\pm10$ and so $\alpha = -12 \quad \text{or} \quad \alpha = 8.$

Squaring sometimes introduces false solutions, so we should check that each of our solutions satisfies $\sqrt{(\alpha+2)^2+4}=2\sqrt{26}$. They both do.