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.
