Question

The co-ordinates of base $BC$ of an equilateral triangle $ABC$ are given by $B(1,3)$ and $C(-2,7)$. What could be the possible co-ordinates of the vertex $A$?

Answer 1

Since the triangle is equilateral, each side is $BC=5$.
$\therefore P=AD=5\sin {60}^o$
$\therefore P=\frac{\sqrt{3}}{2}5.$ Also slop of $BC$ is $-\frac{4}{3}$.
$\therefore$ Slop of $DA$ is $\frac{3}{4}=\tan \theta$
$\therefore \cos \theta =\frac{4}{5},\sin \theta =\frac{3}{5}$.
Since $DA$ can be written in parametric from as
$\frac{x+\frac{1}{2}}{4/5}=\frac{y-5}{3/5}=r=\pm p=\pm \frac{\sqrt{3}}{2}.5$.
We have chosen $r=\pm p$ as the point could be on either side of line $BC$. Hence co-ordinates of $A$ are
$(-\frac{1}{2}+2\surd 3,5+3\frac{\surd 3}{2})$ and $(-\frac{1}{2}-2\surd 3,5-3\frac{\surd 3}{2})$.