Question

Prove that the co-ordinates of the third vertex of an equilateral triangle whose two vertices are at $(3,4)$ and $(-2,3)$ are $(\frac{1+\sqrt{3}}{2},\frac{7-5\sqrt{3}}{2})$ or $(\frac{1-\sqrt{3}}{2},\frac{7+5\sqrt{3}}{2})$

Answer 1

${PA}^2={PB}^2=26={AB}^2$

${PA}^2={PB}^2$ imply $5h+k=6$

${PA}^2=26$ imply

${(h-3)}^2+{(k-4)}^2=26$

Eliminate $k$ between $\left(1\right)$ and $\left(2\right)$,

${2h}^2-2h-1=0$

$\therefore h=\frac{1+\sqrt{3}}{2}$ or $\frac{1+\sqrt{3}}{2}$

$\therefore k=\frac{7-5\sqrt{3}}{2}$ or $\frac{7+5\sqrt{3}}{2}$.