Question

If (-4,0) and (4,0) are two vertices of an equilateral triangle, find the coordinates of its third vertex.

Answer 1

Length of $AB=8$ units

In equilateral triangle all sides are equal

$AC=BC=AB$

$BC=AC$

$\sqrt{(h-4{)}^2+b^2}=\sqrt{(h+4{)}^2+b^2}$

Squaring both sides

$(h-4{)}^2+b^2=(h+4{)}^2+b^2$

$h=0$

$AC=AB$

$\sqrt{(h+4{)}^2+b^2}=8$

Squaring both sides

$(h+4{)}^2+b^2=64$

$b^2=48$

$b=\pm \sqrt{48}$

Third vertex $=(0,\sqrt{48})or(0,-\sqrt{48})$