Question

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates that the origin lies in the (i) interior, (ii) exterior of the triangle.

Answer 1

Given vertex

$A(-4,3)$ and $B(4,3)$

Let $C(x,y)$

Hence triangle is equilateral

$AC=BC$

$\sqrt{(x+4{)}^2+(y-3{)}^2}=\sqrt{(x-4{)}^2+(y-3{)}^2}$

$(x+4{)}^2+(y-3{)}^2=(x-4{)}^2+(y-3{)}^2$

$x^2+16+8x=x^2+16-8x$

$x=0$

Hence $C(0,y)$

Now by distance formula $AB=8$

$AC=8$

$\sqrt{(4{)}^2+(y-3{)}^2}=8$

$16+y^2+9-6y=64$

$y^2-6y-39=0$

$y=\frac{6\pm \sqrt{36+156}}{2}$

$y=\frac{6\pm \sqrt{192}}{2}$

$y=3\pm 4\sqrt{3}$

Point $C(0,3-4\sqrt{3})$ for origin lies interior

and $C(0,3+4\sqrt{3})$ for origin lies exterior