Question

If $(-5,4)$ and $(5,4)$ are tow vertices of an equilateral triangle, then find coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer 1

Let vertix be $(x,y)$

Distance between $(x,y)$ & $(-5,4)$ is $=\sqrt{(x+5{)}^2+(y-4{)}^2}------\left(1\right)$

Distance between $(x,y)$ & $(5,4)$ is $=\sqrt{(x-5{)}^2+(y-4{)}^2}-------\left(2\right)$

Distance between $(-5,4)$ & $(5,4)$ is $=\sqrt{(5+5{)}^2+(4-4{)}^2}=\sqrt{{10}^2}=10$

then $\left(1\right)=\left(2\right)$

$(x+5{)}^2=(x-5{)}^2$

$x^2+25+10x=x^2+25-10x$

$20x=0,x=0$

Now $\left(1\right)=10$

$(x+5{)}^2+(y-4{)}^2=100$

$(0+5{)}^2+(y-4{)}^2=100$

$(y-4{)}^2=100-25$

$(y-4{)}^2=75$

$y-4=\sqrt{75}=5\sqrt{3}$

$y=4\pm 5\sqrt{3}$

So required vertex is $(0,4\pm 5\sqrt{3})$