Locus of vertice of equilateral triangle. Here is a question: 

On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.

I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question? 

The midpoint between A and B is $$M=(\frac{a}{2},\frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$\vec{n} = (\frac{b}{\sqrt{a^2+b^2}},\frac{a}{\sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $\ell = \sqrt{a^2+b^2}$ is $$h = \frac{\sqrt{3}}{2} \sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $\vec{n}$
$$ C = M + \vec{n} h = ( \frac{a}{2}, \frac{b}{2} ) + \sqrt{3} ( \frac{b}{2}, \frac{a}{2} ) = ( \frac{a+b\sqrt{3}}{2}, \frac{b+a\sqrt{3}}{2} )  $$
The locus is $C=(x,y)$ which from the above $a=2x - b\sqrt{3}$ and
 $$ y = \frac{b+(2x - b\sqrt{3})\sqrt{3}}{2} $$
 $$\boxed{y = x\sqrt{3}  - b}$$
