Locus of vertice of equilateral triangle.

Question

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?

Answer 1

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 $$ y = \frac{b+(2x - b\sqrt{3})\sqrt{3}}{2} $$ $$\boxed{y = x\sqrt{3} - b}$$

Answer 2

Let those lines be $p$ and $q$ (and suppose $B$ is on $q$).

We see that $C$ is rotation of $A$ for $60^{\circ}$ around $B$.

So the locus of $C$ is two perpendicular lines (i.e. lines $p$ and $q$ rotated around $B$ for $60^{\circ}$) and one of them (the map of $q$) goes through $B$.