Question: ABC is an equilateral triangle with A(0, 0) and B(a, 0) (a > 0). L, M, and N are the foot of the perpendiculars drawn from a point P to the sides AB, BC, and CA, respectively. If P lies inside the triangle and satisfies the condition PL² = PM. PN, then find the locus of P.
I solved the question using this method below but got stuck at how the modulus sign is opened with positive, and I don't understand the reason "P lies below both the lines". Please elaborate on the reasoning.
As in $P$ is outside the two straight lines $AC$ and $BC$. Furthermore, when substituting the $x$-coordinate of $P$ (i.e. $h$) into each straight line equation to find $y$, the $y$-coordinate of $P$ (i.e. $k$) is less than that resultant $y$.
For example, $AC$ has the equation $y-\sqrt 3 x = 0$. Substituting the $x=h$ would give $y=\sqrt 3 h$, but $P$ is "below" line $AC$, and so $k < \sqrt 3 h$:
$$\begin{align*} k-\sqrt3 h &< 0\\ \left\lvert k-\sqrt3 h\right\rvert &= -\left(k-\sqrt3 h\right) \end{align*}$$