Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

(N/A) Given lines are $3x - 2y - 5 = 0$ $(1)$ and $3x + 2y - 5 = 0$ $(2)$.
Let $(h, k)$ be any point whose distances from lines $(1)$ and $(2)$ are equal.
The distance of $(h, k)$ from line $(1)$ is $\frac{|3h - 2k - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3h - 2k - 5|}{\sqrt{13}}$.
The distance of $(h, k)$ from line $(2)$ is $\frac{|3h + 2k - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3h + 2k - 5|}{\sqrt{13}}$.
Since the distances are equal,we have $\frac{|3h - 2k - 5|}{\sqrt{13}} = \frac{|3h + 2k - 5|}{\sqrt{13}}$,which implies $|3h - 2k - 5| = |3h + 2k - 5|$.
This gives two cases:
Case $1$: $3h - 2k - 5 = 3h + 2k - 5 \implies -2k = 2k \implies 4k = 0 \implies k = 0$.
Case $2$: $3h - 2k - 5 = -(3h + 2k - 5) \implies 3h - 2k - 5 = -3h - 2k + 5 \implies 6h = 10 \implies h = \frac{5}{3}$.
Thus,the locus of the point $(h, k)$ is $y = 0$ or $x = \frac{5}{3}$,both of which represent straight lines.