Question

Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines 3x-4y+10 =0 and 5x-12y-10=0 are equal.

Answer 1

Let that point be (h,k) such that  the perpendiculars drawn from it to the two straight lines 3 x-4 y+10 =0 and 5 x-12 y-10=0 are equal.

Length of perpendicular from point (p,q) to line a x + b y + c=0 is given by

   = $\frac{a p + b q + c}{\sqrt{a^2+b^2}}$

So, if you Convert the above condition to find the locus of point (p,q)

→ $\frac{ 3 p - 4 q + 10}{\sqrt{3^2+ 4^2}}=\pm \frac{5 p - 12 q -10}{\sqrt{5^2+12^2}}\\\\ 13 (3 p - 4 q + 10)= 5 (5 p - 12 q - 10)$ Or→ 13 (3 p - 4 q + 10)= -5 (5 p - 12 q - 10)

→39 p - 52 q + 130 = 25 p - 60 q - 50 Or 39 p - 52 q + 130 = -25 p + 60 q + 50

→ 39 p - 25 p - 52 q + 60 q + 130 + 50=0 or  39 p +25 p - 52 q - 60 q + 130 - 50=0

→ 14 p + 8 q + 180=0 or 64 p - 112 q + 80=0

→ 2×(7 p + 4 q +90) =0  or →16×(4 p - 7 q + 5)=0

→ 7 p + 4 q +90=0 ∨ 4 p - 7 q + 5=0→→Locus of point (p,q)  so that the perpendiculars drawn from it to the two straight lines 3x-4y+10 =0 and 5x-12y-10=0 are equal.