Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
3x - 5y + 13 = 0
3x - 5y - 7 = 0
5x - 3y + 7 = 0
5x + 3y - 1 = 0
Correct answer is D
Midpoint between P(4, 5) and Q(-6, -1) = \(\frac{4 - 6}{2}, \frac{5 - 1}{2} = (-1, 2)\)
Gradient of PQ = \(\frac{5 - (-1)}{4 - (-6)} = \frac{3}{5}\)
Gradient of line perpendicular to PQ = \(\frac{-1}{\frac{3}{5}} = -\frac{5}{3}\)
\(line = \frac{y - 2}{x + 1} = \frac{-5}{3}\)
\(3y - 6 = -5x - 5\)
\(5x + 3y - 6 + 5 = 0 \implies 5x + 3y - 1 = 0\)