Find the locus of points which is equidistant from P(4, 5) and Q(-6, -1).
-
A.
3x – 5y + 13 = 0 -
B.
3x – 5y – 7 = 0 -
C.
5x – 3y + 7 = 0 -
D.
5x + 3y – 1 = 0
Correct Answer: Option D
Explanation
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)