Find the equation of the line that is perpendicular to (2y + 5x – 6 = 0) and bisects the line joining the points P(4, 3) and Q(-6, 1).
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A.
y + 5x + 3 = 0 -
B.
2y – 5x – 9 = 0 -
C.
5y + 2x – 8 = 0 -
D.
5y – 2x – 12 = 0
Correct Answer: Option D
Explanation
(Line: 2y + 5x – 6 = 0)
(2y = 6 – 5x implies y = 3 – frac{5x}{2})
Gradient = (frac{-5}{2})
For the line perpendicular to the given line, Gradient = (frac{-1}{frac{-5}{2}} = frac{2}{5})
The midpoint of P(4, 3) and Q(-6, 1) = ((frac{-6 + 4}{2}, frac{3 + 1}{2}))
= (-1, 2).
Therefore, the line = (frac{y – 2}{x + 1} = frac{2}{5})
(2(x + 1) = 5(y – 2) implies 5y – 2x – 12 = 0)