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).
- y + 5x + 3 = 0
- 2y - 5x - 9 = 0
- 5y + 2x - 8 = 0
-
5y - 2x - 12 = 0
The correct answer is: 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 = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)
\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)