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Find the equation of the line that is perpendicular to \(2y + 5x - 6...

Further Mathematics
WAEC 2013

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).

  • A. y + 5x + 3 = 0
  • B. 2y - 5x - 9 = 0
  • C. 5y + 2x - 8 = 0
  • D. 5y - 2x - 12 = 0
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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 = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)

\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts