A line is perpendicular to \(3x - y + 11 = 0\) and passes through the point (1, -5). Find its equation.

  • A 3y - x -14 = 0
  • B 3x + y + 1 = 0
  • C 3y + x + 1 = 0
  • D 3y + x + 14 = 0
waec 2011furthermathematics

The correct answer is D. 3y + x + 14 = 0

\(3x - y + 11 = 0 \implies y = 3x + 11\)

\(Gradient = 3\)

Gradient of perpendicular line = \(\frac{-1}{3}\)

\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)

\(3(y + 5) = 1 - x\)

\(3y + x + 15 - 1 = 3y + x + 14 = 0\)

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