A line is perpendicular to \(3x - y + 11 = 0\) and passes through
FURTHER MATHEMATICS
WAEC 2011
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
Correct Answer: D. 3y + x + 14 = 0
Explanation
\(3x - y + 11 = 0 \implies y = 3x + 11\)
\(Gradient = 3\)
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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