A line is perpendicular to \(3x - y + 11 = 0\) and passes through the point (1, -5). Find its equation.
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Correct Answer: Option D
Explanation:
\(3x - y + 11 = 0 \implies y = 3x + 11\)
\(Gradient = 3\)
Gradient of perpendicular line = \(\frac{-1}{3}\)
\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)
\(3(y + 5) = 1 - x\)
\(3y + x + 15 - 1 = 3y + x + 14 = 0\)
\(3x - y + 11 = 0 \implies y = 3x + 11\)
\(Gradient = 3\)
Gradient of perpendicular line = \(\frac{-1}{3}\)
\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)
\(3(y + 5) = 1 - x\)
\(3y + x + 15 - 1 = 3y + x + 14 = 0\)