Find the equation of the straight line that passes through (2, -3) and perpendicular to
FURTHER MATHEMATICS
WAEC 2013
Find the equation of the straight line that passes through (2, -3) and perpendicular to the line 3x - 2y + 4 = 0.
- A. 2y - 3x = 0
- B. 3y - 2x + 5 = 0
- C. 3y + 2x + 5 = 0
- D. 2y - 3x - 5 = 0
Correct Answer: C. 3y + 2x + 5 = 0
Explanation
Given line: \(3x - 2y + 4 = 0 \implies 2y = 3x + 4\)
\(y = \frac{3}{2}x + 2\)
\(Gradient (\frac{\mathrm d y}{\mathrm d x}) = \frac{3}{2}\)
Gradient of perpendicular line = \(\frac{-1}{\frac{3}{2}} = \frac{-2}{3}\)
\(\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}\)
\(\frac{y + 3}{x - 2} = \frac{-2}{3} \)
\(3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0\)
= \(3y + 2x + 5 = 0\)
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