Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR
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Correct Answer: Option D
Explanation:
Line PQ : x - 2y + 4 = 0
2y = x + 4 \(\implies y = \frac{x}{2} + 2 \)
Slope = \(\frac{1}{2}\)
Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\)
Line QR: \(y = mx + b\)
\(y = -2x + b\)Â
Point of intersection: (1, -1)
\(-1 = -2(1) + b \implies b = -1 + 2 = 1\)
\(y = -2x + 1 \implies y + 2x - 1 = 0\)
\(QR: 2x + y - 1 = 0\)
Line PQ : x - 2y + 4 = 0
2y = x + 4 \(\implies y = \frac{x}{2} + 2 \)
Slope = \(\frac{1}{2}\)
Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\)
Line QR: \(y = mx + b\)
\(y = -2x + b\)Â
Point of intersection: (1, -1)
\(-1 = -2(1) + b \implies b = -1 + 2 = 1\)
\(y = -2x + 1 \implies y + 2x - 1 = 0\)
\(QR: 2x + y - 1 = 0\)