Two perpendicular lines PQ and QR intersect at (1, -1). I...
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
x + 2y - 1= -0
2x + y - 3 = 0
x - 2y - 3 = 0
2x + y - 1 = 0
Correct answer is D
Line PQ : x - 2y + 4 = 0
2y = x + 4 ⟹y=x2+2
Slope = 12
Slope of the perpendicular line QR: −112=−2
Line QR: y=mx+b
y=−2x+b
Point of intersection: (1, -1)
−1=−2(1)+b⟹b=−1+2=1
y=−2x+1⟹y+2x−1=0
QR:2x+y−1=0
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