Question

The straight line $ 3x+4y-5=0 $ and $ 4x=3y+15 $ intersect at the point P. On these lines the points Q and R are chosen so that PQ = PR. The slopes of the lines QR passing through (1, 2) are

• A. $ -7,1/7 $
• B. $ 7,1/7 $
• C. $ 7,-1/7 $
• D. $ 3,-1/3 $
• E. $ -3,1/3 $

Answer 1

Answer: (A)

The given equations are $ 3x+4y-5=0 $ ...(i) and $ 4x-3y-15=0 $ ... (ii) Since, there lines are perpendicular to each other, so $ \angle QPR $ is right angle and $ PQ=PR $ .
Hence, $ \Delta PQR $ is a right angle isosceles triangle.
$ \therefore $ $ \angle PQR=\angle PRQ=45{}^\circ $
Slope of $ PQ=-\frac{3}{4} $ and slope of
$ PR=\frac{4}{3} $
Let slope of $ QR\text{ }=m $
$ \therefore $ $ \tan 45{}^\circ =\left| \frac{\frac{4}{3}-m}{1+\frac{4}{3}m} \right| $
$ \Rightarrow $ $ m=\frac{1}{7},-7 $