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 $QR=m$
$\therefore$ $\tan 45{}^{\circ }=\left|\frac{\frac{4}{3}-m}{1+\frac{4}{3}m}\right|$
$\Rightarrow$ $m=\frac{1}{7},-7$