A) \[-7,1/7\]
B) \[7,1/7\]
C) \[7,-1/7\]
D) \[3,-1/3\]
E) \[-3,1/3\]
Correct Answer: A
Solution :
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 \[QR\text{ }=m\] \[\therefore \] \[\tan 45{}^\circ =\left| \frac{\frac{4}{3}-m}{1+\frac{4}{3}m} \right|\] \[\Rightarrow \] \[m=\frac{1}{7},-7\]
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