Q. The base $BC$ of $\mathrm{\Delta}\ ABC$ is bisected at $(p,\ q)$ and equation of sides $AB$ and $AC$ are $px + qy = 1$ and $qx + py = 1$ respectively. Then, the equation of the median through $A$ is
A) $(2pq - 1)(px + qy - 1) = \left( p^{2} + q^{2} - 1 \right)(qx + py - 1)$
B) $(qx + qy - 1)(qx + py - 1) = 0$
C) $(px + qy - 1)(qx - py - 1) = 0$
D) None of the above
Explanation
(a)
Any line through $A$ is given by
$(px + qy - 1) + \lambda(qx + py - 1) = 0$
Which is passing through $(p,\ q)$
$Hence,\ \lambda = - \frac{\left( p^{2} + q^{y} - 1 \right)}{2pq - 1}$
Thus, the required line is
$(px + qy - 1) - \frac{\left( p^{2} + q^{2} - 1 \right)}{(2pq - 1)} \bullet (qx + py - 1) = 0$
$\Rightarrow (2pq - 1)(px + qy - 1) - \left( p^{2} + q^{2} - 1 \right)(qx + py - 1) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.