Question

The base $BC$ of $△ABC$ is bisected at $(p,q)$ and equation of sides $AB$ and $AC$ are $px+qy=1$ and $qx+py=1$. Then equation of median through $A$ is

• A. $(2q-1(px+qy)=pq$
• B. $(2pq-1)(px+qy-1)=(p^2+q^2-1)(qx+py-1)$
• C. $(px+qy-1)(qx+py-1)=0$
• D. None of these

Answer 1

Answer: (B)

$(2pq-1)(px+qy-1)=(p^2+q^2-1)(qx+py-1)$

Any line through $A$ is given by $(px+qy-1)+\lambda (qx+py-1)=0$

through $(p,q)$.

Hence $\lambda =\frac{-(p^2+q^2-1)}{(2pq-1)}$

Thus the required line is

$(px+qy-1)-\frac{p^2+q^2-1}{2pq-1}(qx+py-1)=0$