Question 59

Mathematics Line Hard

The lines \(px+qy=1\) and \(qx+py=1\) are respectively the sides AB, AC of the triangle ABC and the base BC is bisected at \((p,q)\). Equation of the median of the triangle through the vertex A is

(A) \((2pq-1)(qx+py-1)-(p^2+q^2-1)(px+qy-1)=0\)
(B) \((2pq-1)(px+qy-1)+(p^2+q^2-1)(qx+py-1)=0\)
(C) \((2pq-1)(px+qy-1)-(p^2+q^2-1)(qx+py-1)=0\)
(D) \((2pq-1)(qx+py-1)+(p^2+q^2-1)(px+qy-1)=0\)
View Dynamic Solution & Explanation
Correct Solution: Option C

Step-by-step Solution:

The median will pass through \(A\) and the midpoint of the opposite side, i.e., \((p, q)\). The general equation of the line that passes through the intersection point of the lines \(AC\) and \(AB\) is: \[ (px + qy - 1) + \lambda (qx + py - 1) = 0 \] As this line passes through the point \((p, q)\), hence: \[ (p^2 + q^2 - 1) + \lambda (2pq - 1) = 0 \Rightarrow \lambda = - \frac{p^2 + q^2 - 1}{(2pq - 1)} \] Putting this value of \(\lambda\), we get the desired equation: \[ (px + qy - 1) - \left( \frac{p^2 + q^2 - 1}{2pq - 1} \right)(qx + py - 1) = 0 \] Or \[ (2pq - 1)(px + qy - 1) - (p^2 + q^2 - 1)(qx + py - 1) = 0 \]