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Question: The base BC of a triangle ABC is bisected at the point (p, q) and the equations to the sides AB and ...

The base BC of a triangle ABC is bisected at the point (p, q) and the equations to the sides AB and AC are respectively px+qy=1p x + q y = 1 and qx+py=1q x + p y = 1 Then the equation to the median through A is .

A

(2pq−1)(px+qy−1)=(p2+q2−1)(qx+py−1)( 2 p q - 1 ) ( p x + q y - 1 ) = \left( p ^ { 2 } + q ^ { 2 } - 1 \right) ( q x + p y - 1 )

B

(p2+q2−1)(px+qy−1)=(2p−1)(qx+py−1)\left( p ^ { 2 } + q ^ { 2 } - 1 \right) ( p x + q y - 1 ) = ( 2 p - 1 ) ( q x + p y - 1 )

C

(pq−1)(px+qy−1)=(p2+q2−1)(qx+py−1)( p q - 1 ) ( p x + q y - 1 ) = \left( p ^ { 2 } + q ^ { 2 } - 1 \right) ( q x + p y - 1 )

D

None of these

Answer

(2pq−1)(px+qy−1)=(p2+q2−1)(qx+py−1)( 2 p q - 1 ) ( p x + q y - 1 ) = \left( p ^ { 2 } + q ^ { 2 } - 1 \right) ( q x + p y - 1 )

Explanation

Solution

Since the median passes through A, the intersection of the given lines. Its equation is given by

(px+qy−1)+λ(qx+py−1)=0( p x + q y - 1 ) + \lambda ( q x + p y - 1 ) = 0, where λ\lambda is some real number. Also, since the median passes through the point (p, q), we have (p2+q2−1)+λ(qp+pq−1)=0\left( p ^ { 2 } + q ^ { 2 } - 1 \right) + \lambda ( q p + p q - 1 ) = 0

⇒ λ=−p2+q2−12pq−1\lambda = - \frac { p ^ { 2 } + q ^ { 2 } - 1 } { 2 p q - 1 } and the equation of median through

A is (px+qy−1)−p2+q2−12pq−1(qx+py−1)=0( p x + q y - 1 ) - \frac { p ^ { 2 } + q ^ { 2 } - 1 } { 2 p q - 1 } ( q x + p y - 1 ) = 0

⇒ (2pq−1)(px+qy−1)=(p2+q2−1)(qx+py−1)( 2 p q - 1 ) ( p x + q y - 1 ) = \left( p ^ { 2 } + q ^ { 2 } - 1 \right) ( q x + p y - 1 ).