Question

The base BC of a triangle ABC is bisected at the point(p, q) and the equation to the side AB & AC are $px+qy=1$ and $qx+py=1$. The equation of the median through A is?

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

Answer 1

Answer: (C)

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

Equation of family of lines through AB and AC.

$⟹(px+qy-1)+\lambda (qx+py-1)=0-\left(1\right)$

Let this be equation of AD

$⟹$ It passes through D(p,q)

$⟹(p^2+q^2-1)+\lambda (pq+pq-1)=0$

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

putting in equation (1)

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

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