Question

The base $B C$ of a $\Delta A B C$ is bisected at the point $(p, q)$

and the equation to the sides $A B$ and $A C$ are $p x+q y=1$ and $q x+p y=1$. The equation of the median through $A$ is:

(A) $q x-p y=0$

(B) $\dfrac{x}{p}+\dfrac{y}{q}=2$

(C) $(2 p q-1)(p x+q y-1)=\left(p^{2}+q^{2}-1\right)(q x+p y-1)$

(D) None of the above

Answer 1

Hint: First of all write the equation of the family of lines passing through AB and AC then from the family of lines, one line is the median and passing through point (p, q). Satisfy this point (p, q) in the family of lines then you will get the equation of the median passing through A.

Complete step-by-step solution -

Let's say the equation of side $A B$ is $L_{1}$ and equation of side $B C$ is $L_{2}$.

We are going to write the family of lines passing through $\mathrm{AB}$ and $\mathrm{AC}$.

$L_{1}+\lambda L_{2}=0$

Substituting the value of $L_{1}$ and $L_{2}$ in the above equation we get,

$p x+q y-1+\lambda(q x+p y-1)=0 \ldots \ldots \ldots \ldots \ldots \ldots . .$ Eq. (1)

Now, in this family of lines equation, one of the family of lines is the

median and passing through point $(p, q)$ so satisfying the point $(p, q)$ in

the above equation will give the value of $\lambda$.

$p^{2}+q^{2}-1+\lambda(q p+p q-1)=0$

$\Rightarrow \lambda(2 p q-1)=1-p^{2}-q^{2}$

$\Rightarrow \lambda=\dfrac{1-p^{2}-q^{2}}{2 p q-1}$

Substituting the value of $\lambda$ in eq. (1) we get,

$p x+q y-1+\dfrac{1-p^{2}-q^{2}}{2 p q-1}(q x+p y-1)=0$

$\Rightarrow(2 p q-1)(p x+q y-1)+\left(1-p^{2}-q^{2}\right)(q x+p y-1)=0$

$\Rightarrow(2 p q-1)(p x+q y-1)=(q x+p y-1)\left(p^{2}+q^{2}-1\right)$

The above equation is the equation of median passing through $\mathrm{A}$.

Hence, the correct option is (c).

Note: There is an alternative way of solving the problem by first getting the intersection point of lines passing through side AB and AC then you will get coordinates of A and we have a point (p, q) also. Now, we can write the equation of a straight line if two points are given.