A $\triangle OPQ$ is formed by the pair of straight lines $x^2 - 4 xy + y^2 = 0$ and the line $PQ$. The equation of $PQ$ is $x + y - 2 = 0$. Find the equation of the median of $\triangle OPQ$ drawn from the origin.
$x^2 - 4xy + y^2 = 0$ $\;\;\; \cdots \; (1)$
represents a pair of straight lines passing through the origin $O \left(0,0\right)$.
Equation $(1)$ can be written as
$x^2 - 4xy + 4y^2 - 3y^2 = 0$
i.e. $\left(x - 2y\right)^2 - \left(\sqrt{3}y\right)^2 = 0$
i.e. $\left(x - 2y + \sqrt{3}y\right) \left(x - 2y - \sqrt{3}y\right) = 0$
$\therefore$ $\;$ Equation $(1)$ represents the pair of lines
$x - y \left(2 - \sqrt{3}\right) = 0$ and $x - y \left(2 + \sqrt{3}\right) = 0$
i.e. $x = y \left(2 - \sqrt{3}\right)$ $\;\;\; \cdots \; (2a)$ and
$x = y \left(2 + \sqrt{3}\right)$ $\;\;\; \cdots \; (2b)$
Equation of line $PQ$ is $\;$ $x + y - 2 = 0$ $\;\;\; \cdots \; (3)$
Solving equations $(2a)$ and $(3)$ simultaneously we have,
$y \left(2 - \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 - \sqrt{3}}$ $\;\;\; \cdots \; (4a)$
Substituting the value of $y$ from equation $(4a)$ in equation $(2a)$ we get,
$x = \left(\dfrac{2}{3 - \sqrt{3}}\right) \times \left(2 - \sqrt{3}\right)$
i.e. $x = \dfrac{2 \times \left(2 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}{\left(3 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}$
i.e. $x = \dfrac{2 \times \left(6 + 2 \sqrt{3} - 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 - \sqrt{3}}{3}$
$\therefore$ $\;$ The point of intersection of equations $(2a)$ and $(3)$ is $A \left(\dfrac{3 - \sqrt{3}}{3}, \dfrac{2}{3 - \sqrt{3}}\right)$
Solving equations $(2b)$ and $(3)$ simultaneously we have,
$y \left(2 + \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 + \sqrt{3}}$ $\;\;\; \cdots \; (4b)$
Substituting the value of $y$ from equation $(4b)$ in equation $(2b)$ we get,
$x = \left(\dfrac{2}{3 + \sqrt{3}}\right) \times \left(2 + \sqrt{3}\right)$
i.e. $x = \dfrac{2 \times \left(2 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}{\left(3 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}$
i.e. $x = \dfrac{2 \times \left(6 - 2 \sqrt{3} + 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 + \sqrt{3}}{3}$
$\therefore$ $\;$ The point of intersection of equations $(2b)$ and $(3)$ is $B \left(\dfrac{3 + \sqrt{3}}{3}, \dfrac{2}{3 + \sqrt{3}}\right)$
Midpoint of $AB$ is $\;$ $M = \left(\dfrac{\dfrac{3 - \sqrt{3} + 3 + \sqrt{3}}{3}}{2}, \dfrac{\dfrac{2}{3 - \sqrt{3}} + \dfrac{2}{3 + \sqrt{3}}}{2}\right)$
i.e. $M = \left(1, \dfrac{6 + 2 \sqrt{3} + 6 - 2 \sqrt{3}}{2 \times \left(3 - \sqrt{3}\right) \left(3 + \sqrt{3}\right)}\right)$
i.e. $M = \left(1, \dfrac{12}{2 \times 6}\right)$ $\implies$ $M = \left(1,1\right)$
$\therefore$ $\;$ Equation of median $MO$ is
$y - 0 = \left(\dfrac{1 - 0}{1 - 0}\right) \left(x - 0\right)$
$\therefore$ $\;$ The required equation of the median of $\triangle OPQ$ drawn from the origin is $\;$ $y = x$