$ riangle \mathrm{OAB}$ is formed by the lines $x^2-4 x y+y^2=0$ and the line $A B$. The equation of line $A B$ is $2 x+3 y-1=0$. Then the equation of the median of the triangle drawn from the origin is

Question

$	riangle \mathrm{OAB}$ is formed by the lines $x^2-4 x y+y^2=0$ and the line $A B$. The equation of line $A B$ is $2 x+3 y-1=0$. Then the equation of the median of the triangle drawn from the origin is

Aicharya · Accepted Answer

Let D be the midpoint of line AB . $\begin{array}{ll} \therefore & \mathrm{A}=\left(x_1, y_1\right) \mathrm{B}=\left(x_2, y_2\right) \\ \therefore & \mathrm{D} \equiv\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \end{array}$ Combined equation of side OA and OB is $x^2-4 x y+y^2=0$ and equation of line AB is $2 x+3 y-1=0$ $\therefore \quad$ Points A, B satisfy $x^2-4 x y+y^2=0$ and $\begin{aligned} & 2 x+3 y-1=0 \\ & \Rightarrow x=\frac{1-3 y}{2} \end{aligned}$
Substituting above value in $x^2-4 x y+y^2=0$ $\begin{aligned} \therefore \quad & \left(\frac{1-3 y}{2}\right)^2-4\left(\frac{1-3 y}{2}\right) y+y^2=0 \\ & \Rightarrow(1-3 y)^2-8 y(1-3 y)+4 y^2=0 \\ & 1-6 y+9 y^2-8 y+24 y^2+4 y^2=0 \\ \therefore \quad & 37 y^2-14 y+1=0 \\ & \text { Sum of roots }=\frac{-\mathrm{b}}{\mathrm{a}}=\frac{14}{37} \\ \therefore \quad & y_1+y_2=\frac{14}{37} \end{aligned}$ $y \text {-coordinate of } \dot{\mathrm{D}}=\frac{y_1+y_2}{2}=\frac{7}{37}$
Since point D lies on line AB $\therefore \quad$ Substituting $y=\frac{7}{37}$ in $2 x+3 y-1=0$ $\begin{aligned} & \Rightarrow 2 x+3\left(\frac{7}{37}\right)-1=0 \\ & \Rightarrow 2 x+\frac{21}{37}-1=0 \\ & \Rightarrow x=\frac{8}{37} \\ \therefore & D \equiv\left(\frac{8}{37}, \frac{7}{37}\right) \end{aligned}$
Equation of median AD is $\begin{aligned} & \frac{x-0}{0-\frac{8}{37}}=\frac{y-0}{0-\frac{7}{37}} \\ & \Rightarrow 7 x-8 y=0 \end{aligned}$

$\triangle \mathrm{OAB}$ is formed by the lines $x^2-4 x y+y^2=0$ and the line $A B$. The equation of line…

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