The Centroid of the triangle formed by the lines $6 x^2+x y-y^2=0$ and $x+3 y-10=0$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 6 x^2+x y-y^2=0 \ & \Rightarrow(3 x-y)(2 x+y)=0 \ & \Rightarrow 3 x-y=0,2 x+y=0 \ & \Rightarrow L_1 \equiv 3 x-y=0, L_2 \equiv 2 x+y=0 	ext { and } L_3 \equiv x+3 y=10 \ & 	ext { point of intersection of } \ & L_1 	ext { and } L_2 	ext { is }(0,0), L_2 	ext { and } L_3 	ext { is }(-2,4), L_3 	ext { and } L_1 	ext { is }(1,3) \ & \Rightarrow 	ext { centroid of triangle } \equiv\left(\frac{0-2+1}{3}, \frac{0+4+3}{3}ight) \equiv\left(\frac{-1}{3}, \frac{7}{3}ight)\end{aligned}$

The Centroid of the triangle formed by the lines $6 x^2+x y-y^2=0$ and $x+3 y-10=0$ is

Solution