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

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Aicharya · Accepted Answer

Lines represented by the equation $6 x^2+x y-y^2=0$ are $y=3 x$ and $y=-2 x$ The co-ordinates of the vertices of the triangle formed by above lines with $x+3 y=10$ are $(0,0),(1,3)$ and $(-2,4)$. $	herefore \quad 	ext { Centroid }=\left(\frac{0+1-2}{3}, \frac{0+3+4}{3}ight)=\left(\frac{-1}{3}, \frac{7}{3}ight)$

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

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