The orthocentre of the triangle formed by the lines $x+y=1$ and $2 y^2-x y-6 x^2=0$ is

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Aicharya · Accepted Answer

Given that, $x+y=1$ and $2 y^2-x y-6 x^2=0$ $\begin{aligned} & \Rightarrow \quad 2 y^2-4 x y+3 x y-6 x^2=0 \\ & \Rightarrow \quad(2 y-3 x)+(y-2 x)=0 \\ & \Rightarrow \quad 2 y+3 x \text { and } y-2 x=0 \end{aligned}$ $\therefore \quad$ Equation of sides are $x+y=1,2 y+3 x=0 \text { and } y-2 x=0$

Solving these equations simultaneously, we get the coordinate of the points $\mathrm{A}(0,0), \mathrm{B}\left(\frac{1}{3}, \frac{2}{3}\right)$ and $\mathrm{C}(-2,3)$ Equation of altitude $\mathrm{AD}$, $x-y=0$... (i) Equation of altitude $\mathrm{CF}$, $x+2 y=\lambda$... (ii) Since, this passes through $(-2,3)$ $\therefore \quad-2+6=\lambda \quad \Rightarrow \lambda=4$ So, equation of altitude $\mathrm{CF}$ $x+2 y=4$ On solving eqs. (i) and (ii), we get $x=\frac{4}{3}, y=\frac{4}{3}$ $\therefore$ Orthocentre of the $\triangle \mathrm{ABC}$ is $\left(\frac{4}{3}, \frac{4}{3}\right)$

The orthocentre of the triangle formed by the lines $x+y=1$ and $2 y^2-x y-6 x^2=0$ is

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