The orthocentre of 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

The given lines are $x+3 y=10$ and $6 x^2+x y-y^2=0$

$\begin{aligned} & 6 x^2+3 x y-2 x y-y^2=0 \\ & 3 x(2 x+y)-y(2 x+y)=0 \\ & 3 x-y=0 \\ & 2 x+y=0 \\ & \end{aligned}$ On solving Eqs. (i) and (ii), we get $x+3 \cdot(3 x)=10$ $\begin{aligned} \Rightarrow & 10 x & =10 \\ \Rightarrow & x & =1\end{aligned}$ and $\begin{array}{r}3 \cdot 1-y=0 \\ y=3\end{array}$ $\therefore$ Coordinates of $B$ are $(1,3)$. On solving Eqs. (ii) and (iii), we get $x=0, y=0$ $\therefore$ Coordinates of $A$ are $(0,0)$. A line perpendicular to $B C$ is $3 x-y=\lambda$ It passes through $(0,0)$, then $0-0=\lambda$ $\Rightarrow \quad \lambda=0$ The line $A D$ is $\quad 3 x-y=0$ A line perpendicular to $A C$ is $x-2 y=\lambda$ It passes through $(1,3)$, then $1-6=\lambda$ $\lambda=-5$ The line $B E$ is $x-2 y=-5$ On solving Fqs. (iv) and (v), we get $x=1, y=3$ Thus, the coordinates of required orthocentre is $(1,3)$.

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

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