The orthocentre of the triangle formed by the | vedclass

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(A) Given lines are $x+y=1$ and $2y^2-xy-6x^2=0$. Factorizing the second equation: $2y^2-4xy+3xy-6x^2=0$ $\Rightarrow 2y(y-2x)+3x(y-2x)=0$ $\Rightarro

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$\left(\frac{4}{3}, \frac{4}{3}\right)$

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(A) Given lines are $x+y=1$ and $2y^2-xy-6x^2=0$. Factorizing the second equation: $2y^2-4xy+3xy-6x^2=0$ $\Rightarrow 2y(y-2x)+3x(y-2x)=0$ $\Rightarrow (2y+3x)(y-2x)=0$. Thus,the sides of the triangle are $L_1: x+y=1$,$L_2: y-2x=0$,and $L_3: 2y+3x=0$. Finding vertices: Intersection of $L_2$ and $L_3$ is $A(0,0)$. Intersection of $L_1$ and $L_2$: $x+2x=1 \Rightarrow x=1/3, y=2/3$,so $B(1/3, 2/3)$. Intersection of $L_1$ and $L_3$: $x+(-3x/2)=1$ $\Rightarrow -x/2=1$ $\Rightarrow x=-2, y=3$,so $C(-2, 3)$. Altitude from $A(0,0)$ to $BC$ $(x+y=1)$: The slope of $BC$ is $-1$,so the altitude slope is $1$. Equation: $y-0=1(x-0) \Rightarrow x-y=0$. Altitude from $C(-2,3)$ to $AB$ $(y-2x=0)$: The slope of $AB$ is $2$,so the altitude slope is $-1/2$. Equation: $y-3 = -1/2(x+2)$ $\Rightarrow 2y-6 = -x-2$ $\Rightarrow x+2y=4$. Solving $x-y=0$ and $x+2y=4$: $y+2y=4$ $\Rightarrow 3y=4$ $\Rightarrow y=4/3$. Thus $x=4/3$. The orthocentre is $\left(\frac{4}{3}, \frac{4}{3}\right)$.

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

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