The orthocentre of the triangle formed by the | vedclass

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(A) The given lines are $x+3y=10$ and $6x^2+xy-y^2=0$.Factorizing the second equation: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightar

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$(1,3)$

Vedclass · Answer

(A) The given lines are $x+3y=10$ and $6x^2+xy-y^2=0$.Factorizing the second equation: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightarrow (3x-y)(2x+y)=0$.The three lines forming the triangle are $L_1: x+3y=10$,$L_2: 3x-y=0$,and $L_3: 2x+y=0$.Solving $L_1$ and $L_2$: $x+3(3x)=10$ $\Rightarrow 10x=10$ $\Rightarrow x=1, y=3$. Vertex $B = (1,3)$.Solving $L_2$ and $L_3$: $x=0, y=0$. Vertex $A = (0,0)$.Solving $L_1$ and $L_3$: $x+3(-2x)=10$ $\Rightarrow -5x=10$ $\Rightarrow x=-2, y=4$. Vertex $C = (-2,4)$.The altitude from $A$ to $BC$ $(x+3y=10)$ has slope $3$. Equation: $y-0=3(x-0) \Rightarrow 3x-y=0$.The altitude from $B$ to $AC$ $(2x+y=0)$ has slope $1/2$. Equation: $y-3=\frac{1}{2}(x-1)$ $\Rightarrow 2y-6=x-1$ $\Rightarrow x-2y=-5$.Solving the altitude equations $3x-y=0$ and $x-2y=-5$: $y=3x$ $\Rightarrow x-2(3x)=-5$ $\Rightarrow -5x=-5$ $\Rightarrow x=1, y=3$.Thus,the orthocentre is $(1,3)$.

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

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