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: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightarrow (3x-y)(2x+y)=0$.So,the lines are $3x-y=0$ and $2x+y=0$.Let the vertices of the triangle be $A, B, C$.Solving $2x+y=0$ and $3x-y=0$ gives $A(0,0)$.Solving $x+3y=10$ and $3x-y=0$ gives $B(1,3)$.Solving $x+3y=10$ and $2x+y=0$ gives $C(-2,4)$.The altitude from $A(0,0)$ to $BC$ $(x+3y=10)$ has slope $3$. Equation: $y-0=3(x-0) \Rightarrow 3x-y=0$.The altitude from $B(1,3)$ to $AC$ $(2x+y=0)$ has slope $1/2$. Equation: $y-3=\frac{1}{2}(x-1) \Rightarrow x-2y+5=0$.Solving $3x-y=0$ and $x-2y+5=0$: $y=3x$ $\Rightarrow x-2(3x)+5=0$ $\Rightarrow -5x=-5$ $\Rightarrow x=1, y=3$.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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