The equation of the pair of lines perpendicul | vedclass

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(D) The given equation is $x^2-5xy+4y^2=0$. Factoring the equation: $x^2-4xy-xy+4y^2=0$ $\Rightarrow x(x-4y)-y(x-4y)=0$ $\Rightarrow (x-y)(x-4y)=0$. T

Vedclass · Accepted Answer

$4x^2+5xy+y^2-21x-12y+27=0$

Vedclass · Answer

(D) The given equation is $x^2-5xy+4y^2=0$. Factoring the equation: $x^2-4xy-xy+4y^2=0$ $\Rightarrow x(x-4y)-y(x-4y)=0$ $\Rightarrow (x-y)(x-4y)=0$. The lines are $L_1: x-y=0$ and $L_2: x-4y=0$. The line perpendicular to $x-y=0$ passing through $(2,1)$ is $1(x-2)+1(y-1)=0 \Rightarrow x+y-3=0$. The line perpendicular to $x-4y=0$ passing through $(2,1)$ is $4(x-2)+1(y-1)=0 \Rightarrow 4x+y-9=0$. The combined equation is $(x+y-3)(4x+y-9)=0$. Expanding this: $4x^2+xy-9x+4xy+y^2-9y-12x-3y+27=0$. Simplifying: $4x^2+5xy+y^2-21x-12y+27=0$.

The equation of the pair of lines perpendicular to the lines represented by $x^2-5xy+4y^2=0$ and passing through the point $(2,1)$ is

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