If one of the pair of lines 4x^2+6xy+ky^2=0 i | vedclass

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(A) The given equations are $4x^2+6xy+ky^2=0$ $(i)$ and $3x^2-5xy+2y^2=0$ (ii).The lines represented by (ii) are $3x^2-3xy-2xy+2y^2=0$ $\Rightarrow 3x

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$2$

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(A) The given equations are $4x^2+6xy+ky^2=0$ $(i)$ and $3x^2-5xy+2y^2=0$ (ii).The lines represented by (ii) are $3x^2-3xy-2xy+2y^2=0$ $\Rightarrow 3x(x-y)-2y(x-y)=0$ $\Rightarrow (3x-2y)(x-y)=0$.Thus,the lines are $y=x$ and $y=\frac{3}{2}x$.If a line $y=mx$ is perpendicular to a line $y=m_1x$,then $m = -\frac{1}{m_1}$.Case $1$: The line $y=x$ is perpendicular to one of the lines of $(i)$. Then the line $y=-x$ must satisfy $(i)$.Substituting $y=-x$ in $(i)$: $4x^2+6x(-x)+k(-x)^2=0$ $\Rightarrow 4-6+k=0$ $\Rightarrow k=2$.Case $2$: The line $y=\frac{3}{2}x$ is perpendicular to one of the lines of $(i)$. Then the line $y=-\frac{2}{3}x$ must satisfy $(i)$.Substituting $y=-\frac{2}{3}x$ in $(i)$: $4x^2+6x(-\frac{2}{3}x)+k(-\frac{2}{3}x)^2=0$ $\Rightarrow 4-4+\frac{4k}{9}=0$ $\Rightarrow k=0$.The possible values of $k$ are $2$ and $0$.The absolute difference is $|2-0|=2$.Twice the absolute difference is $2 	imes 2 = 4$.

If one of the pair of lines $4x^2+6xy+ky^2=0$ is perpendicular to one of the lines represented by $3x^2-5xy+2y^2=0$,then twice the absolute difference of such possible values of $k$ is

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