(C) The given equation of the pair of lines is $x^{2}+kxy+2y^{2}=0$. Since $x+2y=0$ is one of the lines,we can write $x = -2y$. Substituting $x = -2y$

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(C) The given equation of the pair of lines is $x^{2}+kxy+2y^{2}=0$. Since $x+2y=0$ is one of the lines,we can write $x = -2y$. Substituting $x = -2y$ into the equation $x^{2}+kxy+2y^{2}=0$: $(-2y)^{2} + k(-2y)y + 2y^{2} = 0$ $4y^{2} - 2ky^{2} + 2y^{2} = 0$ $6y^{2} - 2ky^{2} = 0$ $2y^{2}(3 - k) = 0$ Since this must hold for all points on the line,we have $3 - k = 0$,which implies $k = 3$.

Class 11 Mathematics Pair of straight lines Equation of pair of straight lines

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If one of the lines given by the equation $x^{2}+kxy+2y^{2}=0$ is $x+2y=0$,then $k=$

MHT CET 2020 All MHT CET

A
$2$
B
$1$
C
$3$
D
$4$

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If one of the lines given by the equation $x^{2}+kxy+2y^{2}=0$ is $x+2y=0$,then $k=$

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