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If one of the lines given by the equation $x^{2}+k x y+2 y^{2}=0$ is $x+2 y=0$, then $\mathrm{k}=$
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3
We have $x^{2}+k x y+2 y^{2}=0$ i.e. $1+k\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^{2}=0$
Slope of line $x+2 y=0$ is $-\frac{1}{2}$
Substituting $\frac{y}{x}=-\frac{1}{2}$, we get
$1+\mathrm{k}\left(-\frac{1}{2}\right)+2\left(-\frac{1}{2}\right)^{2}=0 \Rightarrow 1-\frac{\mathrm{k}}{2}+\frac{1}{2}=0 \Rightarrow \mathrm{k}=3$
Slope of line $x+2 y=0$ is $-\frac{1}{2}$
Substituting $\frac{y}{x}=-\frac{1}{2}$, we get
$1+\mathrm{k}\left(-\frac{1}{2}\right)+2\left(-\frac{1}{2}\right)^{2}=0 \Rightarrow 1-\frac{\mathrm{k}}{2}+\frac{1}{2}=0 \Rightarrow \mathrm{k}=3$
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