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If the equation $a x^{2}+b y^{2}+c x+c y=0, c \neq 0$ represents a pair of lines, then
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$a+b=0$
Given $a x^{2}+b y^{2}+c x+c y=0$ represent pair of line if
$\therefore\left|\begin{array}{ccc}\mathrm{a} & 0 & \frac{\mathrm{c}}{2} \\ 0 & \mathrm{~b} & \frac{\mathrm{c}}{2} \\ \frac{\mathrm{c}}{2} & \frac{\mathrm{c}}{2} & 0\end{array}\right|=0$
$\therefore \quad-\frac{a c^{2}}{4}+\frac{c}{2}\left(\frac{0-b c}{2}\right)=0 \Rightarrow \frac{-a c^{2}}{4}-\frac{b c^{2}}{4}=0$
$-a c^{2}-b c^{2}=0 \quad \Rightarrow \quad c^{2}(a+b)=0$
$\therefore a+b=0 \quad \ldots[\because c \neq 0$, given $]$
$\therefore\left|\begin{array}{ccc}\mathrm{a} & 0 & \frac{\mathrm{c}}{2} \\ 0 & \mathrm{~b} & \frac{\mathrm{c}}{2} \\ \frac{\mathrm{c}}{2} & \frac{\mathrm{c}}{2} & 0\end{array}\right|=0$
$\therefore \quad-\frac{a c^{2}}{4}+\frac{c}{2}\left(\frac{0-b c}{2}\right)=0 \Rightarrow \frac{-a c^{2}}{4}-\frac{b c^{2}}{4}=0$
$-a c^{2}-b c^{2}=0 \quad \Rightarrow \quad c^{2}(a+b)=0$
$\therefore a+b=0 \quad \ldots[\because c \neq 0$, given $]$
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