MHT CET · Maths · Pair of Lines
If the equation \(a x^{2}+b y^{2}+c x+c y=0, c \neq 0\) represents a pair of lines, then
- A \(a+c=0\)
- B \(a+b=0\)
- C \(a-c=0\)
- D \(a-b=0\)
Answer & Solution
Correct Answer
(B) \(a+b=0\)
Step-by-step Solution
Detailed explanation
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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