MHT CET · Maths · Pair of Lines
The equation of the lines passing through the origin and having slopes 3 and \(-\frac{1}{3}\), is
- A \(3 y^{2}+8 x y-3 x^{2}=0\)
- B \(3 x^{2}+8 x y+3 y^{2}=0\)
- C \(3 y^{2}-8 x y-3 x^{2}=0\)
- D \(3 x^{2}+8 x y-3 y^{2}=0\)
Answer & Solution
Correct Answer
(D) \(3 x^{2}+8 x y-3 y^{2}=0\)
Step-by-step Solution
Detailed explanation
Here, \(m_{1}=3, m_{2}=\frac{-1}{3}\)
Hence, the lines are \(y=3 x, y=\frac{-1}{3} x\)
On multiplying both the lines, we get \((y-3 x)(3 y+x)=0\)
\(\Rightarrow \quad 3 x^{2}+8 x y-3 y^{2}=0\)
Hence, the lines are \(y=3 x, y=\frac{-1}{3} x\)
On multiplying both the lines, we get \((y-3 x)(3 y+x)=0\)
\(\Rightarrow \quad 3 x^{2}+8 x y-3 y^{2}=0\)
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