MHT CET · Maths · Pair of Lines
The joint equation of bisectors of the angle between the lines represented by
\(3 x^{2}+2 x y-y^{2}=0\) is
- A \(x^{2}-4 x y-y^{2}=0\)
- B \(x^{2}+4 x y-y^{2}=0\)
- C \(x^{2}-4 x y+y^{2}=0\)
- D \(x^{2}+4 x y+y^{2}=0\)
Answer & Solution
Correct Answer
(A) \(x^{2}-4 x y-y^{2}=0\)
Step-by-step Solution
Detailed explanation
(D)
Equation of lines is \(3 x^{2}+2 x y-y^{2}=0 .\) Comparing with \(a x^{2}+2 h x y+b y^{2}=0\), we write \(a=3 . h=1 . b=-1\)
Combined equation of pair of angle bisectors of given lines is
\(\begin{aligned} & \frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h} \\ \therefore & \frac{x^{2}-y^{2}}{3-(-1)}=\frac{x y}{1} \Rightarrow \frac{x^{2}-y^{2}}{4}=x y \\ & x^{2}-4 x y-y^{2}=0 \end{aligned}\)
Equation of lines is \(3 x^{2}+2 x y-y^{2}=0 .\) Comparing with \(a x^{2}+2 h x y+b y^{2}=0\), we write \(a=3 . h=1 . b=-1\)
Combined equation of pair of angle bisectors of given lines is
\(\begin{aligned} & \frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h} \\ \therefore & \frac{x^{2}-y^{2}}{3-(-1)}=\frac{x y}{1} \Rightarrow \frac{x^{2}-y^{2}}{4}=x y \\ & x^{2}-4 x y-y^{2}=0 \end{aligned}\)
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