MHT CET · Maths · Pair of Lines
The equation of pair of lines \(y=\mathrm{p} x\) and \(y=\mathrm{q} x\) can be written as \((y-\mathrm{p} x)(y-\mathrm{q} x)=0\). Then the equation of the pair of the angle bisectors of the lines \(x^2-4 x y-5 y^2=0\) is
- A \(x^2-3 x y+y^2=0\)
- B \(x^2+4 x y-y^2=0\)
- C \(x^2-3 x y-y^2=0\)
- D \(x^2+3 x y-y^2=0\)
Answer & Solution
Correct Answer
(D) \(x^2+3 x y-y^2=0\)
Step-by-step Solution
Detailed explanation
Equation of angle bisector of two lines whose general equation is \(a x^2+2 \mathrm{~h} x y+b y^2=0\) is
\(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
\(\therefore \quad\) Comparing given equation \(x^2-4 x y-5 y^2=0\) with \(\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0\), we get
\(a=1, b=-5, h=-2\)
\(\therefore \quad\) Equation of angle bisectors is
\(\begin{aligned}
& \frac{x^2-y^2}{1-(-5)}=\frac{x y}{-2} \\
& \Rightarrow x^2-y^2=-3 x y \\
& \Rightarrow x^2+3 x y-y^2=0
\end{aligned}\)
\(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
\(\therefore \quad\) Comparing given equation \(x^2-4 x y-5 y^2=0\) with \(\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0\), we get
\(a=1, b=-5, h=-2\)
\(\therefore \quad\) Equation of angle bisectors is
\(\begin{aligned}
& \frac{x^2-y^2}{1-(-5)}=\frac{x y}{-2} \\
& \Rightarrow x^2-y^2=-3 x y \\
& \Rightarrow x^2+3 x y-y^2=0
\end{aligned}\)
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