MHT CET · Maths · Pair of Lines
The combined equation of two lines through the origin and making an angle of \(45^{\circ}\) with the line \(3 x+y=0\), is
- A \(2 x^2-3 x y-2 y^2=0\)
- B \(2 x^2+3 x y+4 y^2=0\)
- C \(2 x^2+3 x y-2 y^2=0\)
- D \(2 x^2-3 x y+2 y^2=0\)
Answer & Solution
Correct Answer
(C) \(2 x^2+3 x y-2 y^2=0\)
Step-by-step Solution
Detailed explanation
Given line \(3 x+y=0 \Rightarrow\) slope \(=-3\)
Let the slope of required line be \(m\)
\(\begin{aligned}
& \therefore \quad \tan 45^{\circ}=\left|\frac{m+3}{1-3 m}\right| \\
& \quad \Rightarrow 1=\left|\frac{m+3}{1-3 m}\right| \\
& \quad \Rightarrow 2 m^2-3 m-2=0...(i)
\end{aligned}\)
Since the line passes through origin, its equation is \(y=m x\)
\(\Rightarrow \mathrm{m}=\frac{y}{x}\)
Substituting the value of \(m\) in equation (i), we get
\(\begin{aligned}
& 2\left(\frac{y}{x}\right)^2-3\left(\frac{y}{x}\right)-2=0 \\
& \Rightarrow 2 y^2-3 x y-2 x^2=0 \\
& \Rightarrow 2 x^2+3 x y-2 y^2=0
\end{aligned}\)
Let the slope of required line be \(m\)
\(\begin{aligned}
& \therefore \quad \tan 45^{\circ}=\left|\frac{m+3}{1-3 m}\right| \\
& \quad \Rightarrow 1=\left|\frac{m+3}{1-3 m}\right| \\
& \quad \Rightarrow 2 m^2-3 m-2=0...(i)
\end{aligned}\)
Since the line passes through origin, its equation is \(y=m x\)
\(\Rightarrow \mathrm{m}=\frac{y}{x}\)
Substituting the value of \(m\) in equation (i), we get
\(\begin{aligned}
& 2\left(\frac{y}{x}\right)^2-3\left(\frac{y}{x}\right)-2=0 \\
& \Rightarrow 2 y^2-3 x y-2 x^2=0 \\
& \Rightarrow 2 x^2+3 x y-2 y^2=0
\end{aligned}\)
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