The angle between the straight lines \(x^2+4 x y+y^2=0\) is.......
The angle between the straight lines \(x^2+4 x y+y^2=0\) is.......
\(30^{\circ}\)
\(45^{\circ}\)
\(60^{\circ}\)
\(90^{\circ}\)
Solution
Let the angle \((\theta)\) between the straight lines
\(x^2+4 x y+y^2=0\)
Then, \(\tan \theta=\frac{2 \sqrt{(2)^2-1}}{1+1} \quad\left\{\because \tan \theta=\frac{2 \sqrt{h^2-a b}}{(a+b)}\right\}\)
\(=\frac{2 \sqrt{3}}{2}=\sqrt{3} \Rightarrow \theta=60^{\circ}\)