Question

Find acute angle Q between lines 4x² + 5xy + y2 = 0 ?​

Answer 1

Answer:

The answer is $\tan^{-1}\frac{3}{4}=37^\circ$

Step-by-step explanation:

Given lines can be separated as :

$4x^2+5xy+y^2=0\\\Rightarrow (4x^2+4xy)+(xy+y^2)=0\\\Rightarrow 4x(x+y)+y(x+y)=0\\\Rightarrow (x+y)(4x+y)=0$

$\Rightarrow x+y=0$, $4x+y=0$

So slopes of these lines are $m_1=-1,m_2=-4$

Let $\theta$ be the acute angle between these lines , then

                  $\tan\theta=|\frac{m_1-m_2}{1+m_1m_2}|=\frac{-1+4}{1+4}=\frac{3}{4}\\\Rightarrow \theta=\tan^{-1}\frac{3}{4}=37^\circ$