Question

Find the acute angle between the lines, $(a+b)x+(a-b)y=2ab,(a-b)x+(a+b)y=2ab$ where a > b.

Answer 1

angle between two lines $\Rightarrow \tan \theta =\frac{|m_2-m_1|}{|1+m_1{m}_2|}$

where $m_1=$ slope of line $1\left(\right(a+b)x+(a-b)y=2ab)$

$m_2=$ slope of line $2\left(\right(a-b)x+(a+b)y=2ab)$

slope of line $1=\frac{-(1+b)}{(a-b)}=m_1$

slope of line $2=\frac{-(a-b)}{(a+b)}=m_2$

$\therefore m_1\times m_2=\frac{-(a+b)}{(a-b)}\times \frac{-(a-b)}{(a+b)}=1$

$m_2-m_1=\frac{-(a-b)}{(a+b)}+\frac{(a+b)}{(a-b)}$

$\Rightarrow \frac{-(a-b{)}^2+(a+b{)}^2}{a^2-b^2}\Rightarrow \frac{a^2+b^2+2ab-a^2-b^2+2ab}{a^2-b^2}$

$m_2-m_1\Rightarrow \frac{4ab}{a^2-b^2}$

$\therefore \tan \theta =\frac{(m_2-m_1)}{(1+m_1{m}_2)}\Rightarrow \frac{4ab}{a^2-b^2}\times \frac{1}{(1+1)}$

$\tan \theta \Rightarrow \frac{2ab}{a^2-b^2}$

$\therefore$ acute angle between the lines is ${\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right)$