Question

What is the angle between the straight lines $(m^2-mn)y=(mn+n^2)x+n^3$ and $(mn+m^2)y=(mn-n^2)x+m^3$, where $m>n$?

• A. ${\tan }^{-1}\left(\frac{2mn}{m^2+n^2}\right)$
• B. ${\tan }^{-1}\left(\frac{4m^2{n}^2}{m^4-n^4}\right)$
• C. ${\tan }^{-1}\left(\frac{4m^2{n}^2}{m^4+n^4}\right)$
• D. ${45}^o$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{4m^2{n}^2}{m^4-n^4}\right)$

Given $(m^2-mn)y=(mn+n^2)x+n^3$

$(mn+m^2)y=(mn-n^2)x+m^3$

Slope of line $\left(1\right)=\frac{mn+n^2}{m^2-mn}$

Slope of line $\left(2\right)\to \frac{mn-n^2}{m^2+mn}$

By comparing with $y=mx+c$

$\tan \theta =\pm \left(\frac{m_1-m_2}{1+m_1{m}_2}\right),m_1$ and $m_2$ are slope

$\therefore \tan \theta =\frac{(\frac{mn+n^2}{m^2-mn}-\frac{mn-n^2}{m^2+mn})}{\frac{1+(mn+n^2(mn-n^2)}{(m^2-mn)(m^2+mn)}}$

$\tan \theta =\frac{(mn+n^2)(m^2+mn)-(mn-n^2)(m^2-mn)}{m^4-(mn{)}^2+(mn{)}^2-n^4}$

$\tan \theta =\frac{m^{3}n+(mn{)}^2+(mn{)}^2+m{n}^3-m^{3}n+\left(m^2{n}^2\right)+(mn{)}^2-m{n}^3}{m^4-n^4}$

$\tan \theta =\frac{4m^2{n}^2}{m^4-n^4}$

$\therefore \theta ={\tan }^{-1}\left(\frac{4m^2{n}^2}{m^4-n^4}\right)$.