Question

The lines $(a+b)x+(a-b)y-2ab=0,(a-b)x+(a+b)y-2ab=0$ and $x+y=0$ forms an isosceles triangle whose vertical angle is

• A. $\frac{\pi }{2}$
• B. $|{\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right)|$
• C. $|{\tan }^{-1}(\frac{a}{b})|$
• D. $2|{\tan }^{-1}\left(\frac{a}{b}\right)|$

Answer 1

The correct option is B $|{\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right)|$

For $(a+b)x+(a-b)y-2ab=0$
Slope is $m_1=-\frac{a+b}{a-b}$
And for $(a-b)x+(a+b)y-2ab=0$
Slope is $m_2=-\frac{a-b}{a+b}$
Then angle is
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{-\frac{a+b}{a-b}+\frac{a-b}{a+b}}{1+\frac{a+b}{a-b}\frac{a-b}{a+b}}\right|=\left|\frac{a^2+b^2-2ab-a^2-b^2-2ab}{a^2-b^2+a^2-b^2}\right|=\left|\frac{2ab}{a^2-b^2}\right|$