Question

Prove that the straight line $(a+b)x+(a-b)y=2ab,(a-b)x+(a+b)y=2ab$ and $x+y=0$ form an isosceles triangle whose vertical angle is $2ta{n}^{-1}\left(\frac{a}{b}\right)$.

Answer 1

$(a+b)x+(a-b)y=2ab\dots \dots \left(i\right)$

$(a-b)x+(a+b)y=2ab\dots \dots \left(ii\right)$

$x+y=0\dots \dots \left(iii\right)$

Converting all the equation in the form

$y=mx+c$

$y=\frac{-(a+b)x}{a-b}+\frac{2ab}{a+b}$

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

$y=\frac{-(a-b)x}{a+b}+\frac{2ab}{a+b}$

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

$y=-x$

$\Rightarrow m_3=-1$

Thus angle between (i) and (ii)

$tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{-\frac{a+b}{a-b}+\frac{a-b}{a+b}}{1+(\frac{a+b}{a-b}\times \frac{a-b}{a+b})}\right|$

$=\frac{2ab}{b^2-a^2}$

$=\frac{2\frac{a}{b}}{1-{\left(\frac{a}{b}\right)}^2}$

$Tan{\theta }_1=Tan\left(2Ta{n}^{-1}\left(\frac{a}{b}\right)\right)$

${\theta }_1=2Ta{n}^{-1}\left(\frac{a}{b}\right)$

$tan{\theta }_2=\left|\frac{m_2-m_3}{1+m_2{m}_3}\right|$

$=\left|\frac{-\left(\frac{a-b}{a+b}\right)+1}{1+\frac{a-b}{a+b}}\right|$

$=\left|\frac{-a+b+a+b}{a+b+a-b}\right|=\left|\frac{2b}{2a}\right|=\frac{b}{a}$

$tan{\theta }_3=\left|\frac{m_1-m_3}{1+m_1{m}_3}\right|$

$=\left|\frac{-\left(\frac{a+b}{a-b}\right)+1}{1+\frac{a+b}{a-b}}\right|$

$=\left|\frac{-a-b+a-b}{a-b+a+b}\right|=\left|\frac{-2b}{2a}\right|=\frac{b}{a}$

Thus, the vertical angle is $2ta{n}^{-1}\left(\frac{b}{a}\right)$.