Question

Prove that the lines:
$(a+b)x+(a-b)y-2ab=\mathrm{0.......}\left(1\right)$
$(a-b)x+(a+b)y-2ab=\mathrm{0.......}\left(2\right)$
and $x+y=\mathrm{0......}\left(3\right)$ form an isosceles triangle whose vertical angle is $2{\tan }^{-1}\left(a/b\right)$.

Answer 1

Slopes of the given lines are respectively
$-\frac{a+b}{a-b},-\frac{a-b}{a+b},-1$
The triangle will be isosceles if $x+y=0$ is equally inclined to other two whose slopes are written above
$\frac{-1+\left(\frac{a+b}{a-b}\right)}{1+\left(\frac{a+b}{a-b}\right)}=-\frac{-1+\left(\frac{a+b}{a-b}\right)}{1+\frac{a-b}{a+b}}=\tan \theta$
$\therefore \tan \theta =b/a$
$\therefore$ $△$ is isosceles whose vertical angle is
$\pi -2\theta =2(\frac{\pi }{2}-{\tan }^{-1}\frac{b}{a})=2\left({\cot }^{-1}\frac{b}{a}\right)=2{\tan }^{-1}\frac{a}{b}$