There are three equations.
$$(a+b)x+(a-b)y-2ab = 0 \tag1$$ $$(a-b)x+(a+b)y-2ab = 0 \tag2$$ $$x+y = 0 \tag3$$
The question is,
- So that the triangle formed by these equations is an isosceles triangle.
- Show that the angle between the two equal sides of the triangle is $2\tan^{-1}\frac ab$.
What I did:
Part 1.
I could solve (1) and (3) and then (2) and (3) and then calculate two sides of the triangle and show they are equal but that would require much calculations.
So I took a different strategy: Calculate the angle between 1 and 3 and then 1 and 2 and show them to be equal.
But the angle between (1) and (3) is $\tan^{-1}(\pm\frac ba)$ (calculated using gradients of (1) and (3))
And, the angle between (2) and (3) is $\tan^{-1}(\mp\frac ba)$
Which sign should I consider? There will be many cases if different types of values for a and b are taken.
Part 2.
And the angle between (1) and (2) is,
$$\tan^{-1}\left[\pm \frac{\displaystyle\frac{2a}b}{1-\displaystyle\frac{a^2}{b^2}}\right]=2\tan^{-1}\frac ab\text{ or }\pi-2\tan^{-1}\frac ab$$
So I could not calculate one value. How to know which of this two angles is inside the triangle?