Prove that the lines:
(a+b)x+(a−b)y−2ab=0.......(1)
(a−b)x+(a+b)y−2ab=0.......(2)
and x+y=0......(3) form an isosceles triangle whose vertical angle is 2tan−1(a/b).
Slopes of the given lines are respectively
−a+ba−b,−a−ba+b,−1
The triangle will be isosceles if x+y=0 is equally inclined to other two whose slopes are written above
−1+(a+ba−b)1+(a+ba−b)=−−1+(a+ba−b)1+a−ba+b=tanθ
∴tanθ=b/a
∴ △ is isosceles whose vertical angle is
π−2θ=2(π2−tan−1ba)=2(cot−1ba)=2tan−1ab