Prove that the straight lines (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 .

Given:


The given lines are


(a + b) x + (a b) y = 2ab … (1)


(a b) x + (a + b) y = 2ab … (2)


x + y = 0 … (3)


To prove:


The straight lines (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


Assuming:


Let m1, m2 and m3 be the slopes of the lines (1), (2) and (3), respectively.


Explanation:


Now,


Slope of the first line = m1


Slope of the second line = m2


Slope of the third line = m3 = -1


Let θ1 be the angle between lines (1) and (2), θ2 be the angle between lines (2) and (3) and θ3 be the angle between lines (1) and (3).














Here,
θ2 = θ3 and


Hence proved, the given lines form an isosceles triangle whose vertical angle is


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