Find the condition that the points (a, b), (b, a) and (a², – b²) are in a straight line.
Given : (a, b), (b, a) and (a², – b²) are in a straight line.
i.e, they are collinear
Therefore, the area of the triangle formed by the points must be zero.
12|(a²−b³+a²b)–(b²+a³−ab²)|=0
⇒ a² - b³ + a²b – b² – a³ + ab² = 0
⇒ a² – b² – (a³ + b³) + ab (a + b) = 0
⇒ (a + b) [a - b - (a² - ab + b²) + ab] = 0
⇒ (a + b) [(a - b)- (a² - ab + b² - ab)] = 0
⇒ (a + b) [(a - b) - (a - b)²] = 0
⇒ (a + b) (a - b) (1 - a + b) = 0
Therefore, either a + b = 0 or, a – b = 0 or, 1 - a + b = 0.
ie., a + b = 0, a = b and a - b =1.