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Question

Find the condition that the points (a, b), (b, a) and (a², – b²) are in a straight line.

A
a + b = 0
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B
a = b
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C
a + b = 1
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D
a + b = -1
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Solution

The correct options are
A a + b = 0
B a = b

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|(a²b³+a²b)(b²+a³ab²)|=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.


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