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Question
In the following, find the value of ‘a’ for which the given points are collinear
(2, 3), (4, a) and (6, – 3)
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Solution
Let the points be A(2, 3), B(4, a) and C(6, – 3).
Since the given points are collinear.
Area of a triangle = 0
`1/2 [(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]` = 0
`1/2[(2"a" - 12 + 18) - (12 + 6"a" - 6)]` = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
– 4a = 0
⇒ a = `0/4`
= 0
The value of a = 0
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