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Question
In the following, find the value of ‘a’ for which the given points are collinear
(a, 2 – 2a), (– a + 1, 2a) and (– 4 – a, 6 – 2a)
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Solution
Let the points be A(a, 2 – 2a), B(– a + 1, 2a) C(– 4 – a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 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"^2 + (- "a" + 1)(6 - 2"a") + (-4 - "a")(2 - 2"a"))), (-((- "a" + 1)(2 - 2"a") + 2"a"(-4 - "a") + (6 - 2"a")"a"))]` = 0
`1/2[((2"a"^2 - 6"a" + 2"a"^2 + 6 - 2"a" - 8 + 8"a" - 2"a" + 2"a"^2)-), ((-2"a" + 2"a"^2 + 2 - 2"a" - 8"a" - 2"a"^2 + 6"a" - 2"a"^2))]` = 0
`1/2[6"a"^2 - 10"a" + 8"a" - 2 - (2"a"^2 - 4"a"^2 - 12"a" + 6"a" + 2)]` = 0
6a2 – 2a – 2 – (– 2a2 – 6a + 2) = 0
6a2 – 2a – 2 + 2a2 + 6a – 2 = 0
8a2 + 4a – 4 = 0 ...(Divided by 4)
2a2 + a – 1 = 0
2a2 + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0
(a + 1) (2a – 1) = 0
a + 1 = 0 or 2a – 1 = 0
a = – 1 or 2a = 1 ⇒ a = `1/2`
The value of a = – 1 or `1/2`
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