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Let the given points be A(2, 1), B(p, - 1) and C(-1, 3). Here, we have x 1 = 2, y 1 = 1 x 2 = p, y 2 = -1 and x 3 = -1, y 3 = 3 Given three points, will be collinear if, x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - y 2 ) = 0 ⇒ 2(-1 - 3) + p(3 - 1) + (-1)(1 + 1) = 0 ⇒ 2(- 3) + p(2) + (-1 × 2) = 0 ⇒ -6 + 2p - 2 = 0 ⇒ 2p -8 = 0 ⇒ 2p = 8 Hence the value of p = 4

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Coordinate Geometry

Short Answer Type

231. Find a relation between x and y if the points (x, y); (1, 2) and (7, 0) are collinear.

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232. Find a relation between x and y if the points (2, 1), (x, y) and (7, 5) are collinear.

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233. The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ABC is 12 unit², find the value of k.

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234. Find the area of the ∆ABC with vertices A(-5, 7), B(-4, -5), and C(4, 5).

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235. For what value of k are the points (1, 1), (3, k) and (-1, 4) collinear.

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236. For what value of p are the points (2, 1), (p, -1) and (-1, 3) collinear.

Let the given points be A(2, 1), B(p, - 1) and C(-1, 3).
Here, we have
x₁ = 2,    y₁ = 1
x₂ = p,    y₂ = -1
and    x₃ = -1,    y₃ = 3
Given three points, will be collinear if,
x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) = 0
⇒    2(-1 - 3) + p(3 - 1) + (-1)(1 + 1) = 0
⇒    2(- 3) + p(2) + (-1 × 2) = 0
⇒    -6 + 2p - 2 = 0
⇒    2p -8 = 0
⇒    2p = 8
Hence the value of    p = 4

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237. Find the area of the rhombus whose vertices taken in order are the points (3, 0), (4, 5), (-1, 4) and (-2, -1).

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238.

Find a relation between x and y such that the points P(x, y) is equidistant from A(5, 1) and B(-1, 5).

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239.

Find a relation between x and y such that A(x, y) is equidistant from the points (a + b, b - a) and (a - b, a + b).

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240.

Find a relation between x and y such that P(x, y) is equidistant from the points A(-2, 5) and B( 1, -3).

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