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Given points will be collinear, if the area of the triangle formed by there is zero. Here, we have, x 1 = 2, y 1 = 1; x 2 = x, y 2 = y; x 3 = 7, y 3 = 5. Now, x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - 2 )= 0 ⇒ 2(y - 5) + x(5 - 1) + 7( 1 - y) = 0 ⇒ 2(y - 5) + x(4) + 7(1 - y) = 0 ⇒ 2y - 10 + 4x + 7 - 7y = 0 ⇒ 4x - 5y - 3 = 0 ⇒ 4x - 5y = 3

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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.

Given points will be collinear, if the area of the triangle formed by there is zero.
Here, we have,
x₁ = 2,    y₁ = 1;
x₂ = x,    y₂ = y;
x₃ = 7,    y₃ = 5.
Now,
x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ -₂)= 0
⇒    2(y - 5) + x(5 - 1) + 7( 1 - y) = 0
⇒    2(y - 5) + x(4) + 7(1 - y) = 0
⇒    2y - 10 + 4x + 7 - 7y = 0
⇒    4x - 5y - 3 = 0
⇒    4x - 5y = 3

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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.

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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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