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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

171.

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

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

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174. 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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175. 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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176. Find the distance between the points
(cos ө - sin ө), (-cos ө, sin ө)

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177. Find the distance between the points
Let the given points be A (a cos ө, 0) and B(0, a sin ө)

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178. Find the distance between the points
Let the given points be A(0, 0) and B(6, 8) we know that distance between two points A(x₁, y₁) and B(x₂, y₂) is given by

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179. Find the co-ordinates of the mid-point of the line segment joining (-5, 4) and (6, -7).

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180. Find the co-ordinates of the mid-point of the line segment joining (4, 7) and (6, 9).

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