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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

161.

Find the co-ordinates of the mid-points of the line joining the points (-8, 4) and (6, -2).

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162. Find the area of triangle formed by the points P(2, 1), Q(6, 1) and R (2,4).

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163. Find the area of the triangle formed by the points A (5, 2), B (4, 7) and C (7, -4)

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164. (-2, 2), (x. 8) and (6, y) are three concyclic points, whose centre is (2. 5), find the possible values of x and y.

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165. Find the vertices of the triangle, the mid-points of whose sides are (3, 1), (5, 6) and (-3, 2).

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166. If the mid-point of the line joining (3, 4) and (13, 7) is (x, y) and 2x + 2y 1 = 0, find the value of k.

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167. If P(2, -1), Q(3, 4). R( 2, 3) and S(-3, -2) are four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

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168. The points (3, -4) and (-6, 2) are the extremities of a diagonal of a parallelogram if the third vertex is (-1, 3), find the co-ordinates of the fourth vertex.

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169. A, B are two points (3, 4) and (5, 2). find the point P such that PA = PB and area of ∆PAB equal to 10.

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170. 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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