1.

Find the value of p which the points (1,2) (p, p) and (-3,-4) are collinear​

Answer»

Solution :-

Given

(1,2) (p, p) and (-3, - 4) are COLLINEAR

Here, x₁ = 1, y₁ = 2, x₂ = p, y₂ = p, x₃ = - 3, y₃ = - 4

If the points are collinear Area of the TRIANGLE = 0

⇒ Area of the triangle = 0

⇒ 1/2 | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) | = 0

⇒ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0

⇒ 1 [p - ( - 4 ) ] + p(-4 - 2) + (-3)(2 - p) = 0

⇒ 1( p + 4 ) + p(-6) - 3(2 - p) = 0

⇒ p + 4 - 6p - 6 + 3P = 0

⇒ - 2p - 2 = 0

⇒ - 2p = 2

⇒ p = 2/ - 2 = - 1

Therefore the VALUE of p is -1.


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