Question

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

Answer 1

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.

Answer 2

Let the given points be A(1,2),B(p,p) and C(-3,-4) which are collinear

If the points are collinear,

Slope of AB = Slope of BC

$\longrightarrow \: \sf \: \dfrac{p - 2}{ p - 1} = \dfrac{ - 4 - p}{ - 3 - p} \\ \\ \longrightarrow \: \sf \: - (p - 2)(p + 3) = - (p - 1)(p + 4) \\ \\ \longrightarrow \: \sf \: - ( {p}^{2} - 2p + 3p - 6) = - ( {p}^{2} - p + 4p - 4) \\ \\ \longrightarrow \: \sf \: p - 6 = 3p - 4 \\ \\ \longrightarrow \: \sf \: 2p = - 2 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: p = - 1}}$

The points would be (1,2),(-1,-1) and (-3,-4)