Question

For what value of p. are points (2.1). (p. -1) and (-1.3) collinear?​

Answer 1

$\underline{\boxed{\bold{\bigstar \; Question \; \bigstar}}}$

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➠ For what value of "p" are points (2,1) , (p,-1) and (-1,3) collinear ?

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$\underline{\boxed{\bold{\bigstar \; Important \; Information \; \bigstar}}}$

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❏ If three points are collinear then it forms a straight line. In short a common straight line passes from all these 3 points.

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❏ Area of 3 collinear lines = 0. It is because of reason that line never has area as line is not a closed figure.

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• Here is the formula to find Area if three coordinates is given :

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$\boxed { \bold {\longmapsto \; Area = \dfrac{1}{2}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) } }$

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➠ (x₁ , y₁) = (2 , 1)

➠ (x₂ , y₂) = (p , -1)

➠ (x₃ , y₃) = (-1 , 3)

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$\underline{\boxed{\bold{\bigstar \;Answer \; \bigstar}}}$

Let's Start !

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First of all let's find area in terms of "p" .-.

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$\boxed { \bold {\longmapsto \; Area = \dfrac{1}{2}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)] } }$

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$\bold { : \implies \; Area = \dfrac{1}{2}[2(-1-3) + p(3-1) + (-1)(1 +1)] }$

$\bold { : \implies \; Area = \dfrac{1}{2}[2(-4) + p(2) + (-1)(2)] }$

$\bold { : \implies \; Area = \dfrac{1}{2}[-8 + 2p -2] }$

$\bold { : \implies \; Area = \dfrac{1}{2}[ 2p -10] }$

$\bold { : \implies \; Area = p -5 }$

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• Now as we know that , Area of 3 collinear lines = 0.

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$\bold { : \implies \; Area = p -5 }$

$\bold { : \implies \; 0 = p -5 }$

$\bold { \orange { : \implies \; p = 5 } }$

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$\boxed{\boxed{\bold{\therefore Value \; of \; P = 5 \; and \; coordinates \; of \; point = (5,-1) }}}$

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Hope this helps u.../

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