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# For what values of p are the points $\left( {2,1} \right),\left( {p, - 1} \right){\text{ }}and{\text{ }}\left( {1, - 3} \right)$ collinear.

Last updated date: 21st Mar 2023
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Hint- In order to solve such types of questions, we will use the basic property that the given points are collinear if and only if the area of the triangle formed by given points should be zero.

Given points are $(2,1),(p, - 1){\text{ and( - 1,3)}}$
Let $A(2,1),B(p, - 1){\text{ andC( - 1,3)}}$ be the coordinates of the straight line.
$\Delta ABC = 0$
$\therefore \Delta = \left| {\begin{array}{*{20}{c}} 2&1&1 \\ p&{ - 1}&1 \\ { - 1}&3&1 \end{array}} \right|$
$\Delta = \dfrac{1}{2}\left[ {1\left| {\begin{array}{*{20}{c}} p&{ - 1} \\ { - 1}&3 \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 2&1 \\ { - 1}&3 \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} 2&1 \\ p&{ - 1} \end{array}} \right|} \right] \\ = \dfrac{1}{2}\left[ {(3p - 1) - 1(6 + 1) + 1( - 2 - p)} \right] \\ = \dfrac{1}{2}\left[ { - 3p - 1 - 7 - 2 - p} \right] \\ = \dfrac{1}{2}[2p - 10] \\ = p - 5 \\$
$\Rightarrow p - 5 = 0 \\ \Rightarrow p = 5 \\$