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

Answer

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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.

Complete step-by-step answer:

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.

We know that; the given points are said to be collinear if area of

$\Delta ABC = 0$

So, we will proceed further by determining the area of triangle

$\therefore \Delta = \left| {\begin{array}{*{20}{c}}

2&1&1 \\

p&{ - 1}&1 \\

{ - 1}&3&1

\end{array}} \right|$

Finding the determinant of above equation with the help of 3rd column

$

\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 \\

$

Since, area of the triangle should be zero

$

\Rightarrow p - 5 = 0 \\

\Rightarrow p = 5 \\

$

Hence, the value of point p is 5.

Note- To solve these types of questions remember the basic properties of the triangle. In this question we use the basic property of straight line to solve the question i.e. the area of the straight line is zero. We find the area using a matrix formula to determine the area of the triangle.

Complete step-by-step answer:

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.

We know that; the given points are said to be collinear if area of

$\Delta ABC = 0$

So, we will proceed further by determining the area of triangle

$\therefore \Delta = \left| {\begin{array}{*{20}{c}}

2&1&1 \\

p&{ - 1}&1 \\

{ - 1}&3&1

\end{array}} \right|$

Finding the determinant of above equation with the help of 3rd column

$

\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 \\

$

Since, area of the triangle should be zero

$

\Rightarrow p - 5 = 0 \\

\Rightarrow p = 5 \\

$

Hence, the value of point p is 5.

Note- To solve these types of questions remember the basic properties of the triangle. In this question we use the basic property of straight line to solve the question i.e. the area of the straight line is zero. We find the area using a matrix formula to determine the area of the triangle.

Last updated date: 02nd Oct 2023

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