# Which of the following sets of points is collinear?

$

{\text{A}}{\text{. }}\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right) \\

{\text{B}}{\text{. }}\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,1} \right) \\

{\text{C}}{\text{. }}\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right) \\

{\text{D}}{\text{. }}\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right) \\

$

Last updated date: 30th Mar 2023

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Answer

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307.2k+ views

Hint: Here, we will proceed by using the condition for collinearity of three points by making the area of the triangle made by these three points equal to zero. If the area of the triangle obtained by forming the given three points comes out to be zero, then these points are collinear else they are not collinear.

Complete step-by-step answer:

As we know that area of the triangle made by three points A(x1,y1), B(x2,y2) and C(x3,y3) is given by $\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right|$

For these three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear (i.e., lies in a straight line), the area of the triangle should be equal to zero.

i.e., $

\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right| = 0 \\

\Rightarrow \left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right| = 0{\text{ }} \to {\text{(1)}} \\

$

For the three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear, equation (1) should be satisfied

Also we know that by expanding the determinant of any $3 \times 3$ order matrix through first row, we have

$\left| {\begin{array}{*{20}{c}}

{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\

{{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\

{{a_{31}}}&{{a_{32}}}&{{a_{33}}}

\end{array}} \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{22}}{a_{31}}} \right){\text{ }} \to {\text{(2)}}$

For the points given in the first option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&0&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&0&1

\end{array}} \right| = 1\left( {1 - 0} \right) - \left( { - 1} \right)\left( { - 1 - 0} \right) + 1\left( {0 - 0} \right) = 1 - 1 + 0 = 0$

Clearly the LHS of the equation (1) is coming out to be zero which is equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ are collinear.

For the points given in the second option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,1} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&1&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&1&1

\end{array}} \right| = 1\left( {1 - 1} \right) - \left( { - 1} \right)\left( { - 1 - 0} \right) + 1\left( { - 1 - 0} \right) = 0 - 1 - 1 = - 2$

Clearly the LHS of the equation (1) is coming out to be -2 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,1} \right)$ are not collinear.

For the points given in the third option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&0&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&0&1

\end{array}} \right| = 1\left( {1 - 0} \right) - \left( { - 1} \right)\left( { - 1 - 1} \right) + 1\left( {0 - 1} \right) = 1 - 2 - 1 = - 2$

Clearly the LHS of the equation (1) is coming out to be -2 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$ are not collinear.

For the points given in the fourth option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&1&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&1&1

\end{array}} \right| = 1\left( {1 - 1} \right) - \left( { - 1} \right)\left( { - 1 - 1} \right) + 1\left( { - 1 - 1} \right) = 0 - 2 - 2 = - 4$

Clearly the LHS of the equation (1) is coming out to be -4 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$ are not collinear.

Hence, option A is correct.

Note: These types of problems can be solved by using the concept of slope of the line joining two points where slope of the line AB joining two points A(x1,y1) and B(x2,y2) is given by ${m_{{\text{AB}}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$. For three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear, the slope of the line AB should be equal to the slope of the line BC.

Complete step-by-step answer:

As we know that area of the triangle made by three points A(x1,y1), B(x2,y2) and C(x3,y3) is given by $\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right|$

For these three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear (i.e., lies in a straight line), the area of the triangle should be equal to zero.

i.e., $

\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right| = 0 \\

\Rightarrow \left| {\begin{array}{*{20}{c}}

{{x_1}}&{{y_1}}&1 \\

{{x_2}}&{{y_2}}&1 \\

{{x_3}}&{{y_3}}&1

\end{array}} \right| = 0{\text{ }} \to {\text{(1)}} \\

$

For the three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear, equation (1) should be satisfied

Also we know that by expanding the determinant of any $3 \times 3$ order matrix through first row, we have

$\left| {\begin{array}{*{20}{c}}

{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\

{{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\

{{a_{31}}}&{{a_{32}}}&{{a_{33}}}

\end{array}} \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{22}}{a_{31}}} \right){\text{ }} \to {\text{(2)}}$

For the points given in the first option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&0&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&0&1

\end{array}} \right| = 1\left( {1 - 0} \right) - \left( { - 1} \right)\left( { - 1 - 0} \right) + 1\left( {0 - 0} \right) = 1 - 1 + 0 = 0$

Clearly the LHS of the equation (1) is coming out to be zero which is equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ are collinear.

For the points given in the second option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,1} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&1&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

0&1&1

\end{array}} \right| = 1\left( {1 - 1} \right) - \left( { - 1} \right)\left( { - 1 - 0} \right) + 1\left( { - 1 - 0} \right) = 0 - 1 - 1 = - 2$

Clearly the LHS of the equation (1) is coming out to be -2 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,0} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {0,1} \right)$ are not collinear.

For the points given in the third option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&0&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&0&1

\end{array}} \right| = 1\left( {1 - 0} \right) - \left( { - 1} \right)\left( { - 1 - 1} \right) + 1\left( {0 - 1} \right) = 1 - 2 - 1 = - 2$

Clearly the LHS of the equation (1) is coming out to be -2 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,0} \right)$ are not collinear.

For the points given in the fourth option i.e., $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$, the LHS of the equation (1) is given by $\left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&1&1

\end{array}} \right|$.

Now expanding this determinant through the first row using equation (2)

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&{ - 1}&1 \\

{ - 1}&1&1 \\

1&1&1

\end{array}} \right| = 1\left( {1 - 1} \right) - \left( { - 1} \right)\left( { - 1 - 1} \right) + 1\left( { - 1 - 1} \right) = 0 - 2 - 2 = - 4$

Clearly the LHS of the equation (1) is coming out to be -4 which is not equal to the RHS of the equation (1) which means that points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$ doesn’t satisfies equation (1).

Therefore, points $\left( {1, - 1} \right),\left( { - 1,1} \right),\left( {1,1} \right)$ are not collinear.

Hence, option A is correct.

Note: These types of problems can be solved by using the concept of slope of the line joining two points where slope of the line AB joining two points A(x1,y1) and B(x2,y2) is given by ${m_{{\text{AB}}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$. For three points A(x1,y1), B(x2,y2) and C(x3,y3) to be collinear, the slope of the line AB should be equal to the slope of the line BC.

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