
How can you say that points \[\left( {1, - 5} \right),\left( {3, - 2} \right),\left( {7,4} \right)\] are collinear?
Answer
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Hint: The given three points are collinear if the area of the triangle formed by them is equal to zero. The area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]. So, use this concept to reach the solution of the given problem.
Complete step-by-step answer:
Let A \[\left( {{x_1},{y_1}} \right) = \left( {1, - 5} \right)\]
B \[\left( {{x_2},{y_2}} \right) = \left( {3, - 2} \right)\]
C \[\left( {{x_3},{y_3}} \right) = \left( {7,4} \right)\]
To prove that three points are in collinear, the area of the triangle formed by the points A, B, and C is equal to zero.
We know that the area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]
By substituting the above data, we get
\[
\Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 2 - 4} \right) + 3\left( {4 - \left( { - 5}
\right)} \right) + 7\left( { - 5 - \left( { - 2} \right)} \right)} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 9 \right) + 7\left( { - 3}
\right)} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ { - 6 + 27 - 21} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ { - 27 + 27} \right] \\
\therefore \Delta = 0 \\
\]
Hence, the area of the triangle formed by the points A, B, and C is equal to zero.
Thus, we can say that points \[\left( {1, - 5} \right),\left( {3, - 2} \right),\left( {7,4} \right)\] are collinear.
Note: We can prove the collinearity of the three by another method i.e., if the sum of the lengths of any two-line segments among AB, BC, and CA is equal to the length of the remaining line segment, then the points are said to be in collinear.
Complete step-by-step answer:
Let A \[\left( {{x_1},{y_1}} \right) = \left( {1, - 5} \right)\]
B \[\left( {{x_2},{y_2}} \right) = \left( {3, - 2} \right)\]
C \[\left( {{x_3},{y_3}} \right) = \left( {7,4} \right)\]
To prove that three points are in collinear, the area of the triangle formed by the points A, B, and C is equal to zero.
We know that the area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]
By substituting the above data, we get
\[
\Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 2 - 4} \right) + 3\left( {4 - \left( { - 5}
\right)} \right) + 7\left( { - 5 - \left( { - 2} \right)} \right)} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 9 \right) + 7\left( { - 3}
\right)} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ { - 6 + 27 - 21} \right] \\
\Rightarrow \Delta = \dfrac{1}{2}\left[ { - 27 + 27} \right] \\
\therefore \Delta = 0 \\
\]
Hence, the area of the triangle formed by the points A, B, and C is equal to zero.
Thus, we can say that points \[\left( {1, - 5} \right),\left( {3, - 2} \right),\left( {7,4} \right)\] are collinear.
Note: We can prove the collinearity of the three by another method i.e., if the sum of the lengths of any two-line segments among AB, BC, and CA is equal to the length of the remaining line segment, then the points are said to be in collinear.
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