Show that the points (1,-1), (5,2) and (9, 5) are collinear.
Answer
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Hint: Here we will check whether the given points are collinear or not by using the condition of collinearity i.e., sum of length any two segments equal to the length of the remaining line segment.
Complete step-by-step answer:
Three or more points A, B, C ….. are said to be collinear if they lie on a single straight line.
The given points are
\[A = (1, - 1),B = (5,2){\text{ and }}C = (9,5)\]
Distance between any two points with coordinates $(x_1, y_1)$ and $(x_2, y_2)$ is given by
$ \Rightarrow d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_2})}^2}} $
Now, calculating the distance between A & B
$AB = \sqrt {{{(5 - 1)}^2} + {{(2 + 1)}^2}} = \sqrt {16 + 9} = \sqrt {25} = 5$
Now, calculating the distance between B & C
$BC = \sqrt {{{(5 - 9)}^2} + {{(2 - 5)}^2}} = \sqrt {16 + 9} = \sqrt {25} = 5$
Now, calculating the distance between A & C
$AC = \sqrt {{{(1 - 9)}^2} + {{( - 1 - 5)}^2}} = \sqrt {64 + 36} = \sqrt {100} = 10$
Clearly, $AC = AB + BC$
Hence, A, B, C are collinear points.
Note: If the sum of the lengths of any two line segments among AB, BC, and AC is equal to the length of the remaining line segment then the points are collinear otherwise not. Another way to find collinearity is to substitute the coordinates of all the three points in the area of triangle formula. If the area value is 0 then the points are collinear else they are non collinear.
Complete step-by-step answer:
Three or more points A, B, C ….. are said to be collinear if they lie on a single straight line.
The given points are
\[A = (1, - 1),B = (5,2){\text{ and }}C = (9,5)\]
Distance between any two points with coordinates $(x_1, y_1)$ and $(x_2, y_2)$ is given by
$ \Rightarrow d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_2})}^2}} $
Now, calculating the distance between A & B
$AB = \sqrt {{{(5 - 1)}^2} + {{(2 + 1)}^2}} = \sqrt {16 + 9} = \sqrt {25} = 5$
Now, calculating the distance between B & C
$BC = \sqrt {{{(5 - 9)}^2} + {{(2 - 5)}^2}} = \sqrt {16 + 9} = \sqrt {25} = 5$
Now, calculating the distance between A & C
$AC = \sqrt {{{(1 - 9)}^2} + {{( - 1 - 5)}^2}} = \sqrt {64 + 36} = \sqrt {100} = 10$
Clearly, $AC = AB + BC$
Hence, A, B, C are collinear points.
Note: If the sum of the lengths of any two line segments among AB, BC, and AC is equal to the length of the remaining line segment then the points are collinear otherwise not. Another way to find collinearity is to substitute the coordinates of all the three points in the area of triangle formula. If the area value is 0 then the points are collinear else they are non collinear.
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