Question

For which value of 'k' the points (7,−2),(5,1),(3,k) are collinear?

Answer 1

To solve this problem, use the condition that for three points (x1,y1), (x2,y2) and (x3,y3) be collinear if the area of triangle forming by the three coordinates (vertices) is equal to 0. So to find the value of k calculate the area of the triangle and equate it with 0.

Complete Step-by-step solution
Given three points (7,−2),(5,1),(3,k) are collinear.
We know the area of a triangle having vertices (x1,y1), (x2,y2) and (x3,y3) is given by
$\dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_{3}}({y_{1}} - {y_2})]$
Since points are collinear,
So the area of the triangle will be equal to zero.
so,$\dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_{3}}({y_{1}} - {y_2})]$= 0
Take (7,−2) as (x1,y1), (5,1) as (x2​,y2​) and (3,k) as (x3​,y3​), we have
On putting the values of coordinates in above result
We get,
\Rightarrow $$$7\left( {1 - k} \right) + 5\left( {k + 2} \right) + 3\left( { - 2 - 1} \right) = 0$$ \Rightarrow 7 - 7k + 5k + 10 - 9 = 0$$ $\Rightarrow 2k = 8; $\Rightarrow $$$k = 4

Additional information:
1. In general, three points A, B and C are collinear 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, that is,
either $AB + BC = AC$ or $AC + CB = AB$or ${\text{ }}BA + AC = BC$
2. Concept of slope: Let A, B and C be the three points.
If we want A, B and C to be collinear, the following conditions have to be met.
(i) Slope of AB = Slope of BC
(ii) There must be a common point between AB and BC.
(In AB and BC, the common point is B)
If the above two conditions are met, then the three points A, B and C are collinear.

Note: For more than three points we can not use the area of triangle equal to zero condition we can use the slope concept or distance of segments concept. Four points can be collinear if the area of the quadrilateral is zero