Question

For what value of K are the points (k, 2-2k), (-k+1, 2k), (-4-k, 6-2k) are collinear?

Answer 1

Hint: In this question use the concept that if 3 points are collinear then the slope of any 2 pairs of points amongst them must be equal. Use direct formula to find slope if two coordinates $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$. This will help to find the solution.

Complete step-by-step answer:
${\text{A}}\left( {k,2 - 2k} \right),{\text{ B}}\left( { - k + 1,2k} \right),{\text{ and C}}\left( { - 4 - k,6 - 2k} \right)$
Now as we know that if three points are collinear then their slope of any two points must be equal to the other two points.
Therefore slope of AB $ = $ Slope of BC = slope of CA
Collinearity of points: - Collinear points always lie on the same line.
Now we know
Slope between two points ${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$
Consider ${\text{A}}\left( {k,2 - 2k} \right) \equiv \left( {{x_1},{y_1}} \right),{\text{ }}B\left( { - k + 1,2k} \right) \equiv \left( {{x_2},{y_2}} \right),{\text{ }}C\left( { - 4 - k,6 - 2k} \right) \equiv \left( {{x_3},{y_3}} \right)$
Therefore slope of AB${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right){\text{ = }}\dfrac{{2k - \left( {2 - 2k} \right)}}{{ - k + 1 - k}} = \dfrac{{4k - 2}}{{1 - 2k}}$
Therefore slope of BC${\text{ = }}\left( {\dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}} \right){\text{ = }}\dfrac{{6 - 2k - 2k}}{{ - 4 - k - \left( { - k + 1} \right)}} = \dfrac{{6 - 4k}}{{ - 5}}$
Therefore slope of CA${\text{ = }}\left( {\dfrac{{{y_3} - {y_1}}}{{{x_3} - {x_1}}}} \right){\text{ = }}\dfrac{{6 - 2k - \left( {2 - 2k} \right)}}{{ - 4 - k - k}} = \dfrac{4}{{ - 4 - 2k}}$
Points are collinear
Therefore slope of BC $ = $ Slope of CA
$ \Rightarrow \dfrac{{6 - 4k}}{{ - 5}} = \dfrac{4}{{ - 4 - 2k}}$
Now simplify the above equation we have,
$ \Rightarrow \left( {6 - 4k} \right)\left( { - 4 - 2k} \right) = - 20$
$ \Rightarrow - 24 - 12k + 16k + 8{k^2} = - 20$
$ \Rightarrow 8{k^2} + 4k - 4 = 0$
Now divide by 4 throughout we have,
$ \Rightarrow 2{k^2} + k - 1 = 0$
Now factorize the equation we have,
$ \Rightarrow 2{k^2} + 2k - k - 1 = 0$
$ \Rightarrow 2k\left( {k + 1} \right) - 1\left( {k + 1} \right) = 0$
$ \Rightarrow \left( {2k - 1} \right)\left( {k + 1} \right) = 0$
$ \Rightarrow k = \dfrac{1}{2}, - 1$
So for $k = \dfrac{1}{2}$ the slope of AB is (0/0) which is not defined so $k = \dfrac{1}{2}$ is not a valid case.
So this is the required value of k such that the points A, B and C are collinear is (-1)
So this is the required answer.

Note: Collinear means that the points must be lying on the same line. Now there can too be other conditions to prove that points are collinear, like if three points are collinear that is $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$ are collinear than $\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$. Another concept of section formula can also be used if these points are collinear than $\left( {{x_3},{y_3}} \right)$ must divide the line segment of the points $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ in ratio 1: 1.