Question

Find the value of ${\text{k}}$, if the points ${\text{A}}\left( {8,1} \right),{\text{ B}}\left( {3, - 4} \right),{\text{ and C}}\left( {2,k} \right)$ are collinear.

Hint: - If three points are given as A,B,C then they will be collinear if the slopes of the line segment between 2 points are equal. I.e. slope of AB = slope of BC.

Given points are
${\text{A}}\left( {8,1} \right),{\text{ B}}\left( {3, - 4} \right),{\text{ and C}}\left( {2,k} \right)$
Now we know two points are collinear if their slopes are equal
Therefore slope of AB $=$ Slope of BC
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 $A\left( {8,1} \right) \equiv \left( {{x_1},{y_1}} \right),{\text{ }}B\left( {3, - 4} \right) \equiv \left( {{x_2},{y_2}} \right),{\text{ }}C\left( {2,k} \right) \equiv \left( {{x_3},{y_3}} \right)$
Therefore slope of AB${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right){\text{ = }}\dfrac{{ - 4 - 1}}{{3 - 8}} = \dfrac{{ - 5}}{{ - 5}} = 1$
Therefore slope of BC ${\text{ = }}\left( {\dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}} \right){\text{ = }}\dfrac{{k - \left( { - 4} \right)}}{{2 - 3}} = \dfrac{{k + 4}}{{ - 1}} = - k - 4$
Points are collinear
Therefore slope of AB $=$ Slope of BC
$\begin{gathered} \Rightarrow 1 = - k - 4 \\ \Rightarrow k = - 1 - 4 = - 5 \\ \end{gathered}$
So, $k = - 5$ is the required answer.

Note: - If three points are collinear then the area formed by these points should be zero because collinear points always lie on the same line so the area formed by these points is zero. We can also use this property to find the collinearity of the points and the condition is $\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0$, with the help of this we can easily calculate the collinearity of the given points.