Straight Lines
2.0 Condition of collinearity of three points
2.0 Condition of collinearity of three points
Points $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ are collinear if
- Slope of line $AB\ =$ Slope of line $BC\ =$ Slope of line $CA$ i.e., $$\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} = \frac{{{y_3} - {y_1}}}{{{x_3} - {x_1}}}$$
- $\Delta ABC = 0$ i.e., \[\left| {\begin{array}{c}{{x_1}}&{{y_1}}&1 \\{{x_2}}&{{y_2}}&1 \\{{x_3}}&{{y_3}}&1\end{array}} \right| = 0\]
- $AC=AB+BC$ or $AB \sim BC$
- $A$ divides the line segment $BC$ in some ratio
Question 1. For what value of $k$ the points $(k,2-2k),\ (-k+1,2k),\ (-4-k,6-2k)$ are collinear.
Solution: Let $A \equiv (k,2 - 2k),{\text{ }}B \equiv ( - k + 1,2k){\text{ and }}C \equiv ( - 4 - k,6 - 2k)$ are collinear then
$$\begin{equation} \begin{aligned} {\text{Slope of }}AB = {\text{Slope of }}AC \\ \frac{{2k - (2 - 2k)}}{{ - k + 1 - k}} = \frac{{6 - 2k - (2 - 2k)}}{{ - 4 - k - k}} \\ \frac{{4k - 2}}{{ - 2k + 1}} = \frac{4}{{ - 4 - 2k}}{\text{ }}(k \ne \frac{1}{2},\because {\text{Denominator}} \ne 0) \\ (4k - 2)( - 4 - 2k) = 4( - 2k + 1) \\ (2k - 1)( - 2 - k + 1) = 0 \\ \therefore k \ne \frac{1}{2}{\text{ and }}k = - 1 \\\end{aligned} \end{equation} $$
