Question

Using the slope concept check whether the following points are collinear
$\mathrm{A}\left(0,\frac{-1}{4}\right),\mathrm{B}\left(\frac{1}{3},1\right)\mathrm{and}\mathrm{C}\left(\frac{2}{3},\frac{1}{4}\right)$

Answer 1

The given points are $\mathrm{A}\left(0,\frac{-1}{4}\right),\mathrm{B}\left(\frac{1}{3},1\right)\mathrm{and}\mathrm{C}\left(\frac{2}{3},\frac{1}{4}\right)$

Here, x₁ = 0, y₁ = $\frac{-1}{4}$, x₂ = $\frac{1}{3}$, y₂ = 1, x₃ = $\frac{2}{3}$ and y₃ = $\frac{1}{4}$
Now, we will find the slopes.

$$\begin{gathered} \mathrm{Slope}ofAB=\frac{y_2-y_1}{x_2-x_1} \\ =\frac{1-\left(-{\displaystyle \frac{1}{4}}\right)}{{\displaystyle \frac{1}{3}}-0} \\ =\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{1}{3}}} \\ =\frac{15}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{Slope}ofBC=\frac{y_3-y_2}{x_3-x_2} \\ =\frac{{\displaystyle \frac{1}{4}}-1}{{\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \frac{-3}{4}}}{{\displaystyle \frac{1}{3}}} \\ =\frac{-9}{4} \end{gathered}$$

Since the slopes in both cases are not equal, the points are not collinear.