Question

Determine whether the given points are collinear.
(1) A(0,2), B(1,–0.5), C(2,–3)

(2) $\mathrm{P}\left(1,2\right),\mathrm{Q}\left(2,\frac{8}{5}\right),\mathrm{R}\left(3,\frac{6}{5}\right)$

(3) L(1,2), M(5,3) , N(8,6)

Answer 1

(1) A(0,2), B(1,–0.5), C(2,–3)
Slope of AB = $\frac{-0.5-2}{1-0}=\frac{-2.5}{1}=-2.5$
Slope of BC = $\frac{-3-\left(-0.5\right)}{2-1}=\frac{-2.5}{1}=-2.5$
So, the slope of AB = slope of BC.
Point B lies on both the lines.
Hence, the given points are collinear.

(2) $\mathrm{P}\left(1,2\right),\mathrm{Q}\left(2,\frac{8}{5}\right),\mathrm{R}\left(3,\frac{6}{5}\right)$
Slope of PQ = $\frac{{\displaystyle \frac{8}{5}-2}}{2-1}=\frac{{\displaystyle \frac{-2}{5}}}{1}=\frac{-2}{5}$
Slope of QR = $\frac{{\displaystyle \frac{6}{5}-\frac{8}{5}}}{3-2}=\frac{{\displaystyle \frac{-2}{5}}}{1}=\frac{{\displaystyle -2}}{{\displaystyle 5}}$
So, the slope of PQ = slope of QR.
Point Q lies on both the lines.
Hence, the given points are collinear.

(3) L(1,2), M(5,3) , N(8,6)
Slope of LM$=\frac{3-2}{5-1}=\frac{1}{4}$
Slope of MN = $\frac{6-3}{8-5}=\frac{3}{3}=1$
Thus, the slope of LM not equal to slope MN.
So, the given points are not collinear.