Question

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

(2) D(–2, –3), E(1, 0), F(2, 1)

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

(4) P(2, –5), Q(1, –3), R(–2, 3)

(5) R(1, –4), S(–2, 2), T(–3, 4)

(6) A(–4, 4), $\mathrm{K}\left(–2,\frac{5}{2}\right),$ N (4, –2)

Answer 1

(1) A(–1, –1), B(0, 1), C(1, 3)
Slope of AB = $\frac{1-\left(-1\right)}{0-\left(-1\right)}=\frac{2}{1}=2$
Slope of BC = $\frac{3-1}{1-0}=\frac{2}{1}=2$
Slope of AB = Slope of BC = 2
Thus, the given points are collinear.

(2) D(–2, –3), E(1, 0), F(2, 1)
$$\begin{gathered} \mathrm{Slope}of\mathrm{DE}=\frac{0-\left(-3\right)}{1-\left(-2\right)}=\frac{3}{3}=1 \\ \mathrm{Slope}of\mathrm{EF}=\frac{1-0}{2-1}=\frac{1}{1}=1 \end{gathered}$$
Slope of DE = Slope of EF = 1
So, the given points are collinear.

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

(4) P(2, –5), Q(1, –3), R(–2, 3)
$$\begin{gathered} \mathrm{Slope}\mathrm{of}\mathrm{PQ}=\frac{-3-\left(-5\right)}{1-2}=\frac{2}{-1}=-2 \\ \mathrm{Slope}\mathrm{of}\mathrm{QR}=\frac{3-\left(-3\right)}{-2-1}=\frac{6}{-3}=-2 \end{gathered}$$
Slope of PQ = Slope of QR
So, the given points are collinear.

(5) R(1, –4), S(–2, 2), T(–3, 4)
$$\begin{gathered} \mathrm{Slope}\mathrm{of}\mathrm{RS}=\frac{2-\left(-4\right)}{-2-1}=\frac{6}{-3}=-2 \\ \mathrm{Slope}\mathrm{of}\mathrm{ST}=\frac{4-2}{-3-\left(-2\right)}=\frac{2}{-1}=-2 \end{gathered}$$
Slope of RS = Slope of ST
So, the given points are collinear.

(6) A(–4, 4), $\mathrm{K}\left(–2,\frac{5}{2}\right),$ N (4, –2)
$$\begin{gathered} \mathrm{Slope}\mathrm{of}\mathrm{AK}=\frac{{\displaystyle \frac{5}{2}}-4}{-2-\left(-4\right)}=\frac{\frac{-3}{2}}{2}=\frac{-3}{4} \\ \mathrm{Slope}\mathrm{of}\mathrm{KN}=\frac{-2-{\displaystyle \frac{5}{2}}}{4-\left(-2\right)}=\frac{-3}{4} \end{gathered}$$
Slope of AK=Slope of KN
Thus, the given points are collinear.