Question

Using the distance formula, show that the given points are collinear:
(i) (1, −1), (5, 2) and (9, 5) (ii) (6, 9), (0, 1) and (−6, −7)
(iii) (−1, −1), (2, 3) and (8, 11) (iv) (−2, 5), (0, 1) and (2, −3)

Answer 1

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then
$$\begin{gathered} AB=\sqrt{{\left(5-1\right)}^2+{\left(2+1\right)}^2}=\sqrt{4^2+3^2}=\sqrt{25}=5\mathrm{units} \\ BC=\sqrt{{\left(9-5\right)}^2+{\left(5-2\right)}^2}=\sqrt{4^2+3^2}=\sqrt{25}=5\mathrm{units} \\ AC=\sqrt{{\left(9-1\right)}^2+{\left(5+1\right)}^2}=\sqrt{8^2+6^2}=\sqrt{100}=10\mathrm{units} \end{gathered}$$
$\therefore AB+BC=\left(5+5\right)\mathrm{units}=10\mathrm{units}=AC$
Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then
$$\begin{gathered} AB=\sqrt{{\left(0-6\right)}^2+{\left(1-9\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}=\sqrt{100}=10\mathrm{units} \\ BC=\sqrt{{\left(-6-0\right)}^2+{\left(-7-1\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}=\sqrt{100}=10\mathrm{units} \\ AC=\sqrt{{\left(-6-6\right)}^2+{\left(-7-9\right)}^2}=\sqrt{{\left(-12\right)}^2+{\left(-16\right)}^2}=\sqrt{400}=20\mathrm{units} \end{gathered}$$
$\therefore AB+BC=\left(10+10\right)\mathrm{units}=20\mathrm{units}=AC$
Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then
$$\begin{gathered} AB=\sqrt{{\left(2+1\right)}^2+{\left(3+1\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}=\sqrt{25}=5\mathrm{units} \\ BC=\sqrt{{\left(8-2\right)}^2+{\left(11-3\right)}^2}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=\sqrt{100}=10\mathrm{units} \\ AC=\sqrt{{\left(8+1\right)}^2+{\left(11+1\right)}^2}=\sqrt{{\left(9\right)}^2+{\left(12\right)}^2}=\sqrt{225}=15\mathrm{units} \end{gathered}$$
$\therefore AB+BC=\left(5+10\right)\mathrm{units}=15\mathrm{units}=AC$
Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then
$$\begin{gathered} AB=\sqrt{{\left(0+2\right)}^2+{\left(1-5\right)}^2}=\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2}=\sqrt{20}=2\sqrt{5}\mathrm{units} \\ BC=\sqrt{{\left(2-0\right)}^2+{\left(-3-1\right)}^2}=\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2}=\sqrt{20}=2\sqrt{5}\mathrm{units} \\ AC=\sqrt{{\left(2+2\right)}^2+{\left(-3-5\right)}^2}=\sqrt{{\left(4\right)}^2+{\left(-8\right)}^2}=\sqrt{80}=4\sqrt{5}\mathrm{units} \end{gathered}$$
$\therefore AB+BC=\left(2\sqrt{5}+2\sqrt{5}\right)\mathrm{units}=4\sqrt{5}\mathrm{units}=AC$
Hence, the given points are collinear.