Question

Using distance formula prove that the following points are collinear:
(i) A(4, –3, –1), B(5, –7, 6) and C(3, 1, –8)
(ii) P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
(iii) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)

Answer 1

(i) AB = $\sqrt{{\left(5-4\right)}^2+{\left(-7+3\right)}^2+{\left(6+1\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(1\right)}^2+{\left(-4\right)}^2+{\left(7\right)}^2} \\ =\sqrt{1+16+49} \\ =\sqrt{66} \end{gathered}$$

BC = $\sqrt{{\left(3-5\right)}^2+{\left(1+7\right)}^2+{\left(-8-6\right)}^2}$
= $\sqrt{{\left(-2\right)}^2+{\left(8\right)}^2+{\left(-14\right)}^2}$
$$\begin{gathered} =\sqrt{4+64+196} \\ =\sqrt{264} \\ =2\sqrt{66} \end{gathered}$$

AC = $\sqrt{{\left(3-4\right)}^2+{\left(1+3\right)}^2+{\left(-8+1\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-1\right)}^2+{\left(4\right)}^2+{\left(-7\right)}^2} \\ =\sqrt{1+16+49} \\ =\sqrt{66} \end{gathered}$$

$$\begin{gathered} \mathrm{Here},\mathrm{AB}+\mathrm{AC}=\sqrt{66}+\sqrt{66} \\ =2\sqrt{66} \\ =\mathrm{BC} \end{gathered}$$
Hence, the points are collinear.

(ii) PQ = $\sqrt{{\left(1-0\right)}^2+{\left(4-7\right)}^2+{\left(-5+7\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(1\right)}^2+{\left(-3\right)}^2+{\left(2\right)}^2} \\ =\sqrt{1+9+4} \\ =\sqrt{14} \end{gathered}$$

QR = $\sqrt{{\left(-1-1\right)}^2+{\left(10-4\right)}^2+{\left(-9+5\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-2\right)}^2+{\left(6\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{4+36+16} \\ =\sqrt{56} \\ =2\sqrt{14} \end{gathered}$$

PR$=\sqrt{{\left(-1-0\right)}^2+{\left(10-7\right)}^2+{\left(-9+7\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-1\right)}^2+{\left(3\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{1+9+4} \\ =\sqrt{14} \end{gathered}$$

$$\begin{gathered} \mathrm{Here},\mathrm{PQ}+\mathrm{PR}=\sqrt{14}+\sqrt{14} \\ =2\sqrt{14} \\ =\mathrm{QR} \end{gathered}$$
Hence, the points are collinear.

(iii) AB = $\sqrt{{\left(-1-3\right)}^2+{\left(0+5\right)}^2+{\left(8-1\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-4\right)}^2+{\left(5\right)}^2+{\left(7\right)}^2} \\ =\sqrt{16+25+49} \\ =\sqrt{90} \\ =3\sqrt{10} \end{gathered}$$

BC = $\sqrt{{\left(7+1\right)}^2+{\left(-10-0\right)}^2+{\left(-6-8\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(8\right)}^2+{\left(-10\right)}^2+{\left(-14\right)}^2} \\ =\sqrt{64+100+196} \\ =\sqrt{360} \\ =6\sqrt{10} \end{gathered}$$

AC = $\sqrt{{\left(7-3\right)}^2+{\left(-10+5\right)}^2+{\left(-6-1\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(4\right)}^2+{\left(-5\right)}^2+{\left(-7\right)}^2} \\ =\sqrt{16+25+49} \\ =\sqrt{90} \\ =3\sqrt{10} \end{gathered}$$

$$\begin{gathered} \mathrm{Here},\mathrm{AB}+\mathrm{AC}=3\sqrt{10}+3\sqrt{10} \\ =6\sqrt{10} \\ =\mathrm{BC} \end{gathered}$$
Hence, the points are collinear.