Question

Show that the following points are collinear:

(i) A(6, 9) B(0, 1) and C(−6, −7)
(ii) A(−1, −1) B(2, 3) and C(8, 11)
(iii) P(1, 1) Q(−2, 7) and R(3, −3)
(iv) P(2, 0) Q(11, 6) and R(−4, −4)

Answer 1

(i) A(6, 9) B(0, 1) and C(−6, −7)
$$\begin{gathered} AB=\sqrt{{\left(0-6\right)}^2+{\left(1-9\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}=\sqrt{36+64}=\sqrt{100}=10\mathrm{units} \\ BC=\sqrt{{\left(-6-0\right)}^2+{\left(-7-1\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}=\sqrt{36+64}=\sqrt{100}=10\mathrm{units} \\ AC=\sqrt{{\left(-6-6\right)}^2+{\left(-7-9\right)}^2}=\sqrt{{\left(-12\right)}^2+{\left(-16\right)}^2}=\sqrt{144+256}=\sqrt{400}=20\mathrm{units} \\ AB+BC=\left(10+10\right)=20\mathrm{units}=AC \end{gathered}$$

Hence, the points A, B and C are collinear.

(ii) A(−1, −1) B(2, 3) and C(8, 11)
$$\begin{gathered} AB=\sqrt{{\left(2+1\right)}^2+{\left(3+1\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5\mathrm{units} \\ BC=\sqrt{{\left(8-2\right)}^2+{\left(11-3\right)}^2}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=\sqrt{36+64}=\sqrt{100}=10\mathrm{units} \\ AC=\sqrt{{\left(8+1\right)}^2+{\left(11+1\right)}^2}=\sqrt{{\left(9\right)}^2+{\left(12\right)}^2}=\sqrt{81+144}=\sqrt{225}=15\mathrm{units} \\ AB+BC=\left(5+10\right)=15\mathrm{units}=AC \end{gathered}$$
Hence, the points A, B and C are collinear.

(iii) P(1, 1) Q(−2, 7) and R(3, −3)
$$\begin{gathered} PQ=\sqrt{{\left(-2-1\right)}^2+{\left(7-1\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(6\right)}^2}=\sqrt{9+36}=\sqrt{45}=\sqrt{9\times 5}=3\sqrt{5}\mathrm{units} \\ QR=\sqrt{{\left(3+2\right)}^2+{\left(-3-7\right)}^2}=\sqrt{{\left(5\right)}^2+{\left(-10\right)}^2}=\sqrt{25+100}=\sqrt{125}=\sqrt{25\times 5}=5\sqrt{5}\mathrm{units} \\ PR=\sqrt{{\left(3-1\right)}^2+{\left(-3-1\right)}^2}=\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}\mathrm{units} \\ PQ+PR=\left(3\sqrt{5}+2\sqrt{5}\right)=5\sqrt{5}\mathrm{units}=QR \end{gathered}$$

Hence, the points Q, P and R are collinear.

(iv) P(2, 0) Q(11, 6) and R(−4, −4)

$$\begin{gathered} PQ=\sqrt{{\left(11-2\right)}^2+{\left(6-0\right)}^2}=\sqrt{{\left(9\right)}^2+{\left(6\right)}^2}=\sqrt{81+36}=\sqrt{117}=\sqrt{9\times 13}=3\sqrt{13}\mathrm{units} \\ QR=\sqrt{{\left(-4-11\right)}^2+{\left(-4-6\right)}^2}=\sqrt{{\left(-15\right)}^2+{\left(-10\right)}^2}=\sqrt{225+100}=\sqrt{325}=\sqrt{25\times 13}=5\sqrt{13}\mathrm{units} \\ PR=\sqrt{{\left(-4-2\right)}^2+{\left(-4-0\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(-4\right)}^2}=\sqrt{36+16}=\sqrt{52}=\sqrt{4\times 13}=2\sqrt{13}\mathrm{units} \\ PQ+PR=\left(3\sqrt{13}+2\sqrt{13}\right)=5\sqrt{13}\mathrm{units}=QR \end{gathered}$$

Hence, the points Q, P and R are collinear.