Question

Determine whether the points are collinear.
(1) A(1, –3), B(2, –5), C(–4, 7)
(2) L(–2, 3), M(1, –3), N(5, 4)
(3) R(0, 3), D(2, 1), S(3, –1)
(4) P(–2, 3), Q(1, 2), R(4, 1)

Answer 1

(1) A(1, –3), B(2, –5), C(–4, 7)
$$\begin{gathered} AB=\sqrt{{\left(1-2\right)}^2+{\left(-3-\left(-5\right)\right)}^2} \\ =\sqrt{{\left(-1\right)}^2+2^2}=\sqrt{5} \\ BC=\sqrt{{\left(2-\left(-4\right)\right)}^2+{\left(-5-7\right)}^2} \\ =\sqrt{36+144} \\ =\sqrt{180} \\ =6\sqrt{5} \\ AC=\sqrt{{\left(-1-\left(-4\right)\right)}^2+{\left(-3-7\right)}^2} \\ =\sqrt{25+100} \\ =\sqrt{125} \\ =5\sqrt{5} \end{gathered}$$
AB + AC = BC
$\sqrt{5}+5\sqrt{5}=6\sqrt{5}$
Thus, the given points lie on the same line. Hence, they are collinear.

(2) L(–2, 3), M(1, –3), N(5, 4)
$$\begin{gathered} LM=\sqrt{{\left(-2-1\right)}^2+{\left(3-\left(-3\right)\right)}^2} \\ =\sqrt{{\left(-3\right)}^2+6^2}=\sqrt{9+36} \\ =\sqrt{45}=3\sqrt{5} \\ MN=\sqrt{{\left(1-5\right)}^2+{\left(-3-4\right)}^2} \\ =\sqrt{16+49} \\ =\sqrt{65} \\ LN=\sqrt{{\left(-2-5\right)}^2+{\left(3-4\right)}^2} \\ =\sqrt{{\left(-7\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{49+1}=\sqrt{50} \\ =5\sqrt{2} \end{gathered}$$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(3) R(0, 3), D(2, 1), S(3, –1)
$$\begin{gathered} RD=\sqrt{{\left(0-2\right)}^2+{\left(3-1\right)}^2} \\ =\sqrt{{\left(-2\right)}^2+2^2}=\sqrt{4+4} \\ =2\sqrt{2} \\ DS=\sqrt{{\left(2-3\right)}^2+{\left(1-\left(-1\right)\right)}^2} \\ =\sqrt{{\left(-1\right)}^2+4}=\sqrt{1+4} \\ =\sqrt{5} \\ RS=\sqrt{{\left(0-3\right)}^2+{\left(3-\left(-1\right)\right)}^2} \\ =\sqrt{9+16} \\ =\sqrt{25} \\ =5 \end{gathered}$$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(4) P(–2, 3), Q(1, 2), R(4, 1)
$$\begin{gathered} PQ=\sqrt{{\left(-2-1\right)}^2+{\left(3-2\right)}^2} \\ =\sqrt{9+1}=\sqrt{10} \\ QR=\sqrt{{\left(1-4\right)}^2+{\left(2-1\right)}^2} \\ =\sqrt{9+1} \\ =\sqrt{10} \\ PR=\sqrt{{\left(-2-4\right)}^2+{\left(3-1\right)}^2} \\ =\sqrt{36+4} \\ =\sqrt{40} \\ =2\sqrt{10} \end{gathered}$$
PQ + QR = PR
So, the given points lie on the same line. Hence, the given points are collinear.