Question

Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer 1

Let points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) be denoted by P, Q, and R respectively. 
Points P, Q, and R are collinear if they lie on a line.

PQ=\(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}\)
= \(\sqrt{(3)^2+(-1)^2+(-2)^2}\)
= \(\sqrt{9+1+4}\)
=\(\sqrt{14}\)

QR = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\)
= \(\sqrt{(6)^2+(-2)^2+(-4)^2}\)
=\(\sqrt{36+4+16}\)
= \(\sqrt{56}\) =\(2\sqrt{14}\)

PR = \(\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}\)
= \(\sqrt{(9)^2+(-3)^2+(-6)^2}\)
= \(\sqrt{81+9+36}\)
= \(\sqrt{126}\)
= 3\(\sqrt{14}\)

Here, PQ+ QR = \(\sqrt{14}\)+2\(\sqrt{14}\) = 3\(\sqrt{14}\)
Hence, points P(-2, 3, 5), Q(1, 2, 3), and R(7, 0, -1) are collinear.