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Question

Mathematics Question on distance between two points

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

Answer

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

QR = (7−1)2+(0−2)2+(−1−3)2\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}
= (6)2+(−2)2+(−4)2\sqrt{(6)^2+(-2)^2+(-4)^2}
=36+4+16\sqrt{36+4+16}
= 56\sqrt{56} =2142\sqrt{14}

PR = (7+2)2+(0−3)2+(−1−5)2\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}
= (9)2+(−3)2+(−6)2\sqrt{(9)^2+(-3)^2+(-6)^2}
= 81+9+36\sqrt{81+9+36}
= 126\sqrt{126}
= 314\sqrt{14}

Here, PQ+ QR = 14\sqrt{14}+214\sqrt{14} = 314\sqrt{14}
Hence, points P(-2, 3, 5), Q(1, 2, 3), and R(7, 0, -1) are collinear.