Question
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. $\text{PQ}=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}$ $=\sqrt{(3)^3+(-1)^2+(-2)^2}$ $=\sqrt{9+1+4}$ $=\sqrt{14}$ $\text{QR}=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}$ $=\sqrt{(6)^2+(-2)^2+(-4)^2}$ $=\sqrt{36+4+16}$ $=\sqrt{56}$ $=2\sqrt{14}$ $\text{PR}=\sqrt{(7+2)^2+(0+3)^2+(-1-5)^2}$ $=\sqrt{(9)^2+(-3)^2+(-6)^2}$ $=\sqrt{81+9+36}$ $=\sqrt{126}$ $=3\sqrt{14}$ Here, $\text{PQ + QR}=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=\text{PR}$ Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

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