Question

Using distance formula show that the points $P(2,4,6),Q(–2,–2,–2)$ and $R(6,10,14)$ are collinear.

Answer 1

$PQ=\sqrt{(2+2{)}^2+(4+2{)}^2+(6+2{)}^2}=\sqrt{(16+36+64)}=2\sqrt{29}$

$PR=\sqrt{(2-6{)}^2+(4-10{)}^2+(6-14{)}^2}=\sqrt{(16+36+64)}=2\sqrt{29}$

$QR=\sqrt{(-2-6{)}^2+(-2-10{)}^2+(-2-14{)}^2}=\sqrt{(64+144+256)}=4\sqrt{29}$

Since, $QR=PQ+PR$
Therefore, the given points are collinear.