Given: P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
To prove: Points P, Q and R are collinear
Formula used: Points P, Q and R are collinear if PQ + QR = PR or PQ + PR = QR or PR + QR = PQ Distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a-m)^2+(b-n)^2+(c-0)^2}\)
Therefore,
Distance between P(0, 7, -7) and Q(1, 4, -5) is PQ,
= \(\sqrt{(0-1)^2+(7-4)^2+(-7-(-5))^2}\)
= \(\sqrt{(-1)^2+(3)^2+(-2)^2}\)
= \(\sqrt{1+9+4}\)
= \(\sqrt{14}\)
Distance between Q(1, 4, -5) and R(-1, 10, -9) is QR,
= \(\sqrt{(1-(-1))^2+(4-10)^2+(-5-(-9))^2}\)
= \(\sqrt{(2)^2+(-6)^2+(4)^2}\)
= \(\sqrt{4+36+16}\)
= \(\sqrt{56}\)
= \(2\sqrt{14}\)
Distance between P(0, 7, -7) and R(-1, 10, -9) is PR,
= \(\sqrt{(0-(1))^2+(7-10)^2+(-7-(-9))^2}\)
= \(\sqrt{(1)^2+(-3)^2+(2)^2}\)
= \(\sqrt{1+9+4}\)
= \(\sqrt{14}\)
Clearly,
PQ + PR
= \(\sqrt{14}+\sqrt{14}\)
= \(2\sqrt{14}\)
= QR
Hence, P, Q and R are collinear
