# Using distance formula prove that the following points are collinear : P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

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Using distance formula prove that the following points are collinear : P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

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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

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