Question

Using distance formula prove that the following points are collinear

(i) A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

(ii) P(0,7, -7), Q(1, 4, -5) and R(-1, 10, -9)

(iii) A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

Answer 1

(i) A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

AB = $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}$

$=\sqrt{(5-4{)}^2+(-7+3{)}^2+(6+1{)}^2}$

$=\sqrt{(1{)}^2+(-4{)}^2+(7{)}^2}$

$=\sqrt{1+16+49}$

$\sqrt{66}$ units

BC =$\sqrt{(3-5{)}^2+(1+7{)}^2+(8-6{)}^2}$

=$\sqrt{(-2{)}^2+(8{)}^2+(-14{)}^2}$

=$\sqrt{4+64+196}$

=$\sqrt{264}$

=$2\sqrt{266}$

AC = $\sqrt{(3-4{)}^2+(1+3{)}^2+(-8+1{)}^2}$
$=\sqrt{(-1{)}^2+(4{)}^2+(-7{)}^2}$
$=\sqrt{1+16+49}$
$=\sqrt{66}$

Since AC + AB = BC

So, A, B, C are collinear.

(ii) P (0, 7, -7), Q(1, 4, -5) and R (-1, 10, -9)

PQ = $\sqrt{(1-0{)}^2+(4-7{)}^2+(-5+7{)}^2}$

$=\sqrt{(1{)}^2+(-3{)}^2+(2{)}^2}$

$=\sqrt{1+9+4}$

$=\sqrt{14}$ units

QR = $\sqrt{(-1-1{)}^2+(10-4{)}^2+(-9+5{)}^2}$

=$\sqrt{(-2{)}^2+(6{)}^2+(-4{)}^2}$

=$\sqrt{4+36+16}$

=$2\sqrt{14}$ units

PQ = $\sqrt{(-1-0{)}^2+(10-7{)}^2+(-9+7{)}^2}$

=$\sqrt{(-1{)}^2+(3{)}^2+(-2{)}^2}$

=$\sqrt{1+9+4}$

=$\sqrt{14}$

Since PQ + PR =QR

So, P, Q, R are collinear

(iii) A ()3, -5, 1, B(-1, 0, 8) and C(7, -10, -6)

AB = $\sqrt{(-1-3{)}^2+(0+5{)}^2+(8-1{)}^2}$

$=\sqrt{(-4{)}^2+(5{)}^2+(7{)}^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

BC $=\sqrt{(7+1{)}^2+(-10-0{)}^2+(-6-8{)}^2}$

$=\sqrt{(8{)}^2+(-10{)}^2+(-14{)}^2}$

$=\sqrt{64+100+196}$

$=\sqrt{360}$ units

$6\sqrt{10}$ units

CA = $\sqrt{(7-3{)}^2+(-10+5{)}^2+(-6-1{)}^2}$

$=\sqrt{(4{)}^2+(-5{)}^2+(-7{)}^2}$

$=\sqrt{16+25+49}$

$=\sqrt{90}$

$=3\sqrt{10}$ units

Since AB + AC = BC

So, A, B and C are collinear