Question

Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer 1

The given points are, A=( −2,3,5 ) ,  B=( 1,2,3 ) and C=( 7,0,−1 ) .

We need to prove that the given points A , B and C are collinear.

Three points are collinear if they lie on a straight line.

The formula to find the distance d between two points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is,

d= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 

To find the distance AB ,from the given points, the value of x 1 is −2 , x 2 is 1 , y 1 is 3 , y 2 is 2 , z 1 is 5 and z 2 is 3 .

Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1),

AB= ( 1−( −2 ) ) 2 + ( 2−3 ) 2 + ( 3−5 ) 2 = ( 3 2 + ( −1 ) 2 + ( −2 ) 2 ) = ( 9+1+4 ) = 14  units

To find the distance BC between the points B and C, the value of x 1 is 1 , x 2 is 7 , y 1 is 2 , y 2 is 0 , z 1 is 3 and z 2 is −1 .

Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1),

BC= ( 7−1 ) 2 + ( 0−2 ) 2 + ( −1−3 ) 2 = ( 6 2 + ( −2 ) 2 + ( −4 ) 2 ) = ( 36+4+16 ) = 56 = 4×14 = 4 × 14 =2 14  units

To find the distance AC between the points A and C, the value of x 1 is −2 , x 2 is 7 , y 1 is 3 , y 2 is 0 , z 1 is 5 and z 2 is −1 .

Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1),

AC= ( 7−( −2 ) ) 2 + ( 0−3 ) 2 + ( −1−5 ) 2 = ( 9 2 + ( −3 ) 2 + ( −6 ) 2 ) = ( 81+9+36 ) = 126 =3 14  units

Thus, AB+BC=CA .

Therefore, the given points, A , B and C are collinear.