Question

Using section formula prove that the three points a(-2,3,5) , b(1,2,3) c(7,0,-1) are collinear. Also find ratio in which points c divides line segment

Answer 1

Solution:

Given points are A(-2,3,5), B(1,2,3) and C(7,0,-1)

Points A,B,C, are collinear if point" c" divides the AB in ratio externally or internally.

By section formula we have

(x,y,z) = (mx2+nx1/m+n, my2+my1/m+n mz2+nz1/m+n)

here, Let the point B(1,2,3) divide the AC in the ratio of k:1

now, m= k and n = 1 and x1 =-2, y1 = 3, z1 = 5, x2 =1, y2 = 2 and z2 = 3

therefore

by using d=section formula we get B( 7k-2/k+1, 3/k+1, -k+5/k+1) --(1)

now comparing the (1) from co-ordinate of B.

we get  7k-2/k+1 = 1, therefore k = 1/2

3/k+1 = 2 therefore k 1/2 and -k+5/k+1 =3 , therefore k =  1/2

hence we get the ratio 1:2 and it divide the line segment AB in ratio of 1:2.

hence the given points are collinear and the ratio will be 1:2

Answer 2

AnswEr:

Suppose the given points are collinear and C divides AB in the ratio

$\\ \qquad \huge \sf \lambda \: \colon \: 1$

Then, coordinates of C are

$\\ \\ \qquad \sf \: ( \frac{ \lambda \: - 2}{ \lambda \: + 1} \: , \: \frac{2 \: \lambda + 3}{ \lambda + 1} \: , \: \frac{3 \: \lambda + 5}{ \lambda + 1} )$

But, coordinates of C are given as ( 7, 0, -1 ). Therefore,

$\qquad \sf \frac{ \lambda - 2}{ \lambda + 1} = 7 \: , \: \frac{2 \: \lambda + 3}{ \lambda + 1} = 0 \: \: \: and \: \: \: , \\ \\ \qquad \sf \frac{3 \: \lambda + 5}{ \lambda + 1} = - 1$

From each of these equations we obtain

$\huge \sf \qquad \: \lambda \: = \frac{ - 3}{2}$

Therefore, the given points are collinear and C divides AB externally in the ratio 3:2.