Question

using section formula prove that the three points A(-2,3,5) , B(1,2,3) , C(7,0,-1) are collinear​

Answer 1

$\large\underline{\sf{Solution-}}$

Given points are

↝ Coordinates of A (-2, 3, 5)

↝ Coordinates of B (1, 2, 3)

↝ Coordinates of C (7, 0, -1)

Let us assume that B divides AC in the ratio k : 1.

So, we know by Section Formula,

Coordinates of point C which divides the line segment joining the points A (a, b, c) and B (d, e, f) in the ratio m : n is given by

$\boxed{ \tt{ \: C = \bigg(\dfrac{md + na}{m + n}, \dfrac{me + nb}{m + n}, \dfrac{mf + nc}{m + n} \bigg) \: }}$

So, using this, we get

$\rm :\longmapsto\:(1,2,3) = \bigg(\dfrac{7k - 2}{k + 1}, \dfrac{3}{k + 1},\dfrac{ - k + 5}{k + 1} \bigg)$

On comparing the coordinates, we get

$\rm :\longmapsto\:\dfrac{7k - 2}{k + 1} = 1$

$\rm :\longmapsto\:7k - 2 = k + 1$

$\rm :\longmapsto\:7k - k = 2 + 1$

$\rm :\longmapsto\:6k = 3$

$\bf\implies \:k = \dfrac{1}{2}$

Also,

$\rm :\longmapsto\:\dfrac{3}{k + 1} = 2$

$\rm :\longmapsto\:3 = 2k + 2$

$\rm :\longmapsto\:3 - 2 = 2k$

$\rm :\longmapsto\:1 = 2k$

$\bf\implies \:k = \dfrac{1}{2}$

Also,

$\rm :\longmapsto\:\dfrac{ - k + 5}{k + 1} = 3$

$\rm :\longmapsto\: - k + 5 = 3k + 3$

$\rm :\longmapsto\: - k - 3k= - 5 + 3$

$\rm :\longmapsto\: - 4k= - 2$

$\bf\implies \:k = \dfrac{1}{2}$

Hence, we concluded that B divides line segment joining A and C in the ratio 1 : 2.

So, points A, B and C are collinear.