Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.

Asked by Sakshi | 1 year ago |  117

##### Solution :-

Given:

The points A (2, 3, 4), B (-1, 2, -3) and C (-4, 1, -10)

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates.

So, m = k and n = 1

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

Coordinates of C are:

On comparing we get,

$$\dfrac{ [-k + 2] }{[k + 1]}$$ = -4

-k + 2 = -4(k + 1)

-k + 2 = -4k – 4

4k – k = – 2 – 4

3k = -6

k = $$\dfrac{-6}{3}$$

= -2

$$\dfrac{ [2k + 3] }{[k + 1]}$$ = 1

2k + 3 = k + 1

2k – k = 1 – 3

k = – 2

$$\dfrac{ [-3k + 4] }{ [k + 1]}$$ = -10

-3k + 4 = -10(k + 1)

-3k + 4 = -10k – 10

-3k + 10k = -10 – 4

7k = -14

k = $$\dfrac{-14}{7}$$

= -2

The value of k is the same in all three cases.

So, A, B and C are collinear [as k = -2]

We can say that, C divides AB externally in ratio 2:1

Answered by Aaryan | 1 year ago

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