Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, $$\dfrac{1}{3}$$, 2) are collinear.

Asked by Pragya Singh | 11 months ago |  98

##### Solution :-

We are given three points A(2,-3,4) , B(-1,2,1) and C $$(0, \dfrac{1}{3},2)$$

Let P is point which divides the line segment ABin the ratio k:1.

Using section formula,

$$( \dfrac{k(-1)+2}{k+1}, \dfrac{k(2)-3}{k+1}, \dfrac{k(1)+4}{k+1} )$$

The value of k such that the point P coincides with point C will be,

Now, we check if for some value of k, the point coincides with the point C.

Put $$\dfrac{(-k+2)}{(k+1)}$$

-k + 2 = 0

k = 2

When k = 2, then

$$\dfrac{(2k-3)}{(k+1)}=\dfrac{2(2)-3}{(2+1)}$$

$$\dfrac{(4-3)}{3}$$

$$\dfrac{1}{3}$$

And,$$\dfrac{ (k+4)}{(k+1)}=\dfrac{(2+4)}{(2+1)}$$

$$\dfrac{6}{3}$$

= 2

C (0,$$\dfrac{1}{3}$$, 2) is a point which divides AB in the ratio 2: 1 and is same as P.

Hence, A, B, C are collinear.

Answered by Abhisek | 11 months ago

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