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 | 1 year ago |  119

##### 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 | 1 year ago

### Related Questions

#### A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

#### The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates

The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

#### If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.