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 agoA(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 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.

Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.