Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio

(i) 2: 3 internally,

(ii) 2: 3 externally.

Asked by Pragya Singh | 1 year ago |  86

##### Solution :-

Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.

(i) 2: 3 internally

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n is given by:

$$\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}, \dfrac{mz_2+nz_1}{m+n}$$

We have, the point R dividing the line segment joining the points (-2,3,5) and (1,-4,6) internally in the ratio 2:3

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

$$y= \dfrac{2(-4)+3(3)}{2+3}$$

$$z= \dfrac{2(6)+3(5)}{2+3}$$

On solving we get,

$$x=\dfrac{-4}{5}$$

$$x=\dfrac{1}{5}$$

$$x=\dfrac{27}{5}$$

Therefore, the coordinates we obtain are $$(\dfrac{-4}{5}, \dfrac{1}{5}, \dfrac{27}{5})$$

(ii) 2: 3 externally

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m: n is given by:

$$\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}, \dfrac{mz_2+nz_1}{m+n}$$

We have, the point R dividing the line segment joining the points (-2,3,5) and (1,-4,6) externally in the ratio 2:3

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

$$y= \dfrac{2(-4)+3(3)}{2-3}$$

$$z= \dfrac{2(6)+3(5)}{2-3}$$

On solving we get,

x = -8

y = 17

z = 3

Therefore, the coordinates we obtain are, (-8,17,3)

Answered by Abhisek | 1 year ago

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