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Ncert Solutions Maths class 11th Maths Class 11th Introduction to Three Dimensional Geometry

19. A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

[Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by $(\frac{8 k + 2}{k + 1}, \frac{- 3}{k + 1}, \frac{10 k + 4}{k + 1})] .$

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Payal Gupta | Contributor-Level 10

4 months ago

19. Given, x-coordinate of R = 4

Let R divides line segment joining points P(2, –3, 4) and Q(8, 0,10) internally in the ratio k : 1. Then coordinate of R is

$(\frac{k (8) + 1 (2)}{k + 1}, \frac{k (0) + 1 (- 3)}{k + 1}, \frac{k (10) + 1 (4)}{k + 1})$

= $(\frac{8 k + 2}{k + 1}, \frac{- 3}{k + 1}, \frac{10 k + 4}{k + 1})$

Then,

$\frac{8 k + 2}{k + 1}$ = 4

=> $8 k + 2 = 4 (k + 1)$

=> $8 k + 2 = 4 k + 4$

=> $8 k - 4 k = 4 - 2$

=> $4 k = 2$

=> $k = \frac{2}{4}$

=> $k = \frac{1}{2}$

Hence, $y =$ $\frac{- 3}{k + 1}$

= $\frac{- 3}{\frac{1}{2} + 1}$

= $- 3 \div \frac{1 + 2}{2}$

= $- 3× \frac{2}{3}$

= $- 2$

And,

z = $\frac{10 k + 4}{k + 1}$

= $\frac{(10 \frac{1}{2}) + 4}{\frac{1}{2} + 1}$

= $(5 + 4) \div \frac{1 + 2}{2}$

= $9 \frac{2}{3}$

= 6

Therefore, coordinates of R is (4, –2, 6).

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19. A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R. [Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by (8k+2k+1,−3k+1,10k+4k+1)].

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