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NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions

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Question : 10 of 20
Marks: +1, -0
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. ,
Solution:  
(i) Let P(x, y, z) be any point which divides the line segment joining the points A(–2, 3, 5) and B(1, –4, 6) in the ratio 2 : 3 internally.
Then x = 2×1+3×(2)2+3\frac{2 \times 1 + 3 \times (-2)}{2+3} = 265\frac{2-6}{5} = 45\frac{-4}{5} , y = 2×(4)+3×32+3\frac{2 \times (-4) + 3 \times 3}{2+3} = 8+95\frac{-8+9}{5} = 15\frac{1}{5}
z = 2×6+3×52+3\frac{2 \times 6 + 3 \times 5}{2+3} = 12+155\frac{12+15}{5} = 275\frac{27}{5}
∴ Coordinates of P are (45,15,275)\left(-\frac{4}{5},\frac{1}{5},\frac{27}{5}\right)
(ii) Let P(x, y, z) be any point which divides the line segment joining the points A(–2, 3, 5) and B(1, –4, 6) in the ratio 2 : 3 externally.
Then
x = 2×13×(2)23\frac{2 \times 1 - 3 \times (-2)}{2-3} = 2+61\frac{2+6}{-1} = - 8 , y = 2×(4)3×323\frac{2 \times (-4) - 3 \times 3}{2-3} = 891\frac{-8-9}{-1} = 17
z = 2×63×523\frac{2 \times 6 - 3 \times 5}{2-3} = 12151\frac{12-15}{-1} = 3
∴ Coordinates of P are (–8, 17, 3)
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