1 Answer
Solution :-
Given:
The points A (3, 2, -4), B (9, 8, -10) and C (5, 4, -6)
Let C divides AB in ratio k: 1
Three points are collinear if the value of k is the same for x, y and z coordinates.
Then, m = k and n = 1
A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)
Coordinates of C are:
On comparing we get,
\(\dfrac{ [9k + 3] }{[k + 1]}\) = 5
9k + 3 = 5(k + 1)
9k + 3 = 5k + 5
9k – 5k = 5 – 3
4k = 2
k = \( \dfrac{2}{4}\)
= \( \dfrac{1}{2}\)
\(\dfrac{ [8k + 2] }{ [k + 1]}\) = 4
8k + 2 = 4(k + 1)
8k + 2 = 4k + 4
8k – 4k = 4 – 2
4k = 2
k = \( \dfrac{2}{4}\)
= \( \dfrac{1}{2}\)
\(\dfrac{ [-10k – 4] }{[k + 1] }\)= -6
-10k – 4 = -6(k + 1)
-10k – 4 = -6k – 6
-10k + 6k = 4 – 6
-4k = -2
k = \( \dfrac{-2}{-4}\)
= \( \dfrac{1}{2}\)
The value of k is the same in all three cases.
So, A, B and C are collinear [as, k = \( \dfrac{1}{2}\)]
We can say that, C divides AB externally in ratio 1:2
Answered by Aaryan | 1 year agoRelated Questions
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