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Question
Using section formula, show that he points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.
Solution
Let, C divides AB in the ratio \[\lambda\]Then, coordinates of C are\[\left( \frac{\lambda x_2 + x_1}{\lambda + 1}, \frac{\lambda y_2 + y_1}{\lambda + 1}, \frac{\lambda z_2 + z_1}{\lambda + 1} \right)\]
But, the coordinates of C are\[\left( 0, \frac{1}{3}, 2 \right)\]
∴\[\frac{- \lambda + 2}{\lambda + 1}\] \[\frac{2\lambda - 3}{\lambda + 1}\] \[\frac{1}{3}\] \[\frac{\lambda + 4}{\lambda + 1}\]
From each of these equations, we get\[\lambda\] Therefore, the given points are collinear.
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