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Question
Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Solution
Let the distance between the points A(−2, 3, 5), and B(1, 2, 3) be
AB = `sqrt((1 + 2)^2 + (2 - 3)^2 + (3 - 5)^2)`
= `sqrt((3)^2 + (-1)^2 + (-2)^2)`
= `sqrt(9 + 1 + 4)`
= `sqrt14`
BC = `sqrt((7 - 1)^2 + (0 - 2)^2 + (-1 -3)^2)`
= `sqrt((6)^2 + (-2)^2 + (-4)^2)`
= `sqrt(36 + 4 + 16)`
= `sqrt56`
= `2sqrt14`
Distance between points A(−2, 3, 5) and C(7, 0, −1)
AC = `sqrt((7 + 2)^2 + (0 - 3)^2 + (-1 -5)^2)`
= `sqrt((9)^2 + (-3)^2 + (-6)^2)`
= `sqrt(81 + 9 + 36)`
= `sqrt126`
= `3sqrt14`
Now, AB + BC = `sqrt14 + 2sqrt14`
= `3sqrt14`
AC = `3sqrt14`
Here, AB + BC = AC
Hence, points A, B, C are collinear.
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