Example 8
Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear.
Let points be
A (– 4, 6, 10) , B (2, 4, 6) , C (14, 0, – 2)
Point A, B, & C are collinear if point C divides AB in some ratio externally & internally
We know that
Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is
(x, y ,z) = ((mx2 + nx1)/(m + n),(my2 + ny1)/(m + n), (〖𝑚𝑧〗_2 + 〖𝑛𝑧〗_1)/(𝑚 + 𝑛))
Here, let point C(14, 0, – 2) divide A(– 4, 6, 10) , B(2, 4, 6) in the ratio k : 1
Here, x1 = – 4, y1 = 6, z1 = 10
x2 = 2, y2 = 4, z2 = 6
& m = k , n = 1
Putting values
(14, 0, – 2) = ((𝑘(2) + 1(−4))/(𝑘 + 1),(𝑘(4) + 1 (6))/(𝑘 + 1),(𝑘 (6) + 1 (10))/(𝑘 +1))
(14, 0, – 2) = ((2𝑘 − 4 )/( 𝑘 + 1),(4𝑘 + 6)/(k +1),(6𝑘 + 10)/( k + 1))
Comparing x – coordinate
Comparing x – coordinate
14 = (2𝑘 − 4)/(𝑘 + 1)
(k + 1) (14) = 2k – 4
14k + 14 = 2k – 4
14k – 2k = – 4 – 14
12k = – 18
k = (−18)/12
k = (−3)/2
Since k is negative
Point C divides line segment AB externally in the ratio 3 : 2
Thus, A, B & C are collinear

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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