# Example 4

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 4 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear. If three points are collinear, then they lie on a line. Let first calculate distance between the 3 points i.e. PQ. QR and PR Calculating PQ P ( – 2, 3, 5) Q (1, 2, 3) Hence , PQ = x2−x12+y2−y12+z2 −z12 Here , x1 = – 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 PQ = 1−−22+2−32+3−52 = 1+22+2−32+3−52 = 32+−12+(−2)2 = 9+−12+−22 = 9+1+4 = 14 Calculating QR Q ( 1, 2, 3) R (7, 0, –1) QR = x2−x12+y2−y12+z2 −z12 Here x1 = 1, y1 = 2, z1 = 2 x2 = 7, y2 = 0, z2 = – 1 QR = 7−12+0−22+−1−32 = 62+−22+−42 = 36+4+16 = 56 = 14×2×2 = 214 Calculating PR P ( –2, 3, 5) R (7, 0, –1) PR = x2−x12+y2−y12+z2 −z12 Here, x1 = – 2, y1 = 3, z1 = 5 x2 = 7, y2 = 0, z2 = – 1 PR = 7−(−2)2+0−32+−1−52 = 7+22+−32+−62 = 92+9+36 = 81+9+36 = 126 = 14×3×3 = 314 Thus, PQ = 14 , QR = 214 & PR = 314 So, PQ + QR = 14 + 214 = 314 = PR Thus, PQ + QR = PR So, if we draw the points on a graph, with PQ + QR = PR We see that points P,Q,R lie on the same line. Thus, P, Q and R all collinear

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .