Using the distance formula, show that the given points are collinear:
(i) (1, – 1), (5, 2) and (9, 5)
(ii) (6, 9), (0, 1) and (– 6, – 7)
(iii) (– 1, – 1), (2, 3) and (8, 11)
(iv) (– 2, 5), (0, 1) and (2, – 3).
Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C if Slope of AB = slope of BC = slope of AC
then A, B and C are collinear points.
Slope of any two points is given by:
(y2 – y1)/(x2 – x1).
(i) Slope of AB = (2 – (– 1))/(5 – 1) = 3/4
Slope of BC = (5 – 2)/(9 – 5) = 3/4
Slope of AB = slope of BC
Hence collinear.
(ii) Slope of AB = (1 – 9)/(0 – 6) = 8/6 = 4/3
Slope of BC = (– 6 – 0)/(– 7 – 1) = 6/6 = 1
Slope of AC = (– 7 – 9)/(– 6 – 6) = – 16/ – 12 = 4/3
Slope of AB = slope of AC
Hence collinear.
(iii) Slope of AB = ((3 – (– 1))/((2 – (– 1)) = 4/3
Slope of BC = (11 – 2)/(8 – 3) = 9/5 = 1
Slope of AC = ((11 – (– 1))/((8 – (– 1)) = 12/9 = 4/3
Slope of AB = slope of AC
Hence collinear.
(iv) Slope of AB = (1 – 5)/((0 – (– 2)) = – 4/2 = – 2
Slope of BC = (– 3 – 1)/(2 – 0) = – 4/2 = – 2
Slope of AB = slope of AB
Hence collinear.