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Question
Using section formula, show that the points A(7, −5), B(9, −3) and C(13, 1) are collinear
Solution
If three points are collinear, then one of the points divide the line segment joining the other points in the ratio r : 1. If P is between A and B and `"AP"/"PB"` = r
Distance AB = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((9 - 7)^2 + (-3 + 5)^2`
= `sqrt(2^2 + 2^2)`
= `sqrt(4 + 4)`
= `sqrt(8)`
= `2sqrt(2)`
Distance BC = `sqrt((13 - 9)^2 + (1 + 3)^2`
= `sqrt(4^2 + 4^2)`
= `sqrt(16 + 16)`
= `sqrt(32)`
= `sqrt(16 xx 2)`
= `4sqrt(2)`
To find r
r = `"AB"/"BC" = (2sqrt(2))/(4sqrt(2))`
`"AB"/"BC" = 1/2`
The line divides in the ratio 1 : 2
A line divides internally in the ratio m : n
The point P = `(("m"x_2 + "n"x_1)/("m" + "n"), ("m"y_2 + "n"y_1)/("m" + "n"))`
m = 1, n = 2, x1 = 7, x2 = 13, y1 = – 5, y2 = 1
By the given equation,
The point B = `((13 + 14)/5, (1 - 10)/3)`
= `(27/3, (-9)/3)`
= (9, −3)
∴ The three points A, B and C are collinear.
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