Question

Using section formula, show that the points A(7, −5), B(9, −3) and C(13, 1) are collinear

Answer 1

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.