Chapter 3.4, Problem 56HP

To determine

To determine whether the points at $(-2,2)$ , $(2,5)$ and $(6,8)$ are collinear.Justify answer.

Answer to Problem 56HP

points are collinear.

Explanation of Solution

Given information:

Three points are$(-2,2)$ , $(2,5)$ and $(6,8)$

Formula used:

Slope, $m=\frac{y_2-y_1}{x_2-x_1}$

Point-slope form

  $y-y_1=m(x-x_1)$

Calculation:

Points

$(-2,2)$ , $(2,5)$ and $(6,8)$

On taking first two points to find point-slope form of line

$(-2,2)$ , $(2,5)$

Here

  $x_1=-2$

  $y_1=2$

  $x_2=2$

  $y_2=5$

Using slope formula to calculate slope of line

  $m=\frac{5-2}{2-(-2)}$

On solving

  $m=\frac{3}{4}$

Point slope form finding below

Take any one point from first two chosen

  $m=\frac{3}{4}$

Let take $(-2,2)$

Here

  $x_1=-2$

  $y_1=2$

Using point-slope form formula

  $y-2=\frac{3}{4}(x-(-2))$

On solving

  $y-2=\frac{3}{4}(x+2)$

If third point $(6,8)$ lie on the line.

Then it will satisfy the equation.

So

  $8-2=\frac{3}{4}(6+2)$

On solving

  $6=\frac{3}{4}(8)$

On solving

  $6=3(2)$

On solving

  $6=6$ true

Hence, points are collinear.