Exercise 53 Page 144

Let's begin by looking at a generic straight line, one that we know can be written in slope-intercept form, y=mx+b, but we do not know the values of the slope or the y-intercept. Two points on this line can then be labeled as (x_1,y_1) and (x_2,y_2). The equations of the line expressed using these points are then: &( x_1, y_1) ⇒ y_1=m x_1+b &( x_2, y_2) ⇒ y_2=m x_2+b

We can see their relationship in the graph below.

The Slope Formula is Slope=y_2-y_1/x_2-x_1. We can substitute the expressions from the previously stated equations for y_1 and y_2 into the Slope Formula to prove the slope equals m.

Slope=y_2-y_1/x_2-x_1

SubstituteII

y_1= mx_1+b, y_1= mx_1+b

Slope=( mx_2+b)-( mx_1+b)/x_2-x_1

Distr

Distribute -1

Slope=mx_2+b-mx_1-b/x_2-x_1

SubTerms

Subtract terms

Slope=mx_2-mx_1/x_2-x_1

FactorOut

Factor out m

Slope=m(x_2-x_1)/x_2-x_1

MovePartNumLeft

a* b/c=a*b/c

Slope=mx_2-x_1/x_2-x_1

QuotOne

a/a=1

Slope=m