Exercise 21 Page 211

Practice makes perfect

a We are given two points, so we can calculate the slope between these two points using the Slope Formula. We can then compare the slope to the given equations to find a match.

m=y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( 0,0) & ( -1,1)

m=1- 0/- 1- 0

SubTerms

Subtract terms

m=1/- 1

MoveNegDenomToFrac

Put minus sign in front of fraction

m=- 1/1

QuotOne

a/a=1

m=- 1

The slope is - 1. Therefore, the corresponding equation is C, which is the only equation with a matching slope.

b Similar to Part A, we can begin with calculating the slope using the given points.

m=y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( 1,1) & ( -1,-1)

m=- 1- 1/- 1- 1

SubTerms

Subtract terms

m=- 2/- 2

QuotOne

a/a=1

m=1

The slope is 1. This slope can be found in two equations, both A and B. Let's substitute the point (1,1) in the equation A and see if it is a valid solution.

y-10=1(x+2)

SubstituteII

x= 1, y= 1

1-10? =1( 1+2)

AddSubTerms

Add and subtract terms

- 9? =1(3)

Multiply

- 9≠ 3

Because we reached a contradication, we know that the point does not work in equation A. Let's try substituting the same point into equation B so make we can be certain it is our correct answer.

y-0=1(x-0)

SubstituteII

x= 1, y= 1

1-0? =1( 1-0)

SubTerm

Subtract term

1? =1(1)

Multiply

1=1

It worked! The given points match equation B.

c Again, let's substitute the given points into the Slope Formula to calculate the slope.

m=y_2-y_1/x_2-x_1