Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Parallel and Perpendicular Lines
Continue to next subchapter

Exercise 46 Page 335

Start by using the Slope Formula to find the slope of the line.

y=0.25x+1.875

Practice makes perfect
An equation in slope-intercept form follows a specific format. y= mx+b For an equation in this form, m is the slope and b is the y-intercept. Let's use the given points to calculate m. We will start by substituting the points into the Slope Formula.
m=y_2-y_1/x_2-x_1
m=3- 2/4.5- 0.5
m=1/4
m = 0.25
A slope of 0.25 means that for every horizontal step in the positive direction, we take 0.25 vertical steps in the positive direction. Now that we know the slope, we can write a partial version of the equation. y= 0.25 x+b To complete the equation, we also need to determine the y-intercept, b. Since we know that the given points will satisfy the equation, we can substitute one of them into the equation and solve for b. Let's use ( 0.5, 2).
y=0.25x+b
2=0.25( 0.5)+b
â–Ľ
Solve for b
2=0.125+b
2 - 0.125 = 0.125 + b - 0.125
2 - 0.125 = b + 0.125 - 0.125
1.875 = b
b= 1.875
A y-intercept of 1.875 means that the line crosses the y-axis at the point (0,1.875). We can now complete the equation. y= 0.25x+1.875