The Slope and Equation of a Line Help

Find an equation of the line containing the given points.

• (−2, 3) and (10, 15)

  

  We will use x = −2 and y = 3 in y = mx + b to find b .

  3 = 1(−2) + b

  5 = b

  The equation is y = 1 x + 5, or simply y = x + 5.

• and (4, 3)

  

  Using x = 4 and y = 3 in y = mx + b, we have

  

• (0, 1) and (12, 1)

  The y -values are the same, making this a horizontal line. The equation is y = 1.

•  

  

  Fig. 1.8.

  The line in Figure 1.8 crosses the y -axis at 1, so b = 1. From this point, we can go right 2 and up 3 to reach the point (2, 4) on the line. “Right 2” means that the denominator of the slope is 2. “Up 3” means that the numerator of the slope is 3. The slope is , so the equation of the line is .

•  

  

  Fig. 1.9.

  The y-intercept is not easy to determine, but we do have two points. We can either find the slope by using the slope formula, or visually (as we did above). We can find the slope visually by asking how we can go from (−4, 3) to (2, −1): Down 4 (making the numerator of the slope −4) and right 6 (making the denominator 6). If we use the slope formula, we have

  

  Using x = 2 and y = −1 in y = mx + b, we have . From this, we have . The equation is .

•  

  

  Fig. 1.10.

  The line in Figure 1.10 is vertical, so it has the form x = a. All of the x-values are −2, so the equation is x = −2.

• Find the slope of the line 6 x − 2 y = 3.

  6 x − 2 y = 3

  −2 y = −6x + 3

  

  The slope is .

• are negative reciprocals

• are negative reciprocals

  

  Fig. 1.12.

• are negative reciprocals
• 1 and − 1 are negative reciprocals

Determine whether the lines are parallel or perpendicular or neither.

• 4 x − 3 y = −15 and 4 x − 3 y = 6

  

  The lines have the same slope, so they are parallel.

• 3 x − 5 y = 20 and 5 x − 3 y = −15

  

  The slopes are reciprocals of each other but not negative reciprocals, so they are not perpendicular. They are not parallel, either.

• x − y = 2 and x + y = −8

  x − y = 2        x + y = −8

        y = x − 2        y = −x − 8

  The slope of the first line is 1 and the second is −1. Because 1 and −1 are negative reciprocals, these lines are perpendicular.

• y = 10 and x = 3

  The line y = 10 is horizontal, and the line x = 3 is vertical. They are perpendicular.

• Find an equation of the line containing the point (−4, 5) that is parallel to the line y = 2 x + 1.

  The slope of y = 2 x + 1 is 2. This is the same as the line we want, so we will let x = −4, y = 5, and m = 2 in y = mx + b. We get 5 = 2(−4) + b , so b = 13. The equation of the line we want is y = 2x + 13.

• Find an equation of the line with x -intercept 4 that is perpendicular to x − 3y = 12.

  The x -intercept is 4 means that the point (4, 0) is on the line. The slope of the line we want will be the negative reciprocal of the slope of the line x − 3 y = 12. We will find the slope of x − 3 y = 12 by solving for y .

  

  The slope we want is −3, which is the negative reciprocal of . When we let x = 4, y = 0, and m = −3 in y = mx + b , we have 0 = −3(4) + b , which gives us b = 12. The line is y = −3 x + 12.

• Find an equation of the line containing the point (3, −8), perpendicular to the line y = 9.

  The line y = 9 is horizontal, so the line we want is vertical. The vertical line passing through (3, −8) is x = 3.