3.4 The Point-Slope Form of a Line

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300 Chapter 3 Linear Functions Version: Fall 2007 4x + 3y = 12 3y = −4x + 12 y = − 4 x + 4 3 If two lines are perpendicular, then their slopes are negative reciprocals of one another. Therefore, the slope of the line that is perpendicular to the line 4x + 3y = 12 (which has slope −4/3) is m = 3/4. Our second line must pass through the point P (−4, −4). To draw this second line, first plot the point P (−4, −4), then move 4 units to the right and 3 units upward to the point Q(0, −1), as shown in Figure 5(b). The line through the points P and Q is perpendicular to the line 4x + 3y = 12. 3 To determine the equation of the line through the points P and Q, we will use the point-slope form of the line, namely y − y0 = m(x − x0). (16) The slope of the line through points P and Q is m = 3/4. If we use the point P (−4, −4), then (x0, y0) = (−4, −4). Set m = 3/4, x0 = −4, and y0 = −4 in equation (16), obtaining or equivalently, y − (−4) = 3 (x − (−4)), 4 y + 4 = 3 (x + 4). (17) 4 Alternatively, we could use the slope-intercept form of the line. We know that the line through points P and Q in Figure 5(b) crosses the y-axis at Q(0, −1). So, with slope m = 3/4 and y-coordinate of the y-intercept b = −1, the slope-intercept form y = mx + b becomes y = 3 x − 1. 4 On the other hand, if we solve equation (17) for y, y + 4 = 3 (x + 4) 4 y + 4 = 3 x + 3 4 y = 3 x − 1. 4 Note that this is identical to the result found using the slope-intercept form above. 2 If you also remember that the slope of Ax + By = C is m = −A/B, then the slope of 4x + 3y = 12 is m = −A/B = −4/3. 3 It’s a good exercise to measure the angle between the two lines with a protractor. If the angle measures 90 degrees, then you know the lines are truly perpendicular. (18)
Section 3.4 The Point-Slope Form of a Line 301 It is comforting to note that the two forms (point-slope and slope-intercept) give the same result, but how do we determine the most efficient form to use for a particular problem? Here’s a good hint. Determining the Form of the Line to Use. Here is some sound advice when you are trying to determine whether to use the slope-intercept form or the pointslope form of a line. • If you are given the slope and the y-intercept, use the slope-intercept form y = mx + b. • If you are given a point (other than the y-intercept) and the slope, use the point-slope form y − y0 = m(x − x0). Applications of Linear Functions In this section we will look at some applications of linear functions. We begin by developing a function relating Fahrenheit and Celsius temperature. ◮ Example 19. Water freezes at 32 ◦ F and 0 ◦ C. Water boils at 212 ◦ F and 100 ◦ C. F and C are abbreviations for Fahrenheit and Celsius temperature scales, respectively. Assuming a linear relationship, develop a model relating Fahrenheit and Celsius temperature. First, to help keep our focus, we set up a coordinate system on a sheet of graph paper. In Figure 6, we’ve decided to make the Celsius temperature the dependent variable and have assigned the Celsius temperature to the vertical axis. Similarly, we’ve declared the Fahrenheit temperature the independent variable and assigned it to the horizontal axis. 4 Interpret the given data: • Water freezes at 32 ◦ F and 0 ◦ C. This gives us the point (F, C) = (32, 0), which we plot in Figure 6. • Water boils at 212 ◦ F and 100 ◦ C. This gives us the point (F, C) = (212, 100), which we plot in Figure 6. Now we are on familiar ground. We want to find the equation of the line through these two points, which is the same type of problem we tackled in Example 6. First, use the points (32, 0) and (212, 100) to determine the slope of the line. m = ∆C ∆F 100 − 0 100 5 = = = 212 − 32 180 9 . We will now use the point-slope form of the line, y − y0 = m(x − x0) with m = 5/9 and (x0, y0) = (32, 0). Substitute m = 5/9, x0 = 32, and y0 = 0 in y − y0 = m(x − x0) to obtain 4 We could easily reverse these assignments, placing the Fahrenheit temperature on the vertical axis and the Celsius temperature on the horizontal axis. Version: Fall 2007
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3.4 The Point-Slope Form of a Line

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