3.4 The Point-Slope Form of a Line

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294 Chapter 3 Linear Functions The Point-Slope Form of a Line. If line L passes through the point (x0, y0) and has slope m, then the equation of the line is Version: Fall 2007 y − y0 = m(x − x0). (1) This form of the equation of a line is called the point-slope form. To use the point-slope form of a line, follow these steps. Procedure for Using the Point-Slope Form of a Line. When given the slope of a line and a point on the line, use the point-slope form as follows: 1. Substitute the given slope for m in the formula y − y0 = m(x − x0). 2. Substitute the coordinates of the given point for x0 and y0 in the formula y − y0 = m(x − x0). For example, if the line has slope −2 and passes through the point (3, 4), then substitute m = −2, x0 = 3, and y0 = 4 in the formula y − y0 = m(x − x0) to obtain y − 4 = −2(x − 3). ◮ Example 2. Draw the line that passes through the point P (−3, −2) and has slope m = 1/2. Use the point-slope form to determine the equation of the line. First, plot the point P (−3, −2), as shown in Figure 2(a). Starting from the point P (−3, −2), move 2 units to the right and 1 unit up to the point Q(−1, −1). The line through the points P and Q in Figure 2(a) now has slope m = 1/2. P (−3,−2) Q(−1,−1) ∆x=2 y ∆y=1 (a) The line through P (−3, −2) with slope m = 1/2. x Figure 2. y R(0,−0.5) (b) Checking the y-intercept. x
Section 3.4 The Point-Slope Form of a Line 295 To determine the equation of the line in Figure 2(a), we will use the point-slope form of the line y − y0 = m(x − x0). (3) The slope of the line is m = 1/2 and the given point is P (−3, −2), so (x0, y0) = (−3, −2). In equation (3), set m = 1/2, x0 = −3, and y0 = −2, obtaining or equivalently, y − (−2) = 1 (x − (−3)), 2 This is the equation of the line in Figure 2(a). y + 2 = 1 (x + 3). (4) 2 As a check, we’ve estimated the y-intercept of the line in Figure 2(b) as R(0, −0.5). Let’s place equation (4) in slope-intercept form to determine the exact value of the y-intercept. First, distribute 1/2 to get y + 2 = 1 3 x + 2 2 . Subtract 2 from both sides of this last equation. y = 1 3 x + − 2 2 2 Make equivalent fractions with a common denominator and simplify. y = 1 3 4 x + − 2 2 2 y = 1 1 x − 2 2 Comparing equation (5) with y = mx+b gives us b = −1/2. This is the exact y-value of the y-intercept. Note that this result compares exactly with the y-value of point R in Figure 2(b). This is a bit lucky. Don’t expect to get an exact comparison every time. However, if the comparison is not close, look for an error in your work, either in your computations or in your graph. Let’s look at another example. ◮ Example 6. Find the equation of the line passing through the points P (−3, 2) and Q(2, −1). Place your final answer in standard form. Again, to help keep our focus, we draw the line passing through the points P (−3, 2) and Q(2, −1) in Figure 3. (5) Version: Fall 2007
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3.4 The Point-Slope Form of a Line

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