2.1 Equations of Lines

Transcription

1 Section 2.1 Equations of Lines Equations of Lines The Slope-Intercept Form Recall the formula for the slope of a line. Let s assume that the dependent variable is and the independent variable is and we have a line passing through the points P ( 1, 1 ) and Q( 2, 2 ), as shown in Figure 1. Q( 2, 2 ) P ( 1, 1 ) = 2 1 = 2 1 Figure 1. Determining the slope of a line through two points. Note that the change in is = 2 1 and the change in is = 2 1. Thus, the slope of the line is determined b the formula Slope = = (2.1) We can use the slope formula to find the equation of the line with slope m and intercept given b (0, b). In doing so we arrive at the following. The Slope-Intercept Form of a Line. If the line L intercepts the -ais at the point (0, b) and has slope m, then the equation of the line is = m + b. (2.2) This form of the equation of a line is called the slope-intercept form. function defined b the equation The f() = m + b is called a linear function. It is important to note two ke facts about the slope-intercept form = m + b. The coefficient of (the m in = m + b) is the slope of the line. The constant term (the b in = m+b) is the -coordinate of the -intercept (0, b).

2 2 Chapter 2 Let s review an eample of how this form of the line is used. Eample 1 What is the equation of the line having slope 2/3 and -intercept at (0, 3)? Sketch the line on graph paper. The equation of the line is = m + b. (2.3) We re given that the slope m = 2/3 and we re given that the line intercepts the -ais at the point (0, 3). Substitute m = 2/3 and b = 3 in equation (2.3), obtaining = To sketch the graph of the line, first locate the -intercept at P (0, 3), as shown in Figure 2. Starting from the -intercept at P (0, 3), move 3 units to the right and 2 units downward to the point Q(3, 1). The required line passes through the points P and Q. Note that the line intercepts the -ais at 3 and slants downhill, in accordance with the fact that the slope is negative in this eample. P (0,3) =3 Q(3,1) = 2 Figure 2. The line has -intercept at (0, 3) and slope 2/3. Eample 2 Given the graph of the line in Figure 3(a), determine the equation of the line. First, locate the -intercept of the line, which we ve labeled P (0, 1) in Figure 3(b). In the formula = m + b, recall that b represents the -coordinate of the -intercept. Thus, b = 1.

3 Section 2.1 Equations of Lines 3 Secondl, we need to determine the slope of the line. In Figure 3(b), start at the point P, move 2 units to the right, then 3 units upward to the point Q(2, 2). This makes the slope m = = 3 2. Substitute m = 3/2 and b = 1 into the slope-intercept form = m + b to obtain which is the desired equation of the line. = 3 2 1, Q(2,2) P (0, 1) =3 =2 Figure 3. (a) (b) Determining the equation of a line from its graph. Horizontal and Vertical Lines Recall the standard form of the line A + B = C. The Standard Form of a Line. The graph of the equation A + B = C is a line. The form is called the standard form of a line. A + B = C (2.4) The case where A and B are simultaneousl equal to zero is not ver interesting. However, the following two cases are of interest. 1. If we let A = 0 and B 0 in the standard form A + B = C, then B = C, or equivalentl = C/B. Note that this has the form = b, where b is some constant. 2. Similarl, if we let B = 0 and A 0 in the standard form A + B = C, then A = C, or equivalentl, = C/A. Note that this has the form = a, where a is some constant.

4 4 Chapter 2 The lines having the form = a and = b are two of the easiest lines to plot. Let s review an eample of each. Eample 3 Sketch the graph of the equation = 3. The direction sketch the graph of the equation = 3 can be quite veing unless one remembers that a graph of an equation is the set of all points that satisf the equation. Thus, the direction is better posed if we sa sketch the set of all points (, ) that satisf = 3, or equivalentl, sketch the set of all points (, ) that have an -value of 3. Then it is an eas matter to sketch the vertical line shown in Figure 4. = 3 Figure 4. The graph of the equation = 3. Note that each point on the line has an -value equal to 3. Also, note that the slope of this vertical line is undefined. Eample 4 Sketch the graph of the equation = 3. This direction is better posed if we sa sketch the set of all points (, ) that satisf = 3, or equivalentl, sketch the set of all points (, ) that have a -value of 3. Then it is an eas matter to sketch the horizontal line shown in Figure. Note that each point on the line has a -value equal to 3. Also, note that a horizontal line has slope zero. The Point-Slope Form of a Line In the last section, we developed the slope-intercept form of a line ( = m + b). The slope-intercept form of a line is applicable when ou re given the slope and -intercept of the line. However, there will be times when the -intercept is unknown.

5 Section 2.1 Equations of Lines = 3 Figure. The graph of the equation = 3. Q(,) P ( 0, 0 ) Figure 6. slope m. A line through ( 1, 1 ) with Suppose for eample, that ou are asked to find the equation of a line that passes through a particular point P ( 1, 1 ) with slope = m. This situation is pictured in Figure 6. Using the slope formula it can be shown that the line that results from this is given b. The Point-Slope Form of a Line. If line L passes through the point ( 1, 1 ) and has slope m, then the equation of the line is 1 = m( 1 ). (2.) This form of the equation of a line is called the point-slope form.

6 6 Chapter 2 Let s review an eample of how to use this form of the line. Eample 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 7(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 7(a) now has slope m = 1/2. P ( 3, 2) Q( 1, 1) =2 =1 R(0, 0.) (a) The line through P ( 3, 2) with slope m = 1/2. Figure 7. (b) Checking the -intercept. To determine the equation of the line in Figure 7(a), we will use the point-slope form of the line 1 = m( 1 ). (2.6) The slope of the line is m = 1/2 and the given point is P ( 3, 2), so ( 1, 1 ) = ( 3, 2). In equation (2.6), set m = 1/2, 1 = 3, and 1 = 2, obtaining or equivalentl, ( 2) = 1 ( ( 3)), 2 This is the equation of the line in Figure 7(a). + 2 = 1 ( + 3). (2.7) 2 Let s look at another eample. Eample 6 Find the equation of the line passing through the points P ( 3, 2) and Q(2, 1). Place our final answer in slope-intercept form.

7 Section 2.1 Equations of Lines 7 Use the slope formula to determine the slope of the line through the points P ( 3, 2) and Q(2, 1). We ll use the point-slope form of the line m = = ( 3) = 3 1 = m( 1 ). (2.8) Let s use point P ( 3, 2) as the given point ( 1, 1 ). That is, ( 1, 1 ) = ( 3, 2) (note that it doesn t matter which of the given points we chose). Substitute m = 3/, 1 = 3, and 1 = 2 in equation (2.8), obtaining 2 = 3 ( ( 3)). (2.9) This is the equation of the line passing through the points P and Q in point-slope form. We now need to rewrite it in slope-intercept form. If we start with equation (2.9) and distribute the slope, Now add 2 to both sides 2 = 3 ( ( 3)) 2 = 3 9. = = Parallel Lines Recall that slope controls the steepness of a line. Consequentl, if two lines are parallel, the must have the same steepness or slope. Let s look at an eample of parallel lines. Eample 7 Find the equation of the line that passes through the point P ( 2, 2) that is parallel to the line passing through the points Q( 3, 1) and R(2, 1). First, we draw the line through the points Q( 3, 1) and R(2, 1), then plot the point P ( 2, 2), as shown in Figure 8(a). We can use the slope formula to calculate the slope of the line passing through the points Q( 3, 1) and R(2, 1). m = = 1 ( 1) 2 ( 3) = 2

8 8 Chapter 2 T (3,4) P ( 2,2) R(2,1) P ( 2,2) = =2 R(2,1) Q( 3, 1) Q( 3, 1) (a) The line through Q( 3, 1) and R(2, 1). Figure 8. (b) The line through P ( 2, 2) that is parallel to the line through Q and R. We now draw a line through the point P ( 2, 2) that is parallel to the line through the points Q and R. Parallel lines must have the same slope, so we start at the point P ( 2, 2), run units to the right, then rise 2 units up to the point T (3, 4), as shown in Figure 8(b). We seek the equation of the line through the points P and T. We ll use the pointslope form of the line 1 = m( 1 ). (2.10) We ll use ( 1, 1 ) = ( 2, 2) as the given point. The line through P has slope 2/. Let s place the equation (2.11) in slope-intercept form. 2 = 2 ( ( 2)). (2.11) 2 = or if we wished to have it in standard form, = = 14. Perpendicular Lines Suppose that two lines L 1 and L 2 have slopes m 1 and m 2, respectivel. Recall that if L 1 and L 2 are perpendicular, then the product of their slopes is m 1 m 2 = 1. Alternativel,

9 Section 2.1 Equations of Lines 9 the slope of the first line is the negative reciprocal of the second line, and vice-versa; i.e., m 1 = 1/m 2 and m 2 = 1/m 1. Let s look at an eample of perpendicular lines. Eample 8 Find the equation of the line passing through the point P ( 4, 4) that is perpendicular to the line = 12. Let s first determine the slope of the line = 12 b placing this equation in slope-intercept form (i.e., solve the equation = 12 for ) = 12 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 = 12 (which has slope 4/3) is m = 3/4. Our second line must pass through the point P ( 4, 4). Let s use this information to draw the given line 4+3 = 12 and the perpendicular line we seek. S(0,4) R(3,0) Q(0, 1) (a) Plot the and -intercepts of = 12. P ( 4, 4) Figure 9. =4 =3 (b) A line through P ( 4, 4) that is perpendicular to = 12. m= 4/3 To determine the equation of the perpendicular line we seek (though the points P and Q), we will use the point-slope form of the line, namel 1 = m( 1 ). (2.12) The slope of the line we seek is m = 3/4. If we let ( 1, 1 ) = ( 4, 4). Set m = 3/4, 1 = 4, and 1 = 4 in equation (2.12), obtaining ( 4) = 3 ( ( 4)), 4

10 10 Chapter 2 or equivalentl, + 4 = 3 ( + 4). (2.13) 4 Alternativel, we could use the slope-intercept form of the line. We solve equation (2.13) for, + 4 = 3 ( + 4) = (2.14) =