(x 0,y 0 ) (x, y) Figure 2.23: A line through the point (x 0,y 0 ) with slope m.

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1 22 MODULE 2. LINEAR EQUATIONS IN TWO VARIABLES 2c Eauations of a Line Point-Slope Form of a Line In the previous section we learned that if we are provided with the slope of a line and its -intercept, then the equation of the line is = m + b, wherem is the slope of the line and b is the -coordinate of the line s -intercept. However, suppose that the -intercept is unknown? Instead, suppose that we are given a point ( 0, 0 ) on the line and we re also told that the slope of the line is m (see Figure 2.23). ( 0, 0 ) (, ) Slope = m Figure 2.23: A line through the point ( 0, 0 ) with slope m. Let (, ) be an arbitrar point on the line, then use the points ( 0, 0 )and (, ) to calculate the slope of the line. Slope = The slope formula. m = 0 0 Substitute m for the slope. Use (, ) and ( 0, 0 ) to calculate the difference in and the difference in. Clear the fractions from the equation b multipling both sides b the common denominator. [ ] 0 m( 0 )= ( 0 ) 0 Multipl both sides b 0. m( 0 )= 0 Cancel. Thus, the equation of the line is 0 = m( 0 ).

2 2C. EAUATIONS OF A LINE 23 The Point-Slope form of a line. The equation of the line with slope m that passes through the point ( 0, 0 ) is: 0 = m( 0 ) You Tr It! EXAMPLE 1. Draw the line passing through the point ( 3, 4) that has slope 6/, then find its equation. Solution: Plot the point ( 3, 4), move 6 units up, then move units to the right (see Figure 2.24). = =6 -intercept 0.3 ( 3, 4) Figure 2.24: The line passing through ( 3, 4) with slope 6/. Note that we do not know the eact point where the line crosses the -ais. We could estimate the -intercept as (0, 0.3). Then we could use the slopeintercept form = m + b, substitute 6/ form, 0.3 forb, and obtain the following approimation: (2.1) But this result is onl an approimation. Let s use the point-slope form of the line to find the eact equation. 0 = m( 0 ) Point-slope form.

3 24 MODULE 2. LINEAR EQUATIONS IN TWO VARIABLES Substitute 6/ for m. Because the line passes through ( 0, 0 )=( 3, 4), substitute 3 for 0 and 4 for 0. ( 4) = 6 ( ( 3) ) +4= 6 ( +3) Simplif. Substitute: 6/ for m, 3 for 0, and 4 for 0. Thus, the eact equation of the line is +4= 6 ( +3). But how well does this agree with our estimate (2.1)? question, we must solve +4= 6 ( +3)for. To answer this +4= 6 ( +3) +4= = = Equation of the line. Distribute 6/. 4 Subtract 4 from both sides. On the left, simplif. On the right make equivalent fractions with a common denominator. = 6 2 Simplif. Thus, the eact equation of the line is = 6 2. If we divide 2 b, we see that the eact equation of the line an also be epressed in the form = This result tells us that the eact coordinates of the -intercept are (0, 0.4). Note that this agrees closel with the approimation (2.1), giving us great confidence in our answer. Tip. Which form should I use? The form ou should select depends upon the information given. 1. If ou are given the -intercept and the slope, use = m + b. 2. If ou are given a point and the slope, use 0 = m( 0 ). Parallel Lines Slope is a number that measures the steepness of the line. If two lines are parallel (never intersect), the have the same steepness. Thus, if two lines are parallel, the have the same slope.

4 2C. EAUATIONS OF A LINE 2 ou Tr It! EXAMPLE 2. Sketch the line = 3 4 2, then sketch the line passing through the point ( 1, 1) that is parallel to the line = Find the equation of this parallel line. Solution: Note that = is in slope-intercept form = m + b. Hence, its slope is 3/4 and its -intercept is (0, 2). Plot the -intercept (0, 2), move up 3 units, right 4 units, then draw the line (see Figure 2.2). =4 =3 =4 =3 ( 1, 1) (0, 2) Figure 2.2: The line = Figure 2.26: Adding a line parallel to = The second line must be parallel to the first, so it must have the same slope 3/4. Plot the point ( 1, 1), move up 3 units, right 4 units, then draw the line (see the red line in Figure 2.26). To find the equation of the parallel red line in Figure 2.26, use the pointslope form, substitute 3/4 for m, then( 1, 1) for ( 0, 0 ). That is, substitute 1 for 0 and 1 for 0. 0 = m( 0 ) 1= 3 ( ) ( 1) 4 Point-slope form. Substitute: 3/4 for m, 1 for 0, and 1 for 0. 1= 3 ( +1) 4 Simplif. Perpendicular Lines Two lines are perpendicular if the meet and form a right angle (90 degrees). For eample, the lines L 1 and L 2 in Figure 2.27 are perpendicular, but the

5 26 MODULE 2. LINEAR EQUATIONS IN TWO VARIABLES lines L 1 and L 2 in Figure 2.28 are not perpendicular. L 1 L 1 L 2 L 2 Figure 2.27: The lines L 1 and L 2 are perpendicular. The meet and form a right angle (90 degrees). Figure 2.28: The lines L 1 and L 2 are not perpendicular. The do not form a right angle (90 degrees). Before continuing, we need to establish a relation between the slopes of two perpendicular lines. So, consider the perpendicular lines L 1 and L 2 in Figure L 2 m 1 L m 1 Figure 2.29: Perpendicular lines L 1 and L 2. Things to note: 1. If we were to rotate line L 1 ninet degrees counter-clockwise, then L 1 would align with the line L 2, as would the right triangles revealing their slopes. 2. L 1 has slope = m 1 1 = m L 2 has slope = 1 m 1 = 1 m 1.

6 2C. EAUATIONS OF A LINE 27 Slopes of perpendicular lines. If L 1 and L 2 are perpendicular lines and L 1 has slope m 1,theL 2 has slope 1/m 1. That is, the slope of L 2 is the negative reciprocal of L 1. Eamples: To find the slope of a perpendicular line, invert the slope of the first line and negate. If the slope of L 1 is 2, then the slope of the perpendicular line L 2 is 1/2. If the slope of L 1 is 3/4, then the slope of the perpendicular line L 2 is 4/3. You Tr It! EXAMPLE 3. Sketch the line = 2 3 3, then sketch the line through (2, 1) that is perpendicular to the line = Find the equation of this perpendicular line. Solution: Note that = is in slope-intercept form = m+b. Hence, its slope is 2/3 and its -intercept is (0, 3). Plot the -intercept (0, 3), move right 3 units, down two units, then draw the line (see Figure 2.30). =2 =3 (2, 1) =3 (0, 3) = 2 Figure 2.30: The line = Figure 2.31: Adding a line perpendicular to = Because the line = has slope 2/3, the slope of the line perpendicular to this line will be the negative reciprocal of 2/3, namel 3/2. Thus, to draw the perpendicular line, start at the given point (2, 1), move up 3 units, right 2 units, then draw the line (see Figure 2.31). To find the equation of the perpendicular line in Figure 2.31, use the pointslope form, substitute 3/2 for m, then(2, 1) for ( 0, 0 ). That is, substitute 2

7 28 MODULE 2. LINEAR EQUATIONS IN TWO VARIABLES for 0,then1for 0. 0 = m( 0 ) Point-slope form. 1= 3 2 ( 2) Substitute: 3/2 for m, 2for 0, and 1 for 0. Applications Let s look at a real-world application of lines. You Tr It! EXAMPLE 4. Water freezes at 32 F (Fahrenheit) and at 0 C (Celsius). Water boils at 212 F. If the relationship is linear, find an equation that epresses the Celsius temperature in terms of the Fahrenheit temperature. Use the result to find the Celsius temperature when the Fahrenheit temperature is 113 F. Solution: In this eample, the Celsius temperature depends on the Fahrenheit temperature. This makes the Celsius temperature the dependent variable and it gets placed on the vertical ais. This Fahrenheit temperature is the independent variable, so it gets placed on the horizontal ais (see Figure 2.32) C (212, 100) (32, 0) F Figure 2.32: The linear relationship between Celsius and Fahrenheit temperature.

8 2C. EAUATIONS OF A LINE 29 Net, water freezes at 32 Fand0 C, giving us the point (F, C) =(32, 0). Secondl, water boils at 212 F and 100 C, giving us the point (F, C) = (212, 100). Note how we ve scaled the aes so that each of these points fit on the coordinate sstem. Finall, assuming a linear relationship between the Celsius and Fahrenheit temperatures, draw a line through these two points (see Figure 2.32). Calculate the slope of the line. m = C F m = m = m = 9 Slope formula. Use the points (32, 0) and (212, 100). to compute the difference in C and F. Simplif. Reduce. You ma either use (32, 0) or (212, 100) in the point-slope form. The point (32, 0) has smaller numbers, so it seems easier to substitute ( 0, 0 )=(32, 0) and m =/9 into the point-slope form 0 = m( 0 ). 0 = m( 0 ) Point-slope form. 0= 9 ( 32) Substitute: /9 for m, 32for 0, and 0 for 0. = ( 32) 9 Simplif. However, our vertical and horizontal aes are labeled C and F (see Figure 2.32) respectivel, so we must replace with C and with F to obtain an equation epressing the Celsius temperature C in terms of the Fahrenheit temperature F. C = (F 32) (2.2) 9 Finall, to find the Celsius temperature when the Fahrenheit temperature is 113 F, substitute 113 for F in equation (2.2). C = (F 32) Equation (2.2). 9 C = (113 32) Substitute: 113 for F. 9 C = 9 (81) C =4 Subtract. Multipl. Therefore, if the Fahrenheit temperature is 113 F, then the Celsius temperature is 4 C.