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45 Sect 2.5 – Equations of Lines and Linear Models Objective 1: Point-Slope Formula From the last section, we worked with the slope-intercept form of the linear function: Slope-intercept form A linear equation in two variables is said to be written in slope-intercept form if it is in the form y = mx + b where m is the slope of the line and the point (0, b) is the y-intercept. In this section, we want to introduce the point-slope form of a linear function. To see where the point-slope form comes from, let us examine the formula for calculating the slope: m= € € € y 2 − y1 x2 − x1 y 2 − y1 =m x2 − x1 y − y1 =m x− x1 y − y1 (x− x1) x− x1 1 (rewrite the formula) (replace (x2, y2) with the variables x and y) (clear fractions by multiply both sides by (x – x1)) = m(x – x1) (reduce) y – y1 = m(x – x1) This is a point-slope formula. Point-Slope Form € € A linear equation in two variables is said to be written in point-slope form if it is in the form: y – y1 = m(x – x1), where m is the slope and (x1, y1) is any known point on the line. Given the following information, find and write the equation in a) point-slope form, b) slope-intercept form, and c) standard form. Ex. 1 A line with slope of 6 Ex. 2 A line with slope of – 2 3 passing through the point (9, 5). Solution: m = 6; (x1, y1) = (9, 5) passing through the point (– 8, 4). Solution: € 2 m = – ; (x1, y1) = (– 8, 4) y – y1 = m(x – x1) y – y1 = m(x – x1) 3 € 46 2 (x – (– 8)) pt-slope 3 2 y – 4 = – (x + 8) 3 2 16 y–4=– x– 3 3 € 12 +4= + 3 € 2 4 b) y = – x – slope-int 3 € €3 2 4 3(y) = 3 – x – 3 3 3 € a) y – 5 = 6(x – 9) pt-slope a) y – 4 = – y – 5 = 6x – 54 +5= +5 b) y = 6x – 49 slope-int y = 6x – 49 ( – 6x = – 6x – 6x + y = – 49 –1 –1 c) 6x – y = 49 y 2 − y1 x2 − x1 2 2 =– −7 7 m= = = . 3y = – 2x – 4 €+ 2x =€ + 2x c) 2x + 3y = – 4 std form € € std form Ex. 3 A line that passes through the points (2, – 3) and (– 5, – 1). Solution: First, find the slope: Ex. 4 A line that passes through the points (7, – 10) and (5, – 14). Solution: First, find the slope: −1 − (−3) −5−2 € a) y – (– 3) = – € y+3=– 2 7 2 7 −4 −2 y 2 − y1 x2 − x1 = −14−(−10) 5−7 = 2. Now, pick one of the points. € m = 2; (x1, y1) = (7, – 10) y – y1 = m(x – x1) (x – 2) pt-slope a) y – (– 10) = 2(x – 7) pt-slope 4 7 21 7 x+ –3= – € 2 17 b) y = – x – slope-int 7 7 € € 2 17 7y = 7 – x – 7 7 € 7 7y = – 2x – 17 €+ 2x = € + 2x c) 2x + 7y = – 17 std form € € ( m= = Now, pick one of the points. € 2 € m = – ; (x1, y1) = (2, – 3) 7 € y€– y1 = m(x – x1) € ) ( ) ) ( ) y + 10 = 2x – 14 – 10 = – 10 b) y = 2x – 24 slope-int – 2x = – 2x – 2x + y = – 24 –1 –1 c) 2x – y = 24 std form 47 Find the slope and y-intercept and sketch the graph: Ex. 5 8x + 3y = 24 Solution: First solve the equation for y: 8x + 3y = 24 – 8x = – 8x 3y = – 8x + 24 3 3 8 y=– x+8 The slope is – 8 3 3 = −8 3 = "rise" "run" and the y-intercept is (0, 8). Plot the € point (0, 8). Then from that point fall 8 units and run € another € 3 units€to get point. From that new point fall another 8 units and run 3 more units to get the third point. Now, draw the graph. For each graph, write the equation of the line: Ex. 6 Ex. 7 Cost of a TV Cost of a gallon of gas Year Month Solution: Two points on the graph are (2, 1000) and (6, 500). The 500−1000 slope is then = – 125. Solution: Two points on the graph are (0, 3) and (5, 4). The slope is 4−3 then = 0.2. The point The point (0, 1250) is the y-intercept. So, m = – 125 and b = 1250. Thus, the equation of € is y = – 125x + 1250. As the line a linear function, we can write f(x) = – 125x + 1250. (0, 3) is the y-intercept. So, m = 0.2 and b = 3. Thus, the the equation of the line is y = 0.2x + 3. As a linear function, we can write f(x) = 0.2x + 3. 6−2 5−0 € 48 As mentioned before, the two special cases for linear equations were vertical and horizontal lines. Equations of Vertical and Horizontal lines 1) An equation of a vertical line passing through the point (a, b) is x = a. The slope of such line is undefined. 2) An equation of a horizontal line passing through the point (a, b) is y = b. The slope of such line is 0. Use the given information to write the equation of the line: Ex. 8 The slope of the line is 0 and it passes through (– 2, 5). Solution: m = 0, but we do not have a y-intercept. A line with a slope of 0 is a horizontal line. Recall that a horizontal is in the form of y = constant. Since it has to pass through the point (– 2, 5), then y has to be 5. So, the equation is y = 5. Ex. 9 The slope of the line is undefined and it passes through (– 2, 5). Solution: m is undefined and we do not have a y-intercept. A line with a slope that is undefined is a vertical line. Recall that a vertical is in the form of x = constant. Since it has to pass through the point (– 2, 5), then x has to be – 2. So, the equation is x = – 2. Objective 2: Parallel and Perpendicular Lines. Two distinct lines are parallel if and only if they have the same slope. We can denote the slope of a parallel line by m||. Two distinct lines that are neither vertical nor horizontal lines are perpendicular if and only if the product of their slopes is – 1. In other words, the slope of one line is the negative reciprocal of the slope of the line perpendicular to it so long as they are not vertical or horizontal lines. Every vertical line is perpendicular to every horizontal line and every horizontal line is perpendicular to every vertical line. We can denote the slope of a perpendicular line by m⊥. 49 Write the equation of the line in slope-intercept form: Ex. 10 Find the equation of the line passing through (– 2, 6). a) parallel to y = 3x – 2. b) perpendicular to y = 3x – 2. Solution: a) Parallel lines have the same slope, so m|| = 3. y – y1 = m(x – x1) y – 6 = 3(x – (– 2)) = 3(x + 2) = 3x + 6 y – 6 = 3x + 6 +6= +6 y = 3x + 12 b) The slope of the line perpendicular is the negative reciprocal of 1 3. So, m⊥ = – . 3 y – y1 = m(x – x1) 1 y – 6 = – (x – (– 2)) = – y – 6€= – 3 1 x 3 – +6= + € 1 16 y=– x+ 3 3 € € 1 (x 3 2 3 18 3€ + 2) = – € 1 x 3 – 2 3 € € equation of the line passing through (4, 0) Ex. 11 Find the 1 a) €parallel to y = x – 3. € 6 b) perpendicular to y = 1 x 6 – 3. Solution: € lines have the same slope, so m|| = a) Parallel € – x1 ) y – y1 = m(x 1 y – 0 = (x – 4) = 6 y= 1 x 6 – 1 x 6 – 4 6 = 1 x 6 2 3 – 1 . 6 2 3 € b) The slope of the line perpendicular is the negative reciprocal of €1 . So, m € € € € ⊥ = – 6. 6 € € y€– y1 = m(x – x1) y – 0 = – 6(x – 4) = – 6x + 24 y = – 6x + 24 50 Ex. 12 Find the equation of the line passing through (8, – 3) a) parallel to x = 5. b) perpendicular to x = 5. Solution: a) A line parallel to x = 5 is a vertical line and it passes through (8, – 3). So, the equation is x = 8. b) A line perpendicular to x = 5 is a horizontal line and it passes through (8, – 3). So, the equation is y = – 3. Objective 3: Modeling Data that Changes at a Constant Rate. If the data in an application problem changes at nearly a constant rate, we can use a linear function to approximate the data and this gives us a model for the data where the average rate of change is the slope of the line. We will typically draw what is called a trend line to model the data and pick two points that lie the closest on the line to derive the equation of the model. Find the equation of the line. Ex. 13 The Average annual cost of tuition, board, fees, etc at a public four-year institution is given in the graph below where x represents the number of years after 2001 and y represents the cost. $17,860 Cost $12,300 Source: www.collegeboard.org Years after 2001 a) b) Find a linear function that models the data Use the model found in part a to predict the cost in 2016. 51 Solution: a) The two points given on the graph are (1, 12300) and (11, 17860). The slope is y 2 − y1 x2 − x1 = 17860−12300 11−1 = 5560 10 = 556. This means that the cost is increasing at an average rate of $556 a year. To find the equation of the line, we can use the point-slope formula: € €– x ) € y – y1 = m(x 1 y – 12300 = 556(x – 1) y – 12300 = 556x – 556 + 12300 = + 12300 y = 556x + 11744 Thus, the linear function that models the data is f(x) = 556x + 11,744 b) Since x represents the number of years after 2001, then x = 1 represents 2002, x = 2 represents 2003, etc. Thus, x = 15 represents 2016. Evaluating f at x = 15 yields: f(15) = 556(15) + 11,744 = 8340 + 11,744 = 20,084 The model predicts that the cost will be ≈ $20,084 in the year 2016. It is worth noting that if we had use different data points, we would have derived a slightly different answer. Below is a summary of different formulas we have seen involving linear equations. Form Standard Form Ax + By = C Horizontal Line y = constant Vertical Line x = constant Slope-Intercept (Function) Form f(x) = y = mx + b Point-Slope Form y – y1 = m(x – x1) Example 5x – 3y = 17 y = – 2.3 x=3 m = 3, (0, b) = (0, – 5) f(x) = y = 3x – 5 m = – 4, (x1, y1) = (2, 3) y – 3 = – 4(x – 2) Notes A and B are integers & must not both be 0. m = 0, y-int: (0, constant) m is undefined This is not a function. m is the slope, (0, b) is the y-int. m is the slope, (x1, y1) is any point on the line.