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Chapter 1 Linear Equations and Graphs Section 2 Graphs and Lines Table of Content Cartesian Coordinate System Graph of Ax + By = C Slope of a Line Equations of Lines: Special Forms Applications It will be considered one of the most basic geometric figures – a line. Barnett/Ziegler/Byleen College Mathematics 12e 2 Learning Objectives for Section 1.2 Graphs and Lines The student will be able to identify and work with the Cartesian coordinate system. The student will be able to draw graphs for equations of the form Ax + By = C. The student will be able to calculate the slope of a line. The student will be able to graph special forms of equations of lines. The student will be able to solve applications of linear equations. Barnett/Ziegler/Byleen College Mathematics 12e 3 Terms cartesian coordinate system horizontal axis vertical axis coordinate axes x axis y axis quadrants ordered pair Barnett/Ziegler/Byleen College Mathematics 12e abscisa ordinate x coodinate y coodinate origin slope of a line slope-intercept form pont-slope form 4 The Cartesian Coordinate System The Cartesian coordinate system was named after René Descartes. It consists of two real number lines, the horizontal axis (x-axis) and the vertical axis (y-axis) which meet in a right angle at a point called the origin. The two number lines divide the plane into four areas called quadrants. The quadrants are numbered using Roman numerals as shown on the next slide. Each point in the plane corresponds to one and only one ordered pair of numbers (x,y). Two ordered pairs are shown. Barnett/Ziegler/Byleen College Mathematics 12e 5 The Cartesian Coordinate System (continued) II I (3,1) x III (–1,–1) Two points, (–1,–1) and (3,1), are plotted. Four quadrants are as labeled. IV y Barnett/Ziegler/Byleen College Mathematics 12e 6 The Cartesian Coordinate System (continued) y Abscisa 10 II I b Q = (-5,5) -10 5 P = (a,b) Coordenadas Origen 5 -5 Ordenada a x 10 Ejes -5 IV III -10 R = (10,-10) Sistema de Coordenadas Cartesianas (Rectangulares) Barnett/Ziegler/Byleen College Mathematics 12e 7 The Cartesian Coordinate System (continued) If we define an arbirtrary point P, the coordinates are the horizontal and vertical lines through this point. The intersections are at a point with coordinate a and b. Those two numbers, written as the ordered pair (a, b) form the coordinates of the point P. They are called the abscisa and the ordinate of P. The coordinates of a point are also referenced in terms of the axis labels: x coordinate and y coordinates. The point with coordinates (0,0) is called the origin. Barnett/Ziegler/Byleen College Mathematics 12e 8 The Cartesian Coordinate System (continued) The procedure just described assigns to each point P a unique pair of real numbers (a, b). Conversely, if we are given an ordered pair of real numbers (a, b), then, reversing this procedure, we can determine a unique point P in the plane. Thus There is a one-to-one correspondance between the points in a plane and the elements in the set of all ordered pairs of real numbers. This is often referred as the fundamental theorem of analytic geometric. Barnett/Ziegler/Byleen College Mathematics 12e 9 Linear Equations in Two Variables Graphs of Ax + By = C A linear equation in two variables is an equation that can be written in the standard form Ax + By = C, where A, B, and C are constants (A and B not both 0), and x and y are variables. A solution of an equation in two variables is an ordered pair of real numbers that satisfy the equation. For example, (4,3) is a solution of 3x - 2y = 6. The solution set of an equation in two variables is the set of all solutions of the equation. The graph of an equation is the graph of its solution set. Barnett/Ziegler/Byleen College Mathematics 12e 10 Linear Equations in Two Variables Graphs of Ax + By = C Theorem 1 – Graph of a Linear Equation in Two Variables. The graph of any equation of the form Ax + By = C (A and B not both 0) is a line, and any line in a Cartesian coordinate system is the graph of an equation of this form. Barnett/Ziegler/Byleen College Mathematics 12e 11 Linear Equations in Two Variables (continued) If A is not equal to zero and B is not equal to zero, then Ax + By = C can be written as This is known as slope-intercept form. A C y x mx b B B If A = 0 and B is not equal to zero, then the graph is a horizontal line If A is not equal to zero and B = 0, then the graph is a vertical line Barnett/Ziegler/Byleen College Mathematics 12e C y B C x A 12 Using Intercepts to Graph a Line Graph 2x – 6y = 12. Barnett/Ziegler/Byleen College Mathematics 12e 13 Using Intercepts to Graph a Line Graph 2x – 6y = 12. x y 0 –2 y-intercept 6 0 x-intercept 3 –1 check point Barnett/Ziegler/Byleen College Mathematics 12e 14 Special Cases The graph of x = k is the graph of a vertical line k units from the y-axis. The graph of y = k is the graph of the horizontal line k units from the x-axis. Examples: 1. Graph x = –7 2. Graph y = 3 Barnett/Ziegler/Byleen College Mathematics 12e 15 Solutions x = –7 y=4 Barnett/Ziegler/Byleen College Mathematics 12e 16 Slope of a Line Slope of a line: y2 y1 rise m x2 x1 run x1 , y1 rise x2 , y2 run Barnett/Ziegler/Byleen College Mathematics 12e 17 Slope of a Line For a horizontal line, y does not change; its slope is 0. For a vertical line, x does not change; x1 = x2 so its slope is not defined. In general, the slope of a line may be positive, negative, 0, or not defined. See next table. Barnett/Ziegler/Byleen College Mathematics 12e 18 Slope of a Line TABLA 2 – Interpretación Geométrica de la Pendiente Línea Ascendente a medida que x se mueve de izquierda a derecha Descendente a medida que x se mueve de izquierda a derecha Pendiente Ejemplo Positiva Negativa Horizontal 0 Vertical No definida Barnett/Ziegler/Byleen College Mathematics 12e 19 Slope of a Line - Example Finding Slopes. Sketch a line through each pair of points, and find the slope of each line. A) (-3, -2), (3, 4) B) (-1, 3), (2, -3) C) (-2, -3), (3, -3) C) (-2, 4), (-2, -2) Barnett/Ziegler/Byleen College Mathematics 12e 20 Slope of a Line - Example A) B) y x Barnett/Ziegler/Byleen College Mathematics 12e y x 21 Slope of a Line - Example C) D) y y x x Slope is not defined Barnett/Ziegler/Byleen College Mathematics 12e 22 Slope-Intercept Form The equation y = mx+b is called the slope-intercept form of an equation of a line. The letter m represents the slope and b represents the y-intercept. Barnett/Ziegler/Byleen College Mathematics 12e 23 Slope-Intercept Form The constants a and m have the following interpretations: If we let x = 0, then y = b. So the graph of y = mx + b crosses the y axis at (0,b). The constant b is the y intercept. For m, if y = mx + b, then by setting x = 0 and x = 1, we conclude that (0,b) and (1, m + b) are points in the line. The slope of such line is: So m is the slope of the line given by y = mx + b. Barnett/Ziegler/Byleen College Mathematics 12e 24 Find the Slope and Intercept from the Equation of a Line Example: Find the slope and y intercept of the line whose equation is 5x – 2y = 10. Barnett/Ziegler/Byleen College Mathematics 12e 25 Find the Slope and Intercept from the Equation of a Line Example: Find the slope and y intercept of the line whose equation is 5x – 2y = 10. Solution: Solve the equation for y in terms of x. Identify the coefficient of x as the slope and the y intercept as the constant term. 5 x 2 y 10 2 y 5 x 10 5 x 10 5 y x5 2 2 2 Therefore: the slope is 5/2 and the y intercept is –5. Barnett/Ziegler/Byleen College Mathematics 12e 26 Write the Equation of the Line knowing the Slope and the Intercept Example: Find the equation of the line with slope 2/3 and intercept -2. Solution: m = 2/3 and b = -2 so, Barnett/Ziegler/Byleen College Mathematics 12e y = 2/3 x - 2 27 Point-Slope Form The point-slope form of the equation of a line is y y1 m( x x1 ) where m is the slope and (x1, y1) is a given point. It is derived from the definition of the slope of a line: y2 y1 m x2 x1 Barnett/Ziegler/Byleen College Mathematics 12e Cross-multiply and substitute the more general x for x2 28 Example Find the equation of the line through the points (–5, 7) and (4, 16). Barnett/Ziegler/Byleen College Mathematics 12e 29 Example Find the equation of the line through the points (–5, 7) and (4, 16). Solution: 16 7 9 m 1 4 (5) 9 Now use the point-slope form with m = 1 and (x1, x2) = (4, 16). (We could just as well have used (–5, 7)). y 16 1( x 4) y x 4 16 x 12 Barnett/Ziegler/Byleen College Mathematics 12e 30 Summary Equations of a Line Standard Form Ax + By = C A and B not both 0 Slope-intercept form y = mx + b Slope: m; y intercept: b Point-slope form y – y1 = m(x – x1) Slope: m; point: (x1, y1) Horizontal line y=b Slope: 0 Vertical line x=a Slope: Undefined Barnett/Ziegler/Byleen College Mathematics 12e 31 Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years: Barnett/Ziegler/Byleen College Mathematics 12e 32 Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years: Solution: When t = 0, V = 20,000 and when t = 10, V = 2,000. Thus, we have two ordered pairs (0, 20,000) and (10, 2000). We find the slope of the line using the slope formula. The y intercept is already known (when t = 0, V = 20,000, so the y intercept is 20,000). The slope is (2000 – 20,000)/(10 – 0) = –1,800. Therefore, our equation is V(t) = –1,800t + 20,000. Barnett/Ziegler/Byleen College Mathematics 12e 33 Supply and Demand In a free competitive market, the price of a product is determined by the relationship between supply and demand. The price tends to stabilize at the point of intersection of the demand and supply equations. This point of intersection is called the equilibrium point. The corresponding price is called the equilibrium price. The common value of supply and demand is called the equilibrium quantity. Barnett/Ziegler/Byleen College Mathematics 12e 34 Supply and Demand Example Use the barley market data in the following table to find: (a) A linear supply equation of the form p = mx + b (b) A linear demand equation of the form p = mx + b (c) The equilibrium point. Year Supply Mil bu Demand Mil bu Price $/bu 2002 340 270 2.22 2003 370 250 2.72 Barnett/Ziegler/Byleen College Mathematics 12e 35 Supply and Demand Example (continued) (a) To find a supply equation in the form p = mx + b, we must first find two points of the form (x, p) on the supply line. From the table, (340, 2.22) and (370, 2.72) are two such points. The slope of the line is 2.72 2.22 0.5 m 0.0167 370 340 30 Now use the point-slope form to find the equation of the line: p – p1 = m(x – x1) p – 2.22 = 0.0167(x – 340) p – 2.22 = 0.0167x – 5.678 p = 0.0167x – 3.458 Barnett/Ziegler/Byleen College Mathematics 12e Price-supply equation. 36 Supply and Demand Example (continued) (b) From the table, (270, 2.22) and (250, 2.72) are two points on the demand equation. The slope is 2.72 2.22 .5 m 0.025 250 270 20 p – p1 = m(x – x1) p – 2.22 = –0.025(x – 270) p – 2.22 = –0.025x + 6.75 p = –0.025x + 8.97 Barnett/Ziegler/Byleen College Mathematics 12e Price-demand equation 37 Supply and Demand Example (continued) (c) The equilibrium point occurred when the price p is the same, that is, when p = 0.0167x – 3.458 and p = –0.025x + 8.97 are equal. Then, 0.0167x – 3.458 = –0.0250x + 8.97 0.0167x = –0.0250x + 12.428 adding 3.458 in both sides 0.0417x = 12.428 adding 0.0250x in both sides x = 12.428/0.0417 = 298 rounded and p = 1.52 if we substitute x = 298 in any of the two equations [p = 0.0167(298) – 3.458 = 4.9766 – 3.4580] Barnett/Ziegler/Byleen College Mathematics 12e 38 Supply and Demand Example (continued) (c) If we graph the two equations on a graphing calculator, set the window as shown, then use the intersect operation, we obtain: The equilibrium point is approximately (298, 1.52). This means that the common value of supply and demand is 298 million bushels when the price is $1.52. Barnett/Ziegler/Byleen College Mathematics 12e 39 Chapter 1 Linear Equations and Graphs Section 2 Graphs and Lines END Last Update: Juanuary 2013