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The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models

1. The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models Intersection of Straight Lines The Method of Least Squares. Straight Lines and Linear Functions. 1.1. The Cartesian Coordinate System. The Cartesian Coordinate System.

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The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models

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  1. 1 • The Cartesian Coordinate System • Straight Lines • Linear Functions and Mathematical Models • Intersection of Straight Lines • The Method of Least Squares Straight Lines and Linear Functions

  2. 1.1 The Cartesian Coordinate System

  3. The Cartesian Coordinate System • We can represent real numbers geometrically by points on a real number, or coordinate, line: • This line includes allreal numbers. • Exactly one point on the line is associated with each real number, and vice-versa (one dimensional space). Origin Negative Direction Positive Direction p

  4. The Cartesian Coordinate System • The Cartesian coordinate system extends this concept to a plane (two dimensional space) by adding a vertical axis. 4 3 2 1 – 1 –2 –3 –4

  5. The Cartesian Coordinate System • The horizontal line is called the x-axis, and the vertical line is called the y-axis. y 4 3 2 1 – 1 –2 –3 –4 x

  6. The Cartesian Coordinate System • The point where these two lines intersect is called the origin. y 4 3 2 1 – 1 –2 –3 –4 Origin x

  7. The Cartesian Coordinate System • In thex-axis, positive numbers are to the right and negative numbers are to the left of the origin. y 4 3 2 1 – 1 –2 –3 –4 Negative Direction Positive Direction x

  8. The Cartesian Coordinate System • In they-axis, positive numbers are above and negativenumbers are below the origin. y 4 3 2 1 – 1 –2 –3 –4 Positive Direction x Negative Direction

  9. The Cartesian Coordinate System • A point in the plane can now be represented uniquely in this coordinate system by an ordered pair of numbers(x, y). y (–2, 4) 4 3 2 1 – 1 –2 –3 –4 (4, 3) x (3, –1) (–1, –2)

  10. The Cartesian Coordinate System • The axes divide the plane into four quadrants as shown below. y 4 3 2 1 – 1 –2 –3 –4 Quadrant II (–, +) Quadrant I (+, +) x Quadrant III (–, –) Quadrant IV (+, –)

  11. The Distance Formula • The distancebetween any two points in the plane can be expressed in terms of the coordinates of the points. Distance formula • The distance d between two points P1(x1, y1) and P2(x2, y2) in the plane is given by

  12. Examples • Find the distance between the points (–4, 3) and (2, 6). Solution • Let P1(–4, 3) and P2(2, 6) be points in the plane. • We have x1 =–4 y1 =3 x2 =2 y2 =6 • Using the distance formula, we have Example 1, page 4

  13. Examples • Let P(x, y) denote a point lying on the circle with radiusr and centerC(h, k). Find a relationship between x and y. Solution • By the definition of a circle, the distance between P(x, y) and C(h, k) is r. • With the distance formula we get • Squaring both sides gives y P(x, y) C(h, k) r k x h Example 3, page 4

  14. Equation of a Circle • An equation of a circle with centerC(h, k) and radiusr is given by

  15. Examples • Find an equation of the circle withradius2 andcenter(–1, 3). Solution • We use the circle formula with r = 2, h =–1, andk = 3: y (–1, 3) 3 2 x –1 Example 4, page 5

  16. Examples • Find an equation of the circle withradius3 andcenterlocated at the origin. Solution • We use the circle formula with r = 3, h =0, andk = 0: y 3 x Example 4, page 5

  17. 1.2 Straight Lines

  18. Slope of a Vertical Line • Let L denote the unique straight line that passes through the two distinct points (x1, y1) and (x2, y2). • If x1=x2, then L is a vertical line, and the slope is undefined. y L (x1, y1) (x2, y2) x

  19. Slope of a Nonvertical Line • If (x1, y1) and (x2, y2) are two distinct points on a nonvertical lineL, then the slopem of L is given by y L (x2, y2) y2 – y1 = y (x1, y1) x2 – x1 = x x

  20. Slope of a Nonvertical Line • If m > 0, the line slants upwardfromleft to right. y L m = 1 y = 1 x = 1 x

  21. Slope of a Nonvertical Line • If m > 0, the line slants upwardfromleft to right. y L m = 2 y = 2 x = 1 x

  22. Slope of a Nonvertical Line • If m < 0, the line slants downwardfromleft to right. y m = –1 x = 1 y = –1 x L

  23. Slope of a Nonvertical Line • If m < 0, the line slants downwardfromleft to right. y m = –2 x = 1 y = –2 x L

  24. Examples • Sketch the straight line that passes through the point (2, 5) and has slope –4/3. Solution • Plot the point(2, 5). • A slope of –4/3 means that if xincreases by 3, ydecreases by 4. • Plot the resulting point(5, 1). • Draw a line through the two points. y 6 5 4 3 2 1 x = 3 (2, 5) y = –4 (5, 1) x 1 2 3 4 5 6 L

  25. Examples • Find the slopem of the line that goes through the points(–1, 1) and (5, 3). Solution • Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3). • With x1 =–1, y1 = 1, x2 =5, y2 =3, we find Example 2, page 11

  26. Examples • Find the slopem of the line that goes through the points(–2, 5) and (3, 5). Solution • Choose (x1, y1) to be (–2, 5) and (x2, y2) to be (3, 5). • With x1 =–2, y1 = 5, x2 =3, y2 =5, we find Example 3, page 11

  27. Examples • Find the slopem of the line that goes through the points(–2, 5) and (3, 5). Solution • The slope of a horizontal line is zero: y 6 4 3 2 1 (3, 5) (–2, 5) L m = 0 x –2 –1 1 2 3 4 Example 3, page 11

  28. Parallel Lines • Two distinct lines are parallel if and only if their slopes are equal or their slopes are undefined.

  29. Example • Let L1 be a line that passes through the points (–2, 9) and (1, 3), and let L2 be the line that passes through the points (–4, 10) and (3, –4). • Determine whether L1 and L2 are parallel. Solution • The slopem1 of L1 is given by • The slopem2 of L2 is given by • Since m1=m2, the lines L1 and L2 are in fact parallel. Example 4, page 12

  30. Equations of Lines • Let L be a straight lineparallel to the y-axis. • Then Lcrosses the x-axis at some point(a, 0) , with the x-coordinate given by x = a, where a is a real number. • Any other point on L has the form (a, ), where is an appropriate number. • The vertical lineL can therefore be described as x = a y L (a, ) (a, 0) x

  31. Equations of Lines • Let L be a nonvertical line with a slope m. • Let (x1, y1) be a fixed point lying on L, and let (x, y) be a variable point on L distinct from (x1, y1). • Using the slope formula by letting (x, y) =(x2, y2), we get • Multiplying both sides by x – x1 we get

  32. Point-Slope Form • An equation of the line that has slope m and passes through point (x1, y1) is given by

  33. Examples • Find an equation of the line that passes through the point (1, 3) and has slope 2. Solution • Use the point-slope form • Substituting for point(1, 3) and slopem = 2, we obtain • Simplifying we get Example 5, page 13

  34. Examples • Find an equation of the line that passes through the points (–3, 2) and (4, –1). Solution • The slope is given by • Substituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtain Example 6, page 14

  35. Perpendicular Lines • If L1 and L2 are two distinct nonvertical lines that have slopes m1 and m2, respectively, then L1 is perpendicular to L2 (written L1 ┴L2) if and only if

  36. Example • Find the equation of the line L1 that passes through the point (3, 1) and is perpendicular to the line L2 described by Solution • L2 is described in point-slope form, so its slope is m2 = 2. • Since the lines are perpendicular, the slope of L1 must be m1 = –1/2 • Using the point-slope form of the equation for L1 we obtain Example 7, page 14

  37. Crossing the Axis • A straight line L that is neither horizontal nor vertical cuts the x-axis and the y-axis at, say, points (a, 0) and (0, b), respectively. • The numbers a and b are called the x-intercept and y-intercept, respectively, of L. y y-intercept (0, b) x-intercept x (a, 0) L

  38. Slope-Intercept Form • An equation of the line that has slopem and intersects the y-axis at the point(0, b) is given by y = mx + b

  39. Examples • Find the equation of the line that has slope3 and y-intercept of –4. Solution • We substitute m = 3 and b = –4 into y = mx + b and get y = 3x – 4 Example 8, page 15

  40. Examples • Determine the slope and y-intercept of the line whose equation is 3x – 4y = 8. Solution • Rewrite the given equation in the slope-intercept form. • Comparing to y = mx + b, we find that m = ¾ and b = –2. • So, the slope is ¾ and the y-interceptis –2. Example 9, page 15

  41. Applied Example • Suppose an art object purchased for $50,000 is expected to appreciate in value at a constant rate of $5000 per year for the next 5 years. • Write an equation predicting the value of the art object for any given year. • What will be its value3 years after the purchase? Solution • Let x=time (in years) since the object was purchased y=value of object (in dollars) • Then, y = 50,000 when x = 0, so the y-intercept is b =50,000. • Every year the value rises by 5000, so the slope is m = 5000. • Thus, the equation must be y = 5000x + 50,000. • After 3 years the value of the object will be $65,000: y = 5000(3) + 50,000 = 65,000 Applied Example 11, page 16

  42. General Form of a Linear Equation • The equation Ax + By + C = 0 where A, B, and C are constants and A and B are not both zero, is called the general form of a linear equation in the variablesx and y.

  43. General Form of a Linear Equation • An equation of a straight line is a linear equation; conversely, every linear equation represents a straight line.

  44. Example • Sketch the straight line represented by the equation 3x – 4y – 12 = 0 Solution • Since every straight line is uniquely determined by two distinct points, we need find only two such points through which the line passes in order to sketch it. • For convenience, let’s compute the x- and y-intercepts: • Setting y= 0, we find x= 4; so the x-intercept is 4. • Setting x= 0, we find y= –3; so the y-intercept is –3. • Thus, the line goes through the points(4, 0) and(0, –3). Example 12, page 17

  45. Example • Sketch the straight line represented by the equation 3x – 4y – 12 = 0 Solution • Graph the line going through the points (4, 0) and(0, –3). y L 1 –1 –2 –3 –4 (4, 0) x 1 2 3 4 5 6 (0, –3) Example 12, page 17

  46. Equations of Straight Lines Vertical line:x = a Horizontal line:y = b Point-slope form:y – y1 = m(x – x1) Slope-intercept form:y = mx + b General Form:Ax + By + C = 0

  47. 1.3 Linear Functions and Mathematical Models Real - world Mathematical Formulate problem model Solve Test Solution of real - Solution of world Problem mathematical model Interpret

  48. Mathematical Modeling • Mathematics can be used to solve real-world problems. • Regardless of the field from which the real-world problem is drawn, the problem is analyzed using a process called mathematical modeling. • The four steps in this process are: Real-world problem Mathematical model Formulate Solve Test Solution of real- world Problem Solution of mathematical model Interpret

  49. Functions • A function f is a rule that assigns to each value of xone and only one value of y. • The value y is normally denoted by f(x), emphasizing the dependency of y on x.

  50. Example • Let x and y denote the radius and area of a circle, respectively. • From elementarygeometry we have y = px2 • This equation defines y as a function of x, since for each admissible value of x there corresponds precisely one number y = px2 giving the area of the circle. • The area function may be written as f(x) = px2 • To compute the area of a circle with a radius of 5 inches, we simply replacex in the equation by the number 5: f(5) = p(52)= 25p

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