1. 2.1 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

2. 2.1 - 2 2.1 Rectangular Coordinates and Graphs Ordered Pairs The Rectangular Coordinate System The Distance Formula The Midpoint formula Graphing Equations

3. 2.1 - 3 Ordered Pairs An ordered pair consists of two components, written inside parentheses, in which the order of the components is important.

4. 2.1 - 4 Example 1 WRITING ORDERED PAIRS Type of Entertainment Amount Spent VHS rentals/ sales $23 DVD rentals/sales $71 CDs$20 Theme parks$43 Sports tickets$50 Movie tickets$30 Use the table to write ordered pairs to express the relationship between each type of entertainment and the amount spent on it.

5. 2.1 - 5 Example 1 WRITING ORDERED PAIRS Type of Entertainment Amount Spent VHS rentals/ sales $23 DVD rental/sales$71 CDs$20 Theme parks$43 Sports tickets$50 Movie tickets$30 Solution: a. DVD rental sales Use the data in the second row: (DVD rental sales, $71).

6. 2.1 - 6 Example 1 WRITING ORDERED PAIRS Type of Entertainment Amount Spent VHS rentals/ sales $23 DVD rentals/sales $71 CDs$20 Theme parks$43 Sports tickets$50 Movie tickets$30 Solution: b. Movie tickets Use the data in the last row: (Movie tickets, $30).

7. 2.1 - 7 The Rectangular Coordinate System Quadrant I Quadrant II Quadrant III Quadrant IV 0 x-axis y-axis a b P(a, b)

8. 2.1 - 8 The Distance Formula Using the coordinates of ordered pairs, we can find the distance between any two points in a plane. P(– 4, 3) Q(8, 3) R(8, – 2) Horizontal side of the triangle has length Definition of distance. x y

9. 2.1 - 9 The Distance Formula P(– 4, 3) Q(8, 3) R(8, – 2) Vertical side of the triangle has length… x y

10. 2.1 - 10 The Distance Formula P(– 4, 3) Q(8, 3) R(8, – 2) By the Pythagorean theorem, the length of the remaining side of the triangle is x y

11. 2.1 - 11 The Distance Formula P(– 4, 3) Q(8, 3) R(8, – 2) So the distance between (– 4, 3) and (8, – 4) is 13. x y

12. 2.1 - 12 The Distance Formula To obtain a general formula for the distance between two points in a coordinate plane, let P(x 1, y 1 ) and R(x 2, y 2 ) be any two distinct points in a plane. Complete a triangle by locating point Q with coordinates (x 2, y 1 ). The Pythagorean theorem gives the distance between P and R as P(x 1, y 1 ) Q(x 2, y 1 ) R(x 2, y 2 ) x y

13. 2.1 - 13 The Distance Formula Note Suppose that P(x 1, y 1 ) and R(x 2, y 2 ) are two points in a coordinate plane. Then the distance between P and R, written d(P, R) is given by

14. 2.1 - 14 Example 2 USING THE DISTANCE FORMULA Solution: Find the distance between P(– 8, 4) and Q(3, – 2.) Be careful when subtracting a negative number.

15. 2.1 - 15 Using the Distance Formula If the sides a, b, and c of a triangle satisfy a 2 + b 2 = c 2, then the triangle is a right triangle with legs having lengths a and b and hypotenuse having length c?

16. 2.1 - 16 Example 3 DETERMINING WHETHER THREE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE Are points M(– 2, 5), N(12, 3), and Q(10, – 11) the vertices of a right triangle. Solution This triangle is a right triangle if the square of the length of the longest side equals the sum of the squares of the lengths of the other two sides.

17. 2.1 - 17 Example 3 DETERMINING WHETHER THREE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE M(– 2, 5) N(12, 3) Q(10, – 11) x y

18. 2.1 - 18 Example 3 DETERMINING WHETHER THREE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE M(– 2, 5) N(12, 3) Q(10, – 11) x y

19. 2.1 - 19 Example 3 DETERMINING WHETHER THREE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE M(– 2, 5) N(12, 3) Q(10, – 11) x y

20. 2.1 - 20 Example 3 DETERMINING WHETHER THREE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE M(– 2, 5) N(12, 3) R(10, – 11) x y

21. 2.1 - 21 Colinear We can tell if three points are collinear, that is, if they lie on a straight line, using a similar procedure Three points are collinear if the sum of the distances between two pairs of points is equal to the distance between the remaining pair of points.

22. 2.1 - 22 Example 4 DETERMINING WHETHER THREE POINTS ARE COLLINEAR Are the points (– 1, 5), (2,– 4), and (4, – 10) collinear? Solution The distance between (– 1, 5) and (2,– 4) is

23. 2.1 - 23 Example 4 DETERMINING WHETHER THREE POINTS ARE COLLINEAR Are the points (– 1, 5), (2,– 4), and (4, – 10) collinear? Solution The distance between (2,– 4) and (4,– 10) is

24. 2.1 - 24 Example 4 DETERMINING WHETHER THREE POINTS ARE COLLINEAR Are the points (– 1, 5), (2,– 4), and (4, – 10) collinear? Solution The distance between the remaining pair of points (– 1, 5) and (4,– 10) is

25. 2.1 - 25 Midpoint Formula The midpoint of the line segment with endpoints (x 1, y 1 ) and (x 2, y 2 ) has coordinates

26. 2.1 - 26 Example 5 USING THE MIDPOINT FORMULA Use the midpoint formula to determine the following. Solution a. Find the coordinates of M of the segment with endpoints (8, – 4) and (– 6,1). Substitute in the midpoint formula.

27. 2.1 - 27 Example 5 USING THE MIDPOINT FORMULA Use the midpoint formula to determine the following. b. Find the coordinates of the other endpoint B of a segment with one endpoint A(– 6, 12) and midpoint M(8, – 2). Let (x, y) represent the coordinates of B. Use the midpoint formula twice. Solution

28. 2.1 - 28 Example 5 USING THE MIDPOINT FORMULA Let (x, y) represent the coordinates of B. Use the midpoint formula twice. x-value of Ax-value of M Substitute carefully. x-value of Ay-value of M The coordinates of endpoint B are (22, – 16).

29. 2.1 - 29 Example 6 APPLYING THE MIDPOINT FORMULA TO DATA The graph indicates the increase of McDonald’s restaurants worldwide from 20,000 in 1996 to 31,000 in 2006. Use the midpoint formula and the two given endpoints to estimate the number of restaurants in 2001, and compare it to the actual (rounded) figure of 30,000

30. 2.1 - 30 Example 6 APPLYING THE MIDPOINT FORMULA TO DATA 1996 2006 20 25 30 35 Number of McDonald’s Restaurants Worldwide (in thousands) Year

31. 2.1 - 31 Example 6 APPLYING THE MIDPOINT FORMULA TO DATA Solution: 2001 is halfway between 1996 and 2006. Find the coordinates of the midpoint of the segment that has endpoints (1996, 20) and (2006, 31). Our estimate is 25,500, which is well below the actual figure of 30,000.

32. 2.1 - 32 Example 7 FINDING ORDERED PAIRS THAT ARE SOLUTIONS OF EQUATIONS For the equation, find at least three ordered pairs that are solutions Solution: Choose any real number for x or y and substitute in the equation to get the corresponding value of the other variable. a.

33. 2.1 - 33 Example 7 FINDING ORDERED PAIRS THAT ARE SOLUTIONS OF EQUATIONS Solution: a. Let x = – 2. Multiply. Let y = 3. Add 1. Divide by 4. This gives the ordered pairs (– 2, – 9) and (1, 3). Find one more ordered pair that works!

34. 2.1 - 34 Example 7 FINDING ORDERED PAIRS THAT ARE SOLUTIONS OF EQUATIONS Solution b. Given equation Let x = 1 Square both sides One ordered pair is (1, 2). Find two more ordered pair!

35. 2.1 - 35 Example 7 FINDING ORDERED PAIRS THAT ARE SOLUTIONS OF EQUATIONS Solution: c. A table provides an organization method for determining ordered pairs. Five ordered pairs are listed as solutions for this equation. xy – 2– 2 0 – 1– 1– 3– 3 0 – 4– 4 1 – 3– 3 20

36. 2.1 - 36 Graphing an Equation by Point Plotting Step 1 Find the intercepts. Step 2 Find as many additional ordered pairs as needed. Step 3 Plot the ordered pairs from Steps 1 and 2. Step 4 Connect the points from Step 3 with a smooth line or curve.

37. 2.1 - 37 Step 1 Let y = 0 to find the x-intercept, and let x = 0 to find the y-intercept. Example 8 GRAPHING EQUATIONS a. Graph the equation Solution:

38. 2.1 - 38 Step 2 We find some other ordered pairs. Example 8 GRAPHING EQUATIONS a. Graph the equation Solution: Let x = – 2. Multiply. Let y = 3. Add 1. Divide by 4. This gives the ordered pairs (– 2, – 9) and (1, 3).

39. 2.1 - 39 Step 3 Plot the four ordered pairs from Steps 1 and 2. Example 8 GRAPHING EQUATIONS a. Graph the equation Solution: Step 4 Connect the points with a straight line. x y

40. 2.1 - 40 Example 8 GRAPHING EQUATIONS b. Graph the equation Solution: Plot the ordered pairs found in example 7b, and develop a fourth point to confirm the direction the curve will take as x increases. x y

41. 2.1 - 41 Example 8 GRAPHING EQUATIONS c. Graph the equation Solution: xy – 2– 2 0 – 1– 1– 3– 3 0 – 4– 4 1 – 3– 3 20 This curve is called a parabola. x y