1.3, EXERCISES page 29 1. The coordinates of A are (3,3) and it is located in Quadrant I. 2. The coordinates of B are (-5,2) and it is located in Quadrant II. 3. The coordinates of C are (2,-2) and it is located in Quadrant IV. 4. The coordinates of D are (-2,5) and it is located in Quadrant II. 5. The coordinates of E are (-4,-6) and it is located in Quadrant III. 6. The coordinates of F are (8,-2) and it is located in Quadrant IV. 7. A 8. (-5,4) 9. E, F, and G. 10. E 11. F 12. D
For Exercises 13-20, refer to the following figure.
21. Using the distance formula, we find that (4 - 1) 2 + (7 - 3) 2 = 32 + 4 2 = 25 = 5 .
22. Using the distance formula, we find that (4 - 1) 2 + (4 - 0) 2 = 32 + 4 2 = 25 = 5 . 23. Using the distance formula, we find that (4 - ( -1)) 2 + (9 - 3) 2 = 52 + 62 = 25 + 36 = 61 . 24. Using the distance formula, we find that
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(10 - ( -2)) 2 + (6 - 1) 2 = 12 2 + 52 = 144 + 25 = 169 = 13. 25. The coordinates of the points have the form (x, 6). Since the points are 10 units away from the origin, we have (x 0)2 + (6 0)2 = 102 x2 = 64, or x = 8. Therefore, the required points are ( 8,6) and (8,6). 26. The coordinates of the points have the form (3, y). Since the points are 5 units away from the origin, we have (3 0)2 + (y 0)2 = 52, y2 = 16, or x = 4. Therefore, the required points are (3,4) and (3,4). 27. The points are shown in the diagram that follows.
To show that the four sides are equal, we compute the following: d ( A, B) = ( -3 - 3) 2 + (7 - 4) 2 = ( -6) 2 + 32 = 45 d ( B, C ) = [( -6 - ( -3)]2 + (1 - 7) 2 = ( -3) 2 + ( -6) 2 = 45 d (C , D) = [0 - ( -6)]2 + [( -2) - 1]2 = (6) 2 + ( -3) 2 = 45 d ( A, D) = (0 - 3) 2 + ( -2 - 4) 2 = (3) 2 + ( -6) 2 = 45 . Next, to show that ABC is a right triangle, we show that it satisfies the Pythagorean Theorem. Thus, d ( A, C ) = ( -6 - 3) 2 + (1 - 4) 2 = ( -9) 2 + ( -3) 2 = 90 = 3 10 and [d ( A, B)]2 + [d ( B, C )]2 = 90 = [d ( A, C )]2 . Similarly, d ( B , D) = 90 = 3 10 ,
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so BAD is a right triangle as well. It follows that B and D are right angles, and we conclude that ADCB is a square. 28. The triangle is shown in the figure that follows.
To prove that ABC is a right triangle, we show that [d ( A, C )]2 = [d ( A, B)]2 + [d ( B, C )]2 and the result will then follow from the Pythagorean Theorem. Now, [d ( A, C )]2 = ( -5 - 5) 2 + [2 - ( -2)]2 = 100 + 16 = 116 . Next, we find [d ( A, B )]2 + [d ( B, C )]2 = [ -2 - ( -5)]2 + (5 - 2) 2 + [5 - ( -2)]2 + ( -2 - 5) 2 = 9 + 9 + 49 + 49 = 116, and the result follows. 29. The equation of the circle with radius 5 and center (2,-3) is given by ( x - 2) 2 + [ y - (-3)]2 = 52 or ( x - 2) 2 + ( y + 3) 2 = 25. 30. The equation of the circle with radius 3 and center (-2,-4) is given by [ x - (-2)]2 + ( y + 4) 2 = 9 or ( x + 2) 2 + ( y + 4) 2 = 9 31. The equation of the circle with radius 5 and center (0, 0) is given by ( x - 0) 2 + ( y - 0) 2 = 52 or x 2 + y 2 = 25 32. The distance between the center of the circle and the point (2,3) on the circumference of the circle is given by
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d = (3 - 0) 2 + (2 - 0) 2 = 13 . Therefore r = 13 and the equation of the circle centered at the origin and passing through (2,3) is x 2 + y 2 = 13. 33. The distance between the points (5,2) and (2,-3) is given by d = (5 - 2) 2 + (2 - (-3)) 2 = 32 + 52 = 34. Therefore r = 34 and the equation of the circle passing through (5,2) and (2,-3) is ( x - 2) 2 + [ y - (-3)]2 = 34 , or ( x - 2) 2 + ( y + 3) 2 = 34. 34. The equation of the circle with center (- a, a) and radius 2a is given by [ x - (- a)]2 + ( y - a) 2 = (2a ) 2 , or ( x + a ) 2 + ( y - a ) 2 = 4a 2 .
35. Referring to the diagram on page 30 of the text, we see that the distance from A to B is given by d ( A, B) = 4002 + 3002 = 250,000 = 500. The distance from B to C is given by d ( B , C ) = ( -800 - 400) 2 + (800 - 300) 2 = ( -1200) 2 + (500) 2 = 1,690,000 = 1300. The distance C from to D is given by d (C , D) = [ -800 - ( -800)]2 + (800 - 0) 2 = 0 + 8002 = 800 . The distance from D to A is given by d ( D, A) = [( -800) - 0]2 + (0 - 0) = 640000 = 800. Therefore, the total distance covered on the tour, is d ( A, B) + d ( B, C ) + d (C , D) + d ( D, A) = 500 + 1300 + 800 + 800 = 3400, or 3400 miles.
36. Suppose that the furniture store is located at the origin O so that your house is located at A(20,-14). as shown in the figure that follows.
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Since d (O, A) = 202 + ( -14) 2 = 596 = 24.4 your house is located within a 25-mile radius of the store, you will not incur a delivery charge. 37. Referring to the following diagram,
we see that the distance he would cover if he took Route (1) is given by d ( A, B) + d ( B, D) = 4002 + 3002 + (1300 - 400) 2 + (1500 - 300) 2 = 250,000 + 2,250,000 = 500 + 1500 = 2000, or 2000 miles. On the other hand, the distance he would cover if he took Route (2) is given by d ( A, C ) + d (C , D) = 8002 + 15002 + (1300 - 800) 2 = 2,890,000 + 250,000 = 1700 + 500 = 2200, or 2200 miles. Comparing these results, we see that he should take Route (1). 38. Calculations to determine the cost of shipping by freight train: (0.11)(2000)(100) = 22,000 or $22,000.
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Calculations to determine the cost of shipping by truck:
(0.105)(2200)(100) = 23,100, or $23,100. Comparing these results, we see that the automobiles should be shipped by freight train. Net Savings: 23,100 22,000 = 1100, or $1100. 39. Calculations to determine VHF requirements: d = 252 + 352 = 625 + 1225 = 1850 43.01. Models B through D satisfy this requirement.
Calculations to determine UHF requirements:
d = 202 + 32 2 = 400 + 1024 = 1424 = 37.74 Models C through D satisfy this requirement. Therefore, Model C will allow him to receive both channels at the least cost. 40. Length of cable required on land: d ( S , Q) = 10, 000 - x
Length of cable required under water:
d (Q, M ) = ( x 2 - 0) + (0 - 3000) 2 = x 2 + 30002
Cost of laying cable:
1.5(10, 000 - x) + 2.5 x 2 + 30002 If x = 2500, then the total cost is given by 1.5(10000 - 2500) + 2.5 25002 + 30002 21, 012.80 , or $21,012.80. If x = 3000, then 1.5(10000 - 3000) + 2.5 30002 + 30002 21,106.60 , or $21,106.60. 41. a. Let the position of ship A and ship B after t hours be A(0, y) and B(0, y), respectively. Then x = 30t and y = 20t. Therefore, the distance between the two ships is D = (30t ) 2 + (20t ) 2 = 900t 2 + 400t 2 = 10 13t. b. The required distance is obtained by letting t = 2 giving D = 10 13(2) or approximately 72.11 miles. 42. a. Let the positions of ship A and ship B be (0, y) and (x, 0), respectively. Then y = 25(t + 1 ) and x = 20t. The distance D between the two ships is 2
D = ( x - 0) 2 + (0 - y ) 2 = x 2 + y 2 = 400t 2 + 625(t + 1 ) 2 2
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b. The distance between the ships 2 hours after ship A has left port is obtained by letting t = 3 in Equation (1) obtaining 2
D = 400( 3 ) 2 + 625( 3 + 1 ) 2 = 3400 , or approximately 58.31 miles. 2 2 2
43. True. Plot the points.
44. True. Plot the points.
45. False. The distance between P (a, b) and P3 (kc, kd ) is 1
d = (kc - a) 2 + (kd - b) 2 k D= k
2
(c - a ) 2 + (d - b) 2 = k 2 (c - a ) 2 + k 2 (d - b) 2 = [k (c - a)]2 + [k (d - b)]2
2 2 2 2
a2 < a 2 if k > 1. So the radius of the circle with 46. True. kx + ky = a ; x + y = k equation (1) is a circle of radius smaller than a if k > 1 (and centered at the origin). Therefore, it lies inside the circle of radius a with equation x 2 + y 2 = a 2 . 47. Referring to the figure in the text, we see that the distance between the two points is given by the length of the hypotenuse of the right triangle. That is, d = ( x2 - x1 ) 2 + ( y2 - y1 ) 2 48. ( x - h) 2 + ( y - k ) 2 = r 2 ; x 2 - 2 xh + h 2 + y 2 - 2ky + k 2 = r 2 . This has the form x 2 + y 2 + Cx + Dy + E = 0 where c = -2h, D = -2k , and E = h 2 + k 2 - r 2 .
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