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Exercises 21.2, page 577

1. $x + 2 y - 2 = 0$ $x + 2 y - 2 = 0$
3. $2, (0, \frac{5}{2})$ $2, (0, \frac{5}{2})$
5. $4 x - y + 20 = 0$ $4 x - y + 20 = 0$
7. $7 x - 2 y - 24 = 0$ $7 x - 2 y - 24 = 0$
9. $x - y + 19 = 0$ $x - y + 19 = 0$
11. $y = - 2.7$ $y = - 2.7$
13. $x = - 3$ $x = - 3$
15. $x - 3 y - 7 = 0$ $x - 3 y - 7 = 0$
17. $x + y - 7 = 0$ $x + y - 7 = 0$
19. $3 x + y - 18 = 0$ $3 x + y - 18 = 0$
21. $y = 4 x - 8; m = 4, (0, - 8)$ $y = 4 x - 8; m = 4, (0, - 8)$
23. $y = - \frac{3}{5} x + 2; m = - \frac{3}{5}, (0, 2)$ $y = - \frac{3}{5} x + 2; m = - \frac{3}{5}, (0, 2)$
25. $y = \frac{3}{2} x - \frac{1}{2}; m = \frac{3}{2}, (0, - \frac{1}{2})$ $y = \frac{3}{2} x - \frac{1}{2}; m = \frac{3}{2}, (0, - \frac{1}{2})$
27. $y = 3.5 x + 0.5; m = 3.5, (0, 0.5)$ $y = 3.5 x + 0.5; m = 3.5, (0, 0.5)$
29. Parallel
31. Perpendicular
33. Neither
35. Perpendicular
37. $- 2$ $- 2$
39. The slope of the first line is 3. A line perpendicular to it has a slope of $- \frac{1}{3} .$ $- \frac{1}{3} .$ The slope of the second line is $- \frac{k}{3},$ $- \frac{k}{3},$ so $k = 1 .$ $k = 1 .$
41. 5
43. $m_{1} = - \frac{4}{5}, m_{2} = \frac{2}{3}, m_{3} = \frac{2}{3}; m_{4} = - \frac{4}{5}; m_{1} = m_{4}, m_{2} = m_{3}$ $m_{1} = - \frac{4}{5}, m_{2} = \frac{2}{3}, m_{3} = \frac{2}{3}; m_{4} = - \frac{4}{5}; m_{1} = m_{4}, m_{2} = m_{3}$
45. $m_{1} = - \frac{a}{b}, m_{2} = - \frac{a}{b}, m_{1} = m_{2}; m_{3} = - \frac{a}{b}, m_{4} = \frac{b}{a}, m_{3} = 1 / m_{4}$ $m_{1} = - \frac{a}{b}, m_{2} = - \frac{a}{b}, m_{1} = m_{2}; m_{3} = - \frac{a}{b}, m_{4} = \frac{b}{a}, m_{3} = 1 / m_{4}$
47. All parallel, $m = 2$ $m = 2$
49. $y - m x - b = 0$ $y - m x - b = 0$
51. $F = \frac{9}{4} R + 32$ $F = \frac{9}{4} R + 32$
53. $v = 0.607 T + 331$ $v = 0.607 T + 331$
55. $T = \frac{4}{3} x + 3$ $T = \frac{4}{3} x + 3$ ; If the distance increases by 1 cm, the temperature increases by $\frac{4}{3} ° C$ $\frac{4}{3} ° C$ .
57. $w = 30 - 2 t$ $w = 30 - 2 t$
59. $y = 10^{- 5} (2.4 - 5.6 x)$ $y = 10^{- 5} (2.4 - 5.6 x)$
61. $n = \frac{7}{6} t + 10;$ $n = \frac{7}{6} t + 10;$ at 6:30, $n = 10;$ $n = 10;$ at 8.30, $n = 150$ $n = 150$ ; For each minute that passes, $\frac{7}{6}$ $\frac{7}{6}$ more cars pass that point.
63. $p = 9.789 d + 101.286;$ $p = 9.789 d + 101.286;$ 444 kPa; interpolation
65.
67.