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# Algebra: Graphing, Distance between Points and Equations of Line

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Page 187

1. a) 12, b) 14, c) 16, d) 18 (Check for all four of our symmetries SY, SX, SO, SI; consult in WEEK7 NOTES, in COURSE CONTENT. Practice graphing these using the downloaded graphing utility GraphCalc.)

2. a) 20, b) 24
3. 28
4. 32

P. 203

5. a) 22, b) 24
6. a) 26, b) 30
7. a) 36, b) 42

P. 219

8. a) 8, b) 18
9. a) 36, b) 38
10. a) 46, b) 54, c) 58, d) 64, e) 78

##### Solution Summary

Graphing problems involving linear equations are answered.

##### Solution Preview

P187.

12.)
y = 1/2 x + 1 has no symmetry about either x-axis or y-axis or origin.

14.)
y = 2x is unsymmetric about either x-axis or y-axis or origin. --Answer

16.)
|y| = -x
for y> 0 => y = -x
for y < 0 => -y = -x => y = x

18.)
y = -x => symmetric about origin.

20.)
Distance between points (-6,4) and (2,-1)
s = sqrt ((-6 - 2)^2 + (4 - (-1))^2)
s = sqrt((-8)^2 + (5)^2) = sqrt(64 + 25) = sqrt(89) --Answer

24.)
C =(x1,y1) =(0,0); r = 6
eqn of circle,
(x-x1)^2 + (y-y1)^2 = r^2
=> (x-0)^2 + (y-0)^2 = 6^2
=> x^2 + y^2 = 36 --Answer

28.)
C(-1,-3) = (x1, y1) ; r = sqrt(5)
Eqn of circle,
(x - x1)^2 + (y - y1)^2 = r^2
=> (x - (-1))^2 + (y - (-3))^2 = (sqrt(5))^2
=> (x + 1)^2 + (y + 3)^2 = 5
=> x^2 + 2x + 1 + y^2 + 6y + 9 = 5
=> x^2 + Y^2 + 2x + 6y + 5 = 0 --Answer

32.)

P203.)

22.)
slope (m)= -1, y-intercept (c)= 7
Eqn of straight line:
y = mx + c
=> y = -1*x + 7
=> x + y = 7 ...

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