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College Algebra
Self-assessment B – answers
Mathematical Relations
Name: _____________________________________
Problem
1. Name the ordinate of the ordered pair (-5, -7)
Answer
1. -7
2. In which quadrant is the point (3, -8) located>
2. Fourth Quadrant
3. Give the range of the relation below
{(-2, 5 (-5, -4), (3, 0), (2, -1)}
4. Find the x – intercept of the relation 7x – 3y = -42
3. Range = {5, -4, 0, -1}
4. When y = 0, x = -6
Let y = 0, 7x = -42, x = -6
5. Find the point symmetric to (7, 4) with respect to
the y-axis
6. Find the point symmetric to (-2, -5) with respect to
the origin
7. Find the slope of the line 7x -2y = 31
A 7 7
m


B
2 2
8. Find the slope of a line perpendicular to
the line 7x – 2y = 31
5. (-7, 4)
6 (2, 5)
7.
m
7
2
8.
2
7
9. Given that the point (3, -5) belongs to the line
9. (5, 2) or other possible
7x – 2y = 31, find another friendly point. Use the slope answers.
(3+2, -5+7) = (5, 2)
10. Find the center of the circle
10. center: __(3, -7)____
m
x  32  (y  7) 2  49
11. Find the range of the circle
x  3
2
2
 ( y  7)  49
11. center = (3, -7)
radius = 7
Range = [-7-7, -7+7] = [-14, 0]
12. Write the equation of a vertical line through the
point (-2, 7)
13. Write the equation of the circle with radius 1 and
center at (-2, 0)
12.
x  22  (y  0) 2  12
x  22  y 2  1
14. Find the midpoint of the line segment connecting
the point (-8, 5) and (-4, -1)
 x1  x 2 y1  y 2    8  4 5  1 
,
,


  ( 6, 2)
2
2
2 

  2
15. Find an equation of the line with slope m = -3 and
y-intercept b = 17.
14.
x = -2
13
(-6, 2)
15.
y = -3x + 17
Use the slope-intercept form of a line: y = mx + b
y = mx + b, Y = -3x + 17
16. Find the slope m of the line through the points
(-3, 7) and (1, 4).
16.
y  y1 4  7  3
m 2


x 2  x1 1  3
4
m
17. A vertical line has:
a) 0 slope b) positive slope c) negative slope
d) undefined slope
18. Use the point-slope form of a line:
y  y1  m(x  x1 ) to find the equation of the line that
contains the points (-3, 7) and (1, 4)
y  y1 4  7  3
m 2


x 2  x1 1  3
4
y  y1  m( x  x1 )
17.
3
4
d) undefined slope
18.
3x + 4y = 19
3
( x  3)  4( y  7 )  3( x  3)
4
4y  28  3x  9
y7
3x  4y  19
19. Write the equation of the transpose of the relation
x  32  (y  7) 2  49
20. Write the equation of the line that is symmetric to
the line y = 7x + 5 with respect to the y-axis.
Change x to –x and obtain y = 7(-x) + 5
19
x  72  (y  3) 2  49 .
20. y = -7x + 5
Problem 1: Consider the equation 5x + 4y = 31.
a) Does the point ( 3, 4) belongs to the line? Yes or no?
5(3)+4(4) = 15 + 16 = 31, yes
b) Find the slope of the line.
A 5

B
4
c) Find three points of the line
a) ___yes____
b) m 
5
4
m
(3, 4), (3+4, 4-5) =(7, -1), (7+4, -1-5) = (11, -6)
d) Draw the line in the Cartesian System provided below.
c) _( 3, 4) (7, -1), (11, 06) _
y
-1
1 2 3 4
7
11
x
e) On
the graph above, shade the graph of the inequality 5x +4y >31
Problem 2. A circle contains the points (2, -3), (9, 4) and
(-21, 20)
Replacing the x values and y values in the general form of a circle,
x 2  y 2  Dx  Ey  F  0 results in the following system:
 2D  3E  F  13

 9D  4E  F  97
  21D  20E  F  841

a)
Using the Row Reduced Echelon Form (rref in the TI calculator) to solve the system for
D, E and F, we obtain
D = 12, E= -24, F = -109
b) Write the general form of the circle by replacing D, E, F by the values found in part a)
Answer: __ x 2  y 2  12x  24y  109  0 ___________________________________
c) Completing the square: (Fill in the blanks)

 

x 2  y 2  12x  24y  109  0  x 2  12x  y 2  24y  _ 109 _____
 x 2  12x  _ 36 ____  y 2  24y  144  109  36  144
x  _ 6 __ 2  (y  12) 2  289
d) The radius of the circle is _17_________
e) The center of the circle is ___(-6, 12)_________