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```					                                  Mathematical Relations
Summary 2
Mathematical relations: lines and circles

Mathematical relations. A mathematical relation is any set of ordered pairs of real
numbers.
   The domain of a relation is given by the set of all the x values of the ordered pairs in the
relation.
Domain of the relation R  {x | ( x, y)  R}
   The range of a relation is given by the set of all the y values of the ordered pairs in the
relation.
Range of the relation R  { y | ( x, y)  R}
   x-intercept are the points (a, 0) where the graph intersects the x-axis. They are also called
the real zeros of the relation. To find the values of a, let y=0 and solve for x.
   y-intercept, the points (0, b) where the graph intersects the y-axis. To find the values of b,
let x=0 and solve for y.

Some restrictions on the domain and range of relations
The domain and range of functions and relations can be restricted due to the following facts
1. The denominator of a fraction can not be zero
2. The radicand of an even index radical can not be negative
3. Negative numbers can not be used as input of a logarithmic function.
4. The real number a  0 (non-negative)
5. The absolute value of a number, | a | is non-negative.

Ways of expressing a mathematical relation.
A mathematical relation can be expressed in different ways
1. A set of ordered pairs.
2. An algebraic equation or inequality
3. A graph

Transpose of a mathematical relation:
If R is a mathematical relation, the transpose of the relation R T is the new relation that is
obtained by interchanging the x and the y values. That is,
DomR T  range R and Range R T  DomR

Example 1: Consider the relation: R={(0, 7), (2, 3), (4, -1), (5, 6), (0, -3), (3, -5), (6, 0)}
a) Give the domain                                     a) Domain = {0, 2, 4, 5, 3, 6}
b) Give the range                                      b) Range = {7, 3, -1, 6, -3, -5, 0}
c) Give the x-intercepts (when y = 0)                  c) 6
d) Give the y-intercepts (when x = 0)                  d) 7 or -3
e) Give the relative of 3                              e) -5
f) Give the transpose of the relation.
Answer: R T ={(7, 0), (3, 2), (-1, 4), (6, 5), (-3, 0), (-5, 3), (0, 6)}
Note: The domain of the transpose is the same as the range of the relation.
The range of the transpose is the same as the domain of the relation.

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Example 2: R  {(x, y ) | 2x  3y  6}
a) Give the domain                                    a) Domain = (, )
b) Give the range                                     b) Range = (, )
c) Give the x-intercepts (when y = 0)                 c) If y = 0 then x = 3
d) Give the y-intercepts (when x = 0)                 d) If x = 0 then y = 2
e) Give the relative of 3                             e) If x = 3 then y = 0
f) Give the transpose of the relation: R T  {(x, y ) | 3x  2y  6}

Example 3: R  {(x, y ) | (x  2) 2  (y  1) 2  4} }
This is a circle center (2, -1) and radios 2.
a) Give the domain                                 a) Domain = [0, 4]
b) Give the range                                  b) Range = [-3, 1]
c) Give the x-intercepts (when y = 0)              c) If y = 0 then x  2  3
d) Give the y-intercepts (when x = 0)              d) If x = 0 then y = -1
e) Give the relatives of 2                         e) If x = 2 then y -3 or y = 1
f) Give the transpose of the relation: R  {(x, y ) | (x  1) 2  (y  2) 2  4} }

Example 4. Find the intercepts of the graph of the equation x 2  y 2  7x  5y  6  0
Solution:
Let y=0 in the equation x 2  y 2  7 x  5 y  6  0 and obtain
x 2  7 x  6  0  ( x  6)(x  1)  0  x  6 or x  1
Therefore the x-intercepts are 1 and 6.
Let x=0 in the equation x 2  y 2  7 x  5 y  6  0 and obtain
y 2  5 y  6  0  ( y  3)( y  2)  0  y  3 or y  2
Therefore the y-intercepts are -2 and -3.

A circular relation.
A circle is the set of points in a plane equidistant from a fixed point called the center. The
constant distance from any point on the circle to the center is called the radius of the circle.
Equations of a circle:
1.       x 2  y 2  ax  by  c  0 is the general form of a circle.
2.       ( x  h) 2  ( y  k ) 2  r 2 is the standard form of a circle.
Domain: [h – r, h + r] and range: [k – r, k + r]
An equation of the form ( x  h) 2  ( y  k ) 2  r 2 is a circle with center located at (h, k) and
Example 4. The graph of the equation ( x  2) 2  ( y  3) 2  9 is a circle with center at
Domain = [2 – 3, 2 + 3] = [-1, 5] and Range = [-3 – 3, -3 + 3] = [-6, 0]

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Example 5. Find the center, radius, domain and range, of the circle x 2  y 2  10 x  4 y  28  0
Solution: using the technique of completing the square
             
x 2  y 2  10x  4y  28  0  x 2  10x  y 2  4y  28   
x 2  10x  25 y 2  4y  4  28  25  4  x  52  (y  2)2  1
Therefore, the circle is centered at (5, -2) and has radius r=1
Domain = [5 -1, 5 + 1] = [4, 6]      and range = [-2 – 1, -2 +1] = [-3, -1]

Example 6. Find the equation of the circle with radius r = 2 and center (0, 0)

(x  h) 2  (y  k ) 2  r 2  (x  0) 2  (y  0) 2  2 2  x 2  y 2  4

Example 7. Find the equation of the circle with radius r = 1 and center (1, -1)

(x  h) 2  (y  k ) 2  r 2  (x  1) 2  (y  1) 2  12  x  12  y  12  1

Example 8. Find the equation of the circle with radius r = 3 and center (0, -3)

(x  h) 2  (y  k ) 2  r 2  (x  0) 2  (y  3) 2  32  x 2  y  32  9

Example 9. Consider the circle 2(x  3) 2  2y 2  8
a) Change the circle to standard form:
divide by 2
2      2
2(x  3)  2y  8                (x  3) 2  y 2  4
b) Find the radius and center of the circle
radius r = 2 and center (h, k) = (3, 0)
c) Find the domain and range.
Domain: [3 – 2, 3 +2] = [1, 5] and range [0 – 2, 0 + 2] = [ -2, 2]
d) Find the x – intercept.
Let y =0
( x  3) 2  y 2  4  x  32  0  4  x  32  4  x  3  2  x  3  2
 x  3  2 or x  3  2  x  1 or x  5
The x – intercepts are x=1 and x = 5

e) Find the y– intercept.
Let x =0
( x  3) 2  y 2  4   32  y 2  4  9  y 2  4  y 2  5  y    5 not real
 x  3  2 or x  3  2  x  1 or x  5
The y – intercepts are do not exist, the circle misses the y –axis.
f) Find the equation of the transpose of the circle.
the transpose
(x  3) 2  y 2  4                    x 2  (y  3) 2  4

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Linear relations.
A linear relation is any relation of the form Ax + By = C where A, B, C are real number. The
graph of a linear relation is a straight line.
Slope of a linear relation: the slope is a number that measures the degree of steepness or
inclination of the line.
y  y1 change of y rise
slope m  2                         
x 2  x1 change of x run
Important facts of linear relations.
1. A horizontal line has m = 0 slope
2. A vertical line has no slope, that is, the slope of a vertical line is undefined
3. A line going up from left to right has positive slope
4. A line going down from left to right has negative slope
5. Parallel lines have the same slope
If L1 || L2 then m1  m2
6. If two lines are perpendicular, that is, they intersect at a right angle, then their slopes are
negative reciprocals.
1
If L1  L2 then m2         or m1 m2  1
m1
7. The line y=k represents a horizontal line with slope m=0
8. The line x=h represents a vertical line with undefined slope
A              A
9. The slope of the line Ax + By =C is always m             or m 
B            B
(The value of y changes by –A when the value of x changes by B) or (the value of y
changes by A when the value of x changes by –B)
10. The slope of a line is a generator of points.

Example 10. Find the slope, x intercept and y-intercept of each of the following lines
Line              slope                     x-intercept (let y=0)       y – intercept (let x = 0)
1) y =5           m= 0                      x-intercept - none          y-intercept = 5
2) x = -3         undefined                 x-intercept = -3            y-intercept - none
3) 3x – 2y =12          A 3 3             x-intercept = 4             y-intercept =-6
m              
B      2 2
4) 5x + 7y = 70         A 5           5 x-intercept = 14              y-intercept =10
m              
B       7      7
5) -2x + 9y = 36         A  2 2 x-intercept = -18                    y-intercept =4
m                
B        9     9
6) y = -3x +5      m= -3                                   5            y-intercept =5
x-intercept =
3
7) x = y           m= 1                     x-intercept = 0             y-intercept = 0

1               1                      x-intercept = -4              y-intercept = 2
8) y     x2         m
2               2
9) x + 2y =4             1                     x-intercept = 4               y-intercept = 2
m
2

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Example 11. Find the slope of each of the following lines.
1. A line through the point (-2, 3) and (4, -5)                                    Answer:
y  y1  5  3  8  4                                                                4
m 2                                                                            m=
x 2  x 2 4  2   6   3                                                               3
1   2     3 2                                           32
2. A line through the point  ,   and  ,                                       m=
2   5      4 15                                         15
2     2    8

y 2  y1 15      5  15  32
m             
x2  x2     3 1       1 15

4 2       4
3. Any line parallel to the line 5x -6y = 7                                             5 5
5 5                                                                          m=     
m=      . Parallel lines have the same slope                                           6 6
6 6
4. Any line perpendicular to the line 5x -6y = 7                                          6
5 5                                                                 m =
Solution: m =    is the slope of the line 5x -6y = 7                                     5
6 6
The slopes of perpendicular lines have the negative reciprocal slopes
6
Therefore the slope of a line perpendicular to 5x – 6y = 7 is m = 
5

Example 12: The point (-12, 17) belongs to the line 5x + 11y =127. Use the slope to generate three other
points of the line.
A 5    5
The slope of the line 5x + 11y = 127 is m           
B   11  11
Solution. To generate points, start with (-12, 17) and change y by –5 and change x by 11 to obtain
(-12, 17)  (-12+11, 17 -5) = (-1, 12) is another point of the line
(-1, 12)  (-1+11, 12 -5) = (10, 7) is another point of the line
(10, 7)  (10+11, 7 -5) = (21, 2) is another point of the line.
Therefore, three other points are (-1, 12), and (10, 7), and (21, 2)

Example 13: The point (3, 2) belongs to the line 6x -11y =-4. Use the slope to generate three other points
of the line.
A 6      6
The slope of the line 6x -11y = -4 is m              
B    11 11
Solution. To generate points, start with (3, 2) and change y by 6 and change x by 11 to obtain
(3, 2)  (3 + 11, 2+6) = (14, 8) is another point of the line
(14, 8)  (14 + 11, 8 + 6) = (25, 14) is another point of the line
(25, 14)  (25 + 11, 14 + 6) = (36, 20) is another point of the line.
Therefore, three other points are (14, 8), and (25, 14), and (36, 20)

Example 14. The line 2x + 4y = 7 has no points where both x and y are integers, why?
Because for all integer, the left member of the equation 2x + 4y is even and never will equal 7
which is odd.
 2 1         1
The point (2, 0.75) belongs to the line. The slope is          or      . Therefore the point
4     2     2
(2 – 2, 0.75+1)= (0, 1.75) belongs to the line.

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