Linear Equation

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```					Math232 Chapter 1
Linear Functions

Zheng Chen @SUNO
zchen@suno.edu

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Section 1.1 slopes and equations of lines
rectangular coordinate planes

In a plane, a vertical line and a horizotal line intersect at a point.
Each line has its direction, zero and unit. The intesecting point is
their common zero point. Usually, the horizontal line has direction
to the right and called as x  axis, the vertical line has direction to
the up and called as y  axis.
Then this plane is called a rectanglar coordinate plane (or system).

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Rectangle Coordinate of point
Let P be a point on the plane. Through P, draw a vertical line
intercepting x-axis at a point, whose x-value is a. Also draw
a horizontal line and intercepting y -axis at a point, whose
y -value is b. Then notation (a, b) is called the coordinate of P.
We write as P(a, b). Value a is called the x-coordonate of P and
value b is called the y -coordinate of P.

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Examples of Coordinate point
The following are some coordinate point examples

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Whole coordinate plane is divided into 4 quarants.

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Exercise
Graph points
A(4,3) , B(2, 5), C ( 3, 3), P(0,3) , Q( 2, 0), R (0, 7),

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x-Intercept and y-Intercept
A line that intercepts x-axis and y-axis.
Then the x-value of intercepted point on the x-axis is called x-intercept
and the y-value of intercepted point on the y-axis is called y-intercept.

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Exercise
Find x-intercepts and y-intercepts of the following lines

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Slopes of lines
Any non-verical line has a slope value, which is the rate about
the line going up or going down. The basic formula is
rise
slope =
run

run

rise                               rise

run

Note: The slope value could be zero or negative
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Slope definition
Suppose that a line has two points (x1 , y1 ) and (x2 , y2 ), then
we have run x  x2  x1 and the rise y  y2  y1.
y y2  y1
So the slope of the line is m    
x x2  x1

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Example 1. Find the slope of a line that has points (1,  1) and (5, 7).
Consider that (1,  1) is (x1 , y1 ) and (5, 7) is (x2 , y2 ). So the slope is
y 7  (1) 8
m            2
x   5 1   4

Note: If we swap two points, considering that (5, 7) is (x1 , y1 ) and
(1,  1) is (x2 , y2 ), then the slope value is the same.
y (1)  7 8
m               2
x   1 5    4                                                       11
Example 2: Find the slope of a line that has points (1, 4) and (5,  2).
Here we can consider that (1, 4) is (x1 , y1 ) and (5,  2) is (x2 , y2 )
So the slope of line is
y (2)  4 6       3
m                  
x       5 1   4    2

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Example 3: Find the slope of a line that has points (1, 4) and (5, 4).
Here we can consider that (1, 4) is (x1 , y1 ) and (5, 4) is (x2 , y2 )
So the slope of line is
y 4  4 0
m              0
x 5  1 4

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Example 4: Find the slope of a line that has points (2,  1) and (2, 4).
Here we can consider that (2,  1) is (x1 , y1 ) and (2, 4) is (x2 , y2 )
So the slope of line is
y 4  (1) 5
m                  ?!  for a vertical line, the slope is undefined.
x     22       0

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Questions

1. What can we say more about
the line when it has positive
slope?
2. similarly, What can we say
more about the line when it has
negative slope?

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y-intercepts of lines
Any non-verical line has a y -intercept value, which is the y
coordinate value of the point on the line that intercepts the y-axis.
In the following, the red line has y-intercept  3,
the blue line has y -intercept  1.

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The graph of a linear equation
Any linear equation in two variables has a line as its graph.
Example: draw the graph of 4x +7y = 28
Solution:
If value x = a and y = b satisfying this equation, then point (a, b)
is on this line. Make the following table:
0             7     2

4             0     3

3 point (0, 4) (7, 0) and (2, 3) are on this line.

zchen@suno.edu                                                    17
Slope- intercept Form
Suppose a line has slpoe m and y -intercept b. Let ( x, y ) be an arbitary
point on the line. Then the line has two points (0, b), ( x, y ).
By using slope formula, y  y  b, x  x  0  x
y          y b
m        m            m  y  b  mx  y  mx  b
x            x
Theorem: The slope y-intercept line equation formula is y  mx  b

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Example 1: Find the slope y-intercept equation if m  2, b  1
Solution:     y  2x 1
Example 1: Find the slope and y-intercept of line 2 x  3 y  15
2 x  15        2
Solution:       3 y  2 x  15  y             y   x5
3            3
2
Slope =              y-intercepy = 5
3
Exercises:
1. Find the slope and y-intercept equationsof lines
a. m  3, b  0                       b. m  2, b  0
c. m  1, b  1                       d . m  0, b  4
1      1
e. m   , b                          f . m  0, b  0
2      2
2. Find the slope and y-intercept of line 2 x  3 y  6
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Point- Slope Form
Suppose a line has slope m and a point( x1 , y1 ). Let ( x, y ) be an arbitary
point on the line. Then from slope formula
y  y1
 m  y  y1  m( x  x1 )
x  x1
Theorem: Suppose a line has slope m and a point( x1 , y1 ). Then the
slope point equation of this line is      y  y1  m( x  x1 )
Examples: Find equations of the following lines and write as the
slope y-intercept format.
1. slope m  2, point (1, 3)
y  3  2( x  1)
y  3  2x  2
y  2x  2  3
y  2x 1                                                          20
2. slope m  2, point (1,3)
y  3  2( x  (1))
y  3  2( x  1)
y  3  2 x  2
y  2 x  2  3
y  2 x  1
Another method:
From formula y  mx  b, we suppose the line equation is
y  2 x  b
Because point (  1,3) is on the line. We plugin x  1 and y  3
3  2  (1)  b
3 2b
3 2  b                b 1
Therefore the equation is       y  2 x  1
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Two points line formula
Suppose that a line has two points (x1 , y1 ) and (x2 , y2 ), then its slope
y2  y1
m
x2  x1
Then from point slope formula, the line equation is
y2  y1
y  y1           ( x  x1 )
x2  x1
Example: Find the equation of a line that has points (1, 4) and (3,  2).
We can pick any point as (x1 , y1 )
(2)  4
y4              ( x  1)             y  4  3( x  1)
3 1
y  4  3 x  3                        y  3 x  3  4
y  3 x  7

zchen@suno.edu                                                                   22
Exercise 1
Find the slope y-intercept equations of the following lines
1. slope m  3, point (1, 2)

2. slope m  2, point (1,3)

3. points (4,5) and (1,8)

4. points (1,5) and (4,8)

5. Find the horizontal line equation that passing point (1,5)

6. Find the vertical line equation that passing point(4,8).
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General form of linear equations

The general line equation is ax  by  c
Case 1: If b  0, then we can solve for y to get slope and y intercept.
Example: 3 x  4 y  8
3
4 y  3 x  8  y   x  2
4
3
slope          and     y intercept  2
4
Case 2 : If b  0, then we get a vertical line
Example: 3 x  12  x  4 this is a vertical line
Case 3 : If a  0, then we get a horizontal line
Example: 4 y  12  y  3, this is a horizontal line
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Graph of a line
To draw a line of an equation, we need to find two different
solutions of the eqaution, then we have two points and can draw the line.
Example 1: To draw a line of equation y  3x  5
Solution:     take x  0, then y  5；
take x  3, then y  4；
Then we have two points (0, 5) and (3, 4) and can draw the line

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Exercises: Draw the lines of equations
1) y  2 x  3

2) 2 x  3 y  18

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Relation of lines
Theorem 1: Two lines are parallel if and only if they have the same
slope values.
For example, Lines y  2 x  1 and y  2 x  7 are parallel
Lines y  2 x  1 and y  3 x  7 are not parallel

Theorem 2: Two lines y  m1 x  b1 and y  m2 x  b2 are perpendicular
if and only if         m1  m2  1
1
or                    m2  
m1
1
For example, Lines y  2 x  1 and y   x  7 are parallel
2

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Example 1. Find a line that passing point (1, 2) and parallel to
line y  3 x  2
Solution: The slope of the line will be the same of line y  3 x  2.
So the slope m  3, using slope point formula
y  2  3( x  1)  y  2  3 x  3  y  3 x  3  2  y  3 x  1
Example 2. Find a line that passing point (1, 2) and perpendicular to
line y  3 x  2
1
Solution: The slope of the line will be m   .
3
Using slope point formula
1                    1    1
y  2   ( x  1)  y  2   x 
3                    3    3
1    1               1    7
 y   x 2 y   x
3   3               3    3
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Example 3. Are points A(1,5), B(2, 4) and C (4, 10) collinear?
Solution: From two points slope formula, the slope of the line AB is
4  5 9
mAB                 3
2  (1) 3
and the slope of the line AC is
10  5 15
mAC                  3
4  (1)    5
Because that they have the same slope values, the line AB and line
AC are the exactly the same line.
Therefore points A(1,5), B(2, 4) and C (4, 10) are collinear.

Note: From geometry, through one point, we only can draw only
one line that is parallel to a given line.

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Exercise 2

1. Find a line that passing point (1,  2) and parallel to line y  2 x  3

2. Find a line that passing point (3, 2) and perpendicular to y  2 x  5

3.Are points A( 1,5), B(3,3) and C (9, 0) collinear?

30

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