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College Algebra
Linear and Quadratic Functions(Chapter2)
1
2
Objectives
Chapter2
Cover the topics in ( Section 2.1:Linear functions)
After completing these sections, you should be able to:
1. Calculate and Interpret the Slope of a Line2. Graph Lines Given a Point and the Slope3. Use the Point-Slope Form of a Line4. Find the Equation of a Line Given Two Points5. Write the Equation of a Line in Slope-Intercept From and in General Form.
3
Slope of a Line
Chapter2
Let and be two distinct points with . The slope m of the non-vertical line L containing P and Q is defined by the formula
11, yxP 22 , yxQ
21 xx
my yx x
x x
2 1
2 11 2
If , L is a vertical line and the slope m of L is undefined (since this results in division by 0).
21 xx
4
Slope of a Line
Chapter2
x x2 1
y y2 1P = ( , )x y1 1
Q = ( , )x y2 2
y
x
Slope can be though of as the ratio of the vertical change ( ) to the horizontal change ( (, often termed “rise over run”.
y y2 1 x x2 1
5
Slope of a Line
Chapter2
Definition of Slope
The slope of the line through the distinct points (x1, y1) and (x2, y2) is
where x2 – x1 = 0.
Definition of Slope
The slope of the line through the distinct points (x1, y1) and (x2, y2) is
where x2 – x1 = 0.
Change in y
Change in x=
Rise
Run=y2 – y1
x2 – x1
(x1, y1)
x1
y1
x2
y2
(x2, y2)Risey2 – y1
Runx2 – x1
x
y
Definition of Slope
6
Slope of a Line
Chapter2
The Possibilities for a Line’s SlopePositive Slope
x
y
m > 0
Line rises from left to right.
Zero Slope
x
y
m = 0
Line is horizontal.m is
undefined
Undefined Slope
x
y
Line is vertical.
Negative Slope
x
y
m < 0
Line falls from left to right.
7
Slope of a Line
Chapter2
P = ( , )x y1 1
Q = ( , )x y1 2
Ly
x
If , then is zero and the slope is undefined. Plotting the two points results in the graph of a vertical line with the equation .
21 xx x x2 1
1xx
8
Slope of a Line
Chapter2
Example(1): Find the slope of the line joining the points (3,8) and (-1,2).
x y x y1 1 2 23 8 1 2, , , ,
my yx x
2 1
2 1
m 2 81 3
64
32
9
Slope of a Line
Chapter2
Example(2): Draw the graph of the line passing through (1,4) with a slope of -3/2.
Step 1: Plot the given point.
Step 2: Use the slope to find another point on the line (vertical change = -3, horizontal change = 2).
y
x
(1,4)
2
-3
(3,1)
10
Slope of a Line
Chapter2
Example(3): Draw the graph of the equation x = 2.
y
x
x = 2
11
Equation of a Line
Chapter2
When writing an equation of a line, keep in mind that you ALWAYS need two pieces of information when you go to write an equation:
1. ANY point on the line 2. Slope
Equation of a Line
Chapter2
Point-Slope Form of an Equation of a Line
An equation of a non-vertical line of slope m that passes through the point (x1, y1) is:
y y m x x 1 1
13
Equation of a Line
Chapter2
Example(4): Find an equation of a line with slope -2 passing through (-1,5).
m x y 2 1 51 1 and , ,
y y m x x 1 1
y x 5 2 1
y x 5 2 2
y x 2 3
14
Equation of a Line
Chapter2
15
Equation of a Line
Chapter2
Slope/Intercept Equation of a Line
In this form, m represents the slope and b represents the y-intercept of the line.
16
Equation of a Line
Chapter2
17
Equation of a Line
Chapter2
Equation of a Horizontal Line
A horizontal line is given by an equation of the form y = b where b is the y-intercept.
18
Equation of a Line
Chapter2
Equation of a Vertical Line
A vertical line is given by an equation of the form x = awhere a is the x-intercept.
19
Equation of a Line
Chapter2
20
Equation of a Line
Chapter2
21
Equation of a Line
Chapter2
22
Equation of a Line
Chapter2
23
Equation of a Line
Chapter2
The equation of a line L is in general form with it is written as
Ax By C 0
where A, B, and C are three real numbers and A and B are not both 0.
The equation of a line L is in slope-intercept form with it is written as
y = mx + b
where m is the slope of the line and (0,b) is the y-intercept.
24
Equation of a Line
Chapter2
Example: Find the slope m and y-intercept (0,b) of the graph of the line 3x - 2y + 6 = 0.
3x - 2y + 6 = 0
-2y = -3x - 6
y x 32
3
m 32
b 3
25
Equation of a Line
Chapter2
Example: Write the point-slope form of the equation of the line passing through (-1,3) with a slope of 4. Then solve the equation for y.
Solution We use the point-slope equation of a line with m = 4, x1= -1, and
y1 = 3.
This is the point-slope form of the equation.y – y1 = m(x – x1)
Substitute the given values.y – 3 = 4[x – (-1)]
We now have the point-slope form of the equation for the given line.
y – 3 = 4(x + 1)
We can solve the equation for y by applying the distributive property.
y – 3 = 4x + 4
y = 4x + 7 Add 3 to both sides.
26
Equation of a Line
Chapter2
Example: Graph the line whose equation is y = 2/3 x + 2.
Solution The equation of the line is in the form y = mx + b. We can find the slope, m, by identifying the coefficient of x. We can find the y-intercept, b, by identifying the constant term.
y = 2/3 x + 2 23
The slope is 2/3.
The y-intercept is 2.
We plot the second point on the line by starting at (0, 2), the first point. Then move 2 units up (the rise) and 3 units to the right (the run). This gives us a second point at (3, 4).
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
We need two points in order to graph the line. We can use the y-intercept, 2, to obtain the first point (0, 2). Plot this point on the y-axis.
27
Equation of a Line
Chapter2
Equation of a Horizontal Line
A horizontal line is given by an equation of the formy = bwhere b is the y-intercept.
Equation of a Vertical Line
A vertical line is given by an equation of the formx = awhere a is the x-intercept.
28
Equation of a Line
Chapter2
29
Equation of a Line
Chapter2
Equations of Lines
Point-slope form: y – y1 = m(x – x1) Slope-intercept form: y = m x + b
Horizontal line: y = b Vertical line: x = a
General form: Ax + By + C = 0
30
Parallel and Perpendicular Lines
Chapter2
Definitions: Parallel Lines
Two lines are said to be parallel if they do not have any points in common.
Two distinct non-vertical lines are parallel if and only if they have the same slope and have different y-intercepts.
31
Parallel and Perpendicular Lines
Chapter2
32
Parallel and Perpendicular Lines
Chapter2
Definitions: Perpendicular Lines
Two lines are said to be perpendicular if they intersect at a right angle.
Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.
121 mm1
2
1
mm
33
Parallel and Perpendicular Lines
Chapter2
34
Parallel and Perpendicular Lines
Chapter2
Example: Find the equation of the line parallel to y = -3x + 5 passing through (1,5).
Since parallel lines have the same slope, the slope of the parallel line is m = -3.
y y m x x 1 1
y x 5 3 1
y x 5 3 3
y x 3 8
35
Parallel and Perpendicular Lines
Chapter2
Example: Find the equation of the line perpendicular to y = -3x + 5 passing through (1,5).
Slope of perpendicular line:
13
13
y y m x x 1 1
y x 513
1
y x 513
13
y x 13
143
36
Parallel and Perpendicular Lines
Chapter2
Slope and Parallel Lines
• If two non-vertical lines are parallel, then they have the same slope.
• If two distinct non-vertical lines have the same slope, then they are parallel.
• Two distinct vertical lines, both with undefined slopes, are parallel.
37
Parallel and Perpendicular Lines
Chapter2
Example: Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.
Solution We are looking for the equation of the line shown on the left. Notice that the line passes through the point (-3, 2). Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.
y = 2x + 1
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1-2
-3
-4-5
(-3, 2)
Rise = 2
Run = 1y – y1 = m(x – x1)
y1 = 2 x1 = -3
38
Parallel and Perpendicular Lines
Chapter2
Solution Parallel lines have the same slope. Because the slope of the given line is 2, m = 2 for the new equation.
-5 -4 -3 -2 -1 1 2 3 4 5
5
4
3
2
1
-1
-2
-3
-4
-5
(-3, 2)
Rise = 2
Run = 1
y = 2x + 1
y – y1 = m(x – x1)
y1 = 2 m = 2 x1 = -3
Example cont.
39
Parallel and Perpendicular Lines
Chapter2
Solution The point-slope form of the line’s equation is
y – 2 = 2[x – (-3)]
y – 2 = 2(x + 3)
Solving for y, we obtain the slope-intercept form of the equation.
y – 2 = 2x + 6 Apply the distributive property.
y = 2x + 8 Add 2 to both sides. This is the slope-intercept form of the equation.
Example cont.
40
Parallel and Perpendicular Lines
Chapter2
Slope and Perpendicular Lines
Slope and Perpendicular Lines• If two non-vertical lines are perpendicular, then the product of their
slopes is –1.• If the product of the slopes of two lines is –1, then the lines are
perpendicular.• A horizontal line having zero slope is perpendicular to a vertical line
having undefined slope.
Slope and Perpendicular Lines• If two non-vertical lines are perpendicular, then the product of their
slopes is –1.• If the product of the slopes of two lines is –1, then the lines are
perpendicular.• A horizontal line having zero slope is perpendicular to a vertical line
having undefined slope.
Two lines that intersect at a right angle (90°) are said to be perpendicular. There is a relationship between the slopes of perpendicular lines.
90°
41
Parallel and Perpendicular Lines
Chapter2
Example: Find the slope of any line that is perpendicular to the line whose equation is 2x + 4y – 4 = 0.
Solution We begin by writing the equation of the given line in slope-intercept form. Solve for y.
2x + 4y – 4 = 0 This is the given equation.
4y = -2x + 4 To isolate the y-term, subtract x and add 4 on both sides.
Slope is –1/2.
y = -2/4x + 1 Divide both sides by 4.
The given line has slope –1/2. Any line perpendicular to this line has a slope that is the negative reciprocal, 2.