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Algebra 2
More About Linear Equations
Lesson 2-4 Part 2
Goals
Goal
• To write and graph the
equation of a line.
• To write equations of
parallel and perpendicular
lines.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Essential Question
Big Idea: Equivalence
• How are the slopes of two lines related?
Vocabulary
• Parallel lines
• Perpendicular lines
Review - Axial Intercepts
A y-intercept is the point where a
graph intersects the y-axis. The
x-coordinate of this point is
always 0.
An x-intercept is the point where
a graph intersects the x-axis. The
y-coordinate of this point is
always 0.
Example: Finding Intercepts
From a Graph
Find the x- and y-intercepts.
The graph intersects the y-axis
at (0, 1).
The y-intercept is (0, 1).
The graph intersects the xaxis at (–2, 0).
The x-intercept is (–2, 0).
Example: Intercepts From Equation
Find the x- and y-intercepts.
5x – 2y = 10
To find the x-intercept, replace y with 0
and solve for x.
5x – 2y = 10
5x – 2(0) = 10
5x – 0 = 10
5x = 10
x=2
The x-intercept is (2, 0).
To find the y-intercept, replace x with 0
and solve for y.
5x – 2y = 10
5(0) – 2y = 10
0 – 2y = 10
–2y = 10
y = –5
The y-intercept is (0, –5).
Your Turn:
Find the x- and y-intercepts.
The graph intersects the yaxis at (0, 3).
The y-intercept is (0, 3).
The graph intersects the xaxis at (–2, 0).
The x-intercept is (–2, 0).
Your Turn:
Find the x- and y-intercepts.
–3x + 5y = 30
To find the x-intercept, replace y with 0
and solve for x.
–3x + 5y = 30
–3x + 5(0) = 30
–3x – 0 = 30
–3x = 30
x = –10
The x-intercept is (–10, 0).
To find the y-intercept, replace x with 0
and solve for y.
–3x + 5y = 30
–3(0) + 5y = 30
0 + 5y = 30
5y = 30
y=6
The y-intercept is (0, 6).
Your Turn:
Find the x- and y-intercepts.
4x + 2y = 16
To find the x-intercept, replace y with 0
and solve for x.
4x + 2y = 16
4x + 2(0) = 16
4x + 0 = 16
4x = 16
x=4
The x-intercept is (4, 0).
To find the y-intercept, replace x
with 0 and solve for y.
4x + 2y = 16
4(0) + 2y = 16
0 + 2y = 16
2y = 16
y=8
The y-intercept is (0, 8).
Graphing Using Standard
Form
For any two points, there is exactly one line that contains
them both. This means you need only two ordered pairs to
graph a line.
It is often simplest to find the ordered pairs that contain the x
and y intercepts.
The x and y intercepts can easily be found from standard
form. Therefore, to graph a linear equation in standard
form, use the x and y intercepts.
Example: Graphing Standard
Form
Use intercepts to graph the line given by the equation.
3x – 7y = 21
Step 1 Find the intercepts.
x-intercept:
3x – 7y = 21
3x – 7(0) = 21
3x = 21
x=7
y-intercept:
3x – 7y = 21
3(0) – 7y = 21
–7y = 21
y = –3
Example: Continued
x-intercept: x = (7, 0) and y-intercept: y = (0, -3)
Step 2 Graph the line.
Plot (7, 0) and (0, –3).
x
Connect with a straight line.
Example: Graphing Standard
Form
Use intercepts to graph the line given by the equation.
y = –x + 4
Step 1 Write the equation in standard form.
y = –x + 4
+x +x
x+y=4
Add x to both sides.
Example: Continued
x+y=4
Step 2 Find the intercepts.
x-intercept:
x+y=4
x+0=4
x=4
y-intercept:
x+y=4
0+y=4
y=4
Example: Continued
x-intercept: x = (4, 0) and y-intercept: y = (0, 4)
Step 3 Graph the line.
Plot (4, 0) and (0, 4).
Connect with a straight line.
Your Turn:
Use intercepts to graph the line given by the equation.
–3x + 4y = –12
Step 1 Find the intercepts.
x-intercept:
y-intercept:
–3x + 4y = –12
–3x + 4y = –12
–3x + 4(0) = –12
–3x = –12
–3(0) + 4y = –12
4y = –12
x=4
y = –3
Your Turn: Continued
x-intercept: x = (4, 0) and y-intercept: y = (0, -3)
Step 2 Graph the line.
Plot (4, 0) and (0, –3).
Connect with a straight line.
Your Turn:
Use intercepts to graph the line given by the equation.
Step 1 Write the equation in standard form.
Multiply both sides by 3, the LCD of the
fractions, to clear the fraction.
3y = x – 6
–x + 3y = –6
Write the equation in standard form.
Your Turn: Continued
–x + 3y = –6
Step 2 Find the intercepts.
x-intercept:
–x + 3y = –6
–x + 3(0) = –6
–x = –6
y-intercept:
–x + 3y = –6
–(0) + 3y = –6
3y = –6
x=6
y = –2
Your Turn: Continued
x-intercept: x = (6, 0) and y-intercept: y = (0, -2)
Step 3 Graph the line.
Plot (6, 0) and (0, –2).
Connect with a straight line.
Equations of Horizontal
and Vertical Lines
Equations of Horizontal
and Vertical Lines
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y=b
where b is the y-intercept. Note: m = 0.
Equation of a Vertical Line
A vertical line is given by an equation of the form
x=a
where a is the x-intercept. Note: m is undefined.
Y
Example: Horizontal Line
• Let’s find the equation for the line passing through the
points (0,2) and (5,2)
y = mx + b ( Slope Intercept Form ).
Where m is:
Y-axis
m=
DY
DX
=
(2 – 2)
(5 – 0)
=0
(0,2)
(5,2)
DX
DY = 0
X-axis
X
Y
Example: Horizontal Line
• Because the value of m is 0,
y = 0x + 2 becomes
Y-axis
y=2
(A Constant Function)
(0,2)
(5,2)
X-axis
X
Y
Your Turn:
• Find the equation for the lines passing through the
following points.
1.) (3,2) & ( 8,2)
y=2
2.) (-5,4) & ( 10,4)
y=4
3.) (1,-2) & ( 7,-2)
y = -2
4.) (4,3) & ( -2,3)
y=3
X
Y
Example: Vertical Line
• Let’s look at a line
with no y-intercept
b, an x-intercept a,
passing through
(3,0) and (3,7).
Y-axis
(3,7)
(3,0)
X-axis
X
Y
Example: Vertical Line
• The equation for the vertical line is:
x = 3 ( 3 is the X-Intercept of the line).
Because m is:
Y-axis
(3,7)
m=
DY
DX
=
(7 – 0)
(3 – 3)
=
7
0
= Undefined
(3,0)
X-axis
X
Y
Your Turn:
• Find the equation for the lines passing through the
following points.
1.) (3, 5) & (3, -2)
x=3
2.) (-5, 1) & (-5, -1) x = -5
3.) (1, -6) & (1, 8)
x=1
4.) (4, 3) & (4, -4)
x=4
X
Equations of Vertical and
Horizontal lines
An equation of the vertical line through the
point (a, b) is x = a.
An equation of the horizontal line through
the point (a, b) is y = b.
Example: Application
The table shows the rents and selling
prices of properties from a game.
Express the rent as a function of the selling
price.
Let x = selling price and y = rent.
Find the slope by choosing two points. Let
(x1, y1) be (75, 9)
and (x2, y2) be (90, 12).
Selling Price
($)
75
Rent
($)
9
90
12
160
26
250
44
Example: Continued
To find the equation for the rent function, use point-slope
form.
y – y1 = m(x – x1)
Use the data in the first row of
the table.
Simplify.
Example: Continued
Graph the relationship between the selling price and the rent. How
much is the rent for a property with a selling price of $230?
To find the rent for a property, use the graph or substitute its selling
price of $230 into the function.
Substitute.
y = 46 – 6
y = 40
The rent for the property is $40.
Your Turn:
Express the cost as a linear function of
the number of items.
Let x = items and y = cost.
Find the slope by choosing two
points. Let (x1, y1) be (4, 14) and (x2, y2)
be (7, 21.50).
Items
Cost ($)
4
14.00
7
21.50
18
Continued
To find the equation for the number of items, use pointslope form.
y – y1 = m(x – x1)
y – 14 = 2.5(x – 4)
y = 2.5x + 4
Use the data in the first row of the
table.
Simplify.
Continued
Graph the relationship between the number of items and the cost.
Find the cost of 18 items.
To find the cost, use the graph or substitute the number
of items into the function.
y = 2.5(18) + 4
Substitute.
y = 45 + 4
y = 49
The cost for 18 items is $49.
Parallel and
Perpendicular Lines
By comparing slopes, you can determine if the lines are
parallel or perpendicular. You can also write equations of
lines that meet certain criteria.
Geometric Definition
• Parallel Lines - are lines in the same plane that
never intersect.
Line WX is parallel to line YZ.
WX || YZ.
Parallel Lines
These two lines are parallel.
Parallel lines are lines in the
same plane that have no
points in common. In other
words, they do not intersect.
As seen on the graph, parallel lines have the same slope.
For y = 3x + 100, m = 3 and for y = 3x + 50, m = 3.
Algebraic Definition
Geometric Definition
• Perpendicular Lines – lines that intersect to form 90o
angles, or right angles.
Line RS is perpendicular to line TU.
RS __| TU.
Algebraic Definition
Perpendicular lines have slopes that are opposite reciprocals.
Opposite Reciprocals
Helpful Hint
If you know the slope of a line, the slope of a
perpendicular line will be the "opposite
reciprocal.”
Example: Writing Equations of
Parallel Lines
Write an equation in slope-intercept form for the line that passes
through (4, 10) and is parallel to the line described by y = 3x + 8.
Step 1 Find the slope of the line.
The slope is 3.
y = 3x + 8
The parallel line also has a slope of 3.
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.
y – 10 = 3(x – 4)
Substitute 3 for m, 4 for x1,
and 10 for y1.
Example: Continued
Step 3 Write the equation in slope-intercept form.
y – 10 = 3(x – 4)
y – 10 = 3x – 12)
y = 3x – 2
Distribute 3 on the right side.
Add 10 to both sides.
Your Turn:
Write an equation in slope-intercept form for the line that passes
through (5, 7) and is parallel to the line described by y =
x – 6.
Step 1 Find the slope of the line.
y=
x –6
The parallel line also has a slope of
The slope is
.
.
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.
Your Turn: Continued
Step 3 Write the equation in slope-intercept form.
Distribute
on the right side.
Add 7 to both sides.
Example: Writing Equations of
Perpendicular Lines
Write an equation in slope-intercept form for the line that passes
through (2, –1) and is perpendicular to the line described by
y = 2x – 5.
Step 1 Find the slope of the line.
The slope is 2.
y = 2x – 5
The perpendicular line has a slope of
because
Step 2 Write the equation in point-slope form.
y – y1 = m(x – x1)
Use the point-slope form.
Substitute
2 for x1.
for m, –1 for y1, and
Example: Continued
Step 3 Write the equation in slope-intercept form.
Distribute
on the right side.
Subtract 1 from both sides.
Your Turn:
Write an equation in slope-intercept form for the
line that passes through (–5, 3) and is perpendicular
to the line described by y = 5x.
Step 1 Find the slope of the line.
The slope is 5.
y = 5x
The perpendicular line has a slope of
because
.
Step 2 Write the equation in point-slope form.
Use the point-slope form.
y – y1 = m(x – x1)
Your Turn: Continued
Write an equation in slope-intercept form for the line that passes
through (–5, 3) and is perpendicular to the line described by
y = 5x.
Step 3 Write in slope-intercept form.
Distribute
on the right side.
Add 3 to both sides.
Your Turn:
Find the equation in slope-intercept form
of the line that passes through the point
(3, 5) and satisfies the given condition.
a. parallel to the line 2x + 5y = 4
b. perpendicular to the line 2x + 5y = 4
Solution (3, 5) is on the line so we need to find
the slope to use the point-slope form. Write the
equation in the slope-intercept form (solve for y).
2.5 - 52
Continued
Find the equation in slope-intercept form of the
line that passes through the point (3, 5) and
satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
2x  5y  4
5 y  2 x  4
2
4
y  x
5
5
Subtract 2x.
Divide by 5.
2.5 - 53
Continued:
Find the equation in slope-intercept form of the
line that passes through the point (3, 5) and
satisfies the given condition.
a. parallel to the line 2x + 5y = 4
2
4
y  x
5
5
The slope is – 2/5. Since the lines are parallel,
– 2/5 is also the slope of the line whose
equation is to be found.
Solution
2.5 - 54
Continued:
Find the equation in slope-intercept form of the
line that passes through the point (3, 5) and
satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
y  y1  m  x  x1 
2
y  5    x  3
5
2
6
y 5   x 
5
5
Point-slope form
m = – 2/5, x1 = 3, y1
=5
Distributive property
2.5 - 55
Continued:
Find the equation in slope-intercept form of the
line that passes through the point (3, 5) and
satisfies the given condition.
a. parallel to the line 2x + 5y = 4
Solution
2
6
y 5   x 
5
5
2
31
y  x
5
5
Distributive property
Add 5 (25/5).
2.5 - 56
Continued:
Find the equation in slope-intercept form of the
line that passes through the point (3, 5) and
satisfies the given condition.
b. perpendicular to the line 2x + 5y = 4
Solution We know the slope of the line, so the
slope of any line perpendicular to it is 5/2.
y  y1  m  x  x1 
2.5 - 57
Continued:
b. perpendicular to the line 2x + 5y = 4
Solution
y  y1  m  x  x1 
5
y  5   x  3
2
5
15
y 5  x 
2
2
5
5
y  x
2
2
5
m  , x1  3, y1  5
2
Distributive property
Add 5 (10/2).
2.5 - 58
Essential Question
Big Idea: Equivalence
• How are the slopes of two lines related?
• The slopes of two lines in the same plane indicate
how the lines are related. Two lines are parallel if
they have the same slope (and different yintercepts). Two lines are perpendicular if their
slopes are negative reciprocals.
Assignment
• Section 2-4 Part 2, Pg95 – 97; #1 – 5 all, 6 –
44 even.