1. 
Find the x and y intercepts: 3y = 2x + 5
2. 
Find the slope and y intercept of 4x – 3y = 7
3. 
Graph: 4x = 12
4. 
Graph: 3x – 2y = -6
5. 
Graph: y = -½ x+2 Algebra II
1
Writing the
Equation of a Line
Algebra II
How do you know if you have the y
intercept of an equation?
Remember....the y intercept must be (0,#)!
same
Parallel lines have the ___________
slope.
Perpendicular lines have slopes that are
opposite reciprocals
_______________________.
Algebra II
3
¡ 
When you are given slope and yintercept, use slope intercept form.
y = mx + b
¡ 
If you are given anything else, use
point slope form.
y – y1 = m(x – x1)
Algebra II
4
1. Slope is 3/2 and
passes through (0, -1)
y = mx + b
slopeintercept y = 3/2x – 1
form
2. Slope is -1/3 and
passes through (0,4)
y = mx + b
slopeintercept y = -1/3 x + 4
form
-3/2x + y = -1
1/3x + y = 4
-2(-3/2 x + y = -1)
3(1/3x + y = 4)
standard 3x – 2y = 2
form
Algebra II
x + 3y = 12 standard
form
5
3. Slope is -1/2 and
passes through (2,3)
y – y1 = m(x – x1)
y – 3 = -1/2(x – 2)
y – 3 = -1/2x + 1
slope-intercept
form y = -1/2x + 4
½x + y = 4
2( ½x + y = 4)
x + 2y = 8 standard
form
Algebra II
4. Slope is 2/3 and passes
through (-2,4) y – y1 = m(x – x1)
y – 4 = 2/3(x + 2) y – 4 = 2/3x + 4/3 y = 2/3x + 4/3 + 4
y = 2/3x + 16/3
slope-2/3x + y = 16/3 intercept
-3(-2/3x + y = 16/3) form 2x – 3y = -16
standard
form
6
5. Passes through (3,2)
and is parallel to
y = -3x + 2.
m=-3 and point is (3,2)
y – y1 = m(x – x1)
y – 2 = -3(x – 3)
y – 2 = -3x + 9
y = -3x + 11
3x + y = 11
Algebra II
6. Passes through (-7, 1)
and is perpendicular to 2x – y = 3.
m=-1/2 and point: (-7,1)
y – y1 = m(x – x1) y – 1 = -1/2(x + 7)
y – 1 = -1/2x – 7/2 y = -1/2x – 5/2 ½x + y = -5/2
2(½x + y = -5/2)
x + 2y = -5
7
7. Passes through (0,3)
and is perpendicular to
y = -1/3x + 5.
8. Passes through (-2,-5) and parallel to
2y = 4x – 5.
m = 2 , point: (-2, -5)
y – y1 = m(x – x1)
y + 5 = 2(x + 2) y + 5 = 2x + 4 y = 2x – 1 -2x + y = -1
-1(-2x + y = -1)
2x – y = 1
m = 3, point: (0,3)
y = mx + b
y = 3x + 3 -3x + y = 3
-1(3x + y = 3)
3x – y = -3
Algebra II
8
9. Passes through
(-2,-1) and (3,4). m = (4 + 1)/(3 + 2) = 1
Point: (-2,-1)
y – y1 = m(x – x1)
y + 1 = 1(x + 2) y+1=x+2 y = x + 1
-x + y = 1
-1(-x + y = 1)
x – y = -1
Algebra II
10. Passes through (6,8) and
(-4,1)
m=(1-8)/(-4-6)=7/10
Point: (6,8)
y – y1 = m(x – x1)
y – 8 = 7/10(x - 6)
y – 8 = 7/10x - 42/10
y = 7/10x - 42/10 + 8
y = 7/10x + 38/10 -7/10x + y = 38/10
-10(-7/10 x + y = 38/10)
7x – 10y = -38
9
11. Passes through (-2, 3)
and parallel to y = 7
12. Passes through (4, -8)
and perpendicular to x = -2. y = 7 is a horizontal line so
the slope is zero.
x = 2 is a vertical line, so the
slope is undefined. A line parallel to that would
also have a slope of zero,
meaning it also has to be
y = #. The slope of a line that is
perpendicular to that line
would be horizontal,
meaning it would be y = #. What does y = in our
ordered pair? What does y = in our
problem?
y = 3
y = -8
Algebra
II
10
1. In 1970 there were 160 African-American woman in
elected public office in the US. By 1993, the number had
increased to 2332. Write a linear model for the number of
African-American women who were elected to public office
at any given time between 1970 to 1993. Then use the
model to predict the number of African-American women
who will hold office in 2012. ** Let 0 correspond to year 1970**
Gives (0, 160) and (23, 2332); find (42, ?)
m = (2332-160)/(23-0) = 2172/23
y = mx + b
y = 2172/23 x + 160
y = 2172/23(42) + 160
y ≈ 4126.26087
About 4126 African-American women will hold office in 2012.
Algebra II
11
2. In 1991, there were 57 million cats as pets in the US. By
1998, this number was 61 million. Write a linear model for
the number of cats as pets. Then use the model to predict
the number of cats as pets in 2015. ** Let 0 correspond to year 1991**
Gives (0, 57) and (7, 61); find (24, ?)
m = (61-57)/(7-0) = 4/7
y = mx + b
y = 4/7 x + 57
y = 4/7(24) + 57
y ≈ 70.71428571
About 71 million cats will be had as pets in 2015.
Algebra II
12
3. In 1984, American purchased an average of 113 meals or
snacks per person at restaurants. By 1996, this number was
131 meals. Write a linear model for the number of meals or
snacks purchased per person annually. Then use the model
to predict the number of meals or snacks that was purchased
per person in 2006. ** Let 0 correspond to year 1984**
Gives (0, 113) and (12, 131); find (22, ?)
m = (131-113)/(12-0) = 3/2
y = mx + b
y = 3/2 x + 113
y = 3/2(22) + 113
y = 146
_
146 meals or snacks will be purchased per person in 2006
Algebra II
13
1. Write the equation of a line that passes through the
points (-4,14) and (0, 8). (Put answer in Standard form)
2. Write the equation of line that passes through the point
(-3, -4) and is perpendicular to the line 4x + y = 20.
(Put answer in slope intercept form)
3. In 1992, the height of a river from the bottom of a bridge
was 52 feet. Due to erosion, in 1998, the height from the
bottom of the bridge is 55 feet. Write a linear model for the
height of the river from the bridge, then use the model to
predict the distance from the bridge to the river in 2000. Algebra II
14