Algebraically Finiding an Equation
of a Line with a Point and Slope
Algebraically Finding the y-intercept
Colleen wants to know how much a chick weighs when it is
hatched. Colleen tracked one of her chickens and found
it grew steadily by about 5.2 grams each day since it was
born. Nine days after it hatched, the chick weighed 98.4
grams. Algebraically determine how much the chick
weighed the day it was hatched.
When the chick was hatched, it was day 0. Thus, we need to find the y-intercept.
y  mx  b
The growth (slope) is 5.2
grams per day
y  5.2 x  b
98.4  5.2  9  b
98.4  46.8  b The equation now has one
Since the growth is constant,
the situation in linear:
The chick weighed 98.4
grams (y) after 9 days (x)
51.6 grams
46.8 46.8
51.6  b
distinct variable. Solve it.
Algebraically Finding a x-Value
Now Colleen wants to know when the chicken will weigh
140 grams. Algebraically find the answer.
Use the slope and y-intercept to write an equation.
Use the SlopeIntercept form:
y  5.2 x  51.6
The 140 grams represents a y value.
140  5.2x  51.6
The equation now
has one distinct
variable. Solve it.
51.6
88.4  5.2x
5.2
5.2
17  x
51.6
Substitute 140 for y
17 Days
Example
Algebraically find the equation for a line with
a slope of -3, passing through the point
(15,-50).
Find the y-intercept.
Use Slope-Intercept Form:
A point on the graph is x=15
and y=-50
y  mx  b
y  3x  b The slope is -3
50  3 15  b
50  45  b The equation now has one
45
5  b
Substitute back into SlopeIntercept Form:
45
distinct variable. Solve it.
y  3x  5
Perpendicular Lines
A line is perpendicular to another if it meets
or crosses at right angles (90°). For
instance, a horizontal and a vertical line
are perpendicular lines.
Slopes of Perpendicular Lines
Complete the following assuming Line A and Line B are
perpendicular.
2
1. Make Line A have a slope of 3 .
B
-3
2
A
3
2
2.
What is the slope of Line B (the line perpendicular to
3
Line A)?
2

Slopes of Perpendicular Lines
Two lines are perpendicular if their slopes
are opposite reciprocals of each other. In
a
other words, if the slope of a line is b then
b

the perpendicular line has a slope of a .
Example: What is the slope of a line perpendicular to each
equation below.
a. y  5 x  2
slope  5  15
b. y  12 x  7 slope  12
5
5
slope


c. y  10  7 x
7
Per. slope   15
Per. slope   12  2
Per. slope    75 
7
5
Example
Algebraically find the equation of the line that goes through
the point (2,3) and is perpendicular to y = -4x – 2. This y-intercept
Slope of given line  4   14
does not matter.
Per. slope    14 
Find the y-intercept.
Use Slope-Intercept Form:
A point on the graph is x=2
and y=-3
y  mx  b
y  14 x  b
3  14  2   b
3  0.5  b
0.5 0.5
The slope is ¼
The equation now has one
distinct variable. Solve it.
2.5  b
Substitute back into SlopeIntercept Form:
y  14 x  2.5
1
4
Example: Parallel Lines
Algebraically find the equation of the line that goes through
the point (16,4) and is parallel to 3x + 4y = 8.
Find the Slope
3x  4 y  8
3x
3x
4 y  3x  8
4
4 4
y   34 x  2
Find the y-intercept.
Use SlopeIntercept
Form:
A point on
the graph
is x=16
and y=4
Since Parallel Lines have the
same slope, our new equation
also has slope -3/4. (This yintercept does not matter)
Substitute back into Slope-Intercept Form:
y  mx  b
The slope
is -3/4
y   34 x  b
4   34 16   b The
4  12  b equation
now has
12
12
20  b
one distinct
variable.
Solve it.
y   34 x  20