ADVANCED GEOMETRY: REVIEW OF LINEAR FUNCTIONS
(or, TABLES AND GRAPHS AND EQUATIONS, OH MY!)
PART ONE: SLOPE AND Y-INTERCEPT
First of all, let’s start with the most standard of all linear functions, y = x. We should understand that this equation
represents the set of all points (also known as ordered pairs, with the x-value listed first and the y-value listed
second) such that the x and y coordinates are the same. So, a selection of points that satisfy the requirement that
the y – value is the same as the x – value would be the following:
(-3, -3), (0, 0), (2, 2) and (6, 6) just to name a few.
When we look at a plot of these ordered pairs on a coordinate axis system, we see that the image of the set of
points is a straight line through the origin cutting precisely between the x and y axes and tilting upward as we read
the graph from left to right.
This equation y = x is considered the “standard linear equation”. We define the slope of this line to be 1. Any line
that is steeper and rises from left to right will have a greater positive slope. Any line that is less steep but that still
rises from left to right will have a lesser positive slope.
To numerically confirm that the slope of this line is 1, we recall that slope is a ratio of rise to run. Simply select any
two points on the line and calculate the rise as the difference between the y-values, and the run as the difference
between the x-values. Suppose for our line we choose the points (-3, -3) as
( x1 , y1 ) and (2, 2) as ( x2 , y2 ) . Our
calculation of slope would look as follows:
𝑆𝑙𝑜𝑝𝑒 =
∆𝑦
∆𝑥
=
𝑦2 −𝑦1
𝑥2 −𝑥1
=
2  (3) 5
 1
2  (3) 5
Note also that the slope of 1 corresponds to the coefficient of x in the equation y = x. We note that y = x is the
same as writing y = 1x.
Next, consider the equation y = 2x. We should immediately be thinking that the slope of this line is 2, since 2 is the
coefficient of x. Furthermore, we should be thinking that this line is going to be steeper (since 2 is greater than 1)
and that the line is going to rise as we read the graph from left to right (since the slope is positive). We can
confirm all of this by considering a few ordered pairs where the y value is indeed equal to 2 times the x value. See
the table below:
x
-2
0
1
3
Y
-4
0
2
6
If we look at a quick plot of these points, we’ll confirm our thoughts above about this line: namely, that it rises
from left to right and is steeper that y = 1x.
Next, let’s consider the equation y = x + 3. We note that the slope of this line is 1 due to the coefficient of x being
equivalent to 1. (We can rewrite this equation as y = 1x + 3.) So, with the same slope as the standard linear
equation y = x, we know this line will be parallel to y = x. But, note that the y-values are going to be 3 greater with
the new equation (at any given x-value). This suggests that the new graph of y = x + 3 will just lie 3 units above the
standard graph of y = x. To confirm, let’s consider a table of values where the y-values are indeed calculated by
taking the x-value and adding 3.
x
-5
-1
0
2
Y
-2
2
3
5
When we plot these points, we confirm that the line y = x + 3 is indeed parallel to the line y = x, but that the new
graph is just 3 units above y = x.
Recall that the number 3 in this equation also defines the y-intercept. This is the number at which the line
intersects (crosses) the y-axis. We can see that indeed the line crosses the y-axis at y=3.
The previous analyses help us recall that the standard form for a linear equation is y = mx + b. In this equation, the
value of m is a number that tells us about the slope of the line. The value of b is a number that tells us about the
upward (or downward) shift of the line and where the y-intercept will be.
So, to close out these ideas, consider the equation y = -5x + 4. We identify the slope of this line (the coefficient of
x, or in other words the m-value) is -5. This would seem to indicate that the line is going to fall as we read it from
left to right due to the fact at that the slope is negative, and that it will be quite steep since the absolute value of
the slope is 5, which is greater than 1. Next, because of the “+4”, we are thinking that the y-intercept of this line
will be at y = 4. (4 is also our b-value.) Hence overall, we are imagining a steeply falling graph that passes through
the point (0, 4). Let’s confirm this by setting up a table of values where the y-value is indeed calculated by
multiplying the x-value by -5 and then adding 4.
x
-1
0
1
2
Y
9
4
-1
-6
Please confirm the calculations for the y-values by using mental math or a calculator to multiply the x-value by -5
and then to add 4. After confirming the values in the table, a plot of points will confirm our thinking (below).
***Do Exercise #1 at the end of this document now). But not yet #2-4 until reading the next section***
PART TWO: DETERMINING EQUATIONS FOR LINES
Here is a typical text question: Find the equation of the line passing through (-1,5) and (5, 4). When we plot the
points and draw a representation of the line, we should immediately make a couple of conclusions:
First, the line is falling as we read it from left to right. So, the slope is negative.
Second, the line is not very steep at all. We can guess that the slope is small, namely less than 1 in absolute value.
Third, the y-intercept is somewhere between 4 and 5 (since that is where the line crosses the y-axis).
There are two methods for finding the equation. Whether we use method 1 or method 2 (outlined below), the
first step is finding the slope of the line. Finding m is straightforward. Let (-1, 5) be
( x1 , y1 ) and let (5, 4) be
( x2 , y2 ) . Our calculation of slope would look as follows:
𝑆𝑙𝑜𝑝𝑒 =
∆𝑦
∆𝑥
=
𝑦2 −𝑦1
𝑥2 −𝑥1
=
45
1

5  (1)
6
Now, to determine the equation, here are the two methods:

Method 1: Using the Slope-Intercept Equation (y = mx + b)
We know every line has an equation of the form y = mx + b. So, it is just a matter of calculating b since we
know m. We have two points on the line. We can use either point. Let’s use the point (5, 4). This gives
us a location on the line were we know the x-value (5), we know the y-value at that instant (4) and we
know the slope (-1/6). So, let’s substitute and solve for b:
y  mx  b
1
4   (5)  b
6
5
4   b
6
At this point, to solve for b, we need to add 5/6 to both sides:
5
b
6
5
5


6
6
_____________
4
4
 
5
 b
6
So, now we know the final equation: y  
1
5
x4 .
6
6
(This confirms all that we recognized above: The slope is indeed negative; and it is indeed small in
absolute value. We also see that the y-intercept is between 4 and 5 just as we realized.)

Method 2: Using the point-slope equation
y  y1  m( x  x1 )
To use this equation, we define one of our known points (for example (-1, 5) as ( x1 , y1 )
Recall that we have already calculated the slope to be -1/6.
Now, we substitute and solve:
y  y1  m( x  x1 )
1
y  5   ( x  (1))
6
1
y  5   ( x  1)
6
Now, let ' s distribute on the right side :
1
1
y 5   x
6
6
And now add 5 to both sides
1
1
y 5   x
6
6
5
5
_______________
1
5
y   x4
6
6
Note: We end with the same equations as in Method 1.
.
One more example of a typical problem related to equations of lines:
Problem: Find the equation of a line parallel to y = -4x + 1 that passes through the point (-3, 1).
Solution: The “new” line must be parallel to y = -4x + 1. So, the new line must have the same slope as y = -4x + 1.
Since the slope of y = -4x + 1 is -4, that must be the slope of the new line as well.
Now, we know the slope of the new line is -4 and that the new line passes through the point (-3, 1). So, we
can use either method 1 or method 2 outlined above to determine the new line equation.

Let’s use the point-slope equation
y  y1  m( x  x1 )
To use this equation, we define our known point (-3, 1) as
( x1 , y1 ) .
Recall that we know that m = -4 (that is the slope).
Now, we substitute and solve:
y  y1  m( x  x1 )
y  1   4( x  (3))
y  1   4( x  3)
Now, let ' s distribute on the right side :
y  1   4 x  12
And now add 1 to both sides
y  1   4 x  12
1
1
_______________
y
  4 x  11
***Do Exercise 2 and 3 at the end of the document now (but, wait to do #4 until reading more!)***
PART THREE: PERPENDICULAR LINES
Final Idea: Let’s consider two lines that are perpendicular to each other.
Note that one line rises while the other falls as we read the graph from left to right. This means that one
of the lines has positive slope, while the other has negative slope. Next, note that one line is clearly
steeper than the other. Hence, the steep line will have a much greater slope (in absolute value) than the
line that is less steep. You may recall a conclusion from last year that matches our thoughts here:
Namely, perpendicular lines have slopes that are opposite reciprocals of one another.
With a “rise-over-run” investigation, we can define the value of the slope for each line. On the detailed
graph of these lines shown below, note that the steeper line (the one that is rising) has a slope of
The less steep line (that is falling) has a slope of
reciprocals!

1.
2
Indeed, the slopes opposite in sign and
2
.
1
Final Example: Find the line perpendicular to y =
1
x + 3 and passing through the point (2, 4).
3
Solution: We know that the “new” line must have a slope that is the opposite reciprocal of the slope of
the line y =
3
1
1
x + 3. Since the existing line has a slope of , then our new line must have a slope of
,
1
3
3
or in other words
 3.
So, we know the slope of the new line is -3 and we know the new line passes through (2, 4). We can use
either method 1 or method 2 above. This time, let’s use method 1.
Using the Slope-Intercept Equation (y = mx + b)
We know every line has an equation of the form y = mx + b. Since the point (2, 4) has to be on our line,
we need x = 2 when y = 4. We also know the slope (m) is -3. So, let’s substitute and solve for b:
y  mx  b
4  3( 2)  b
4  6  b
At this point, to solve for b, we need to add 6 to both sides:
4  6b
6
6
_____________
10  b
So, now we know the final equation: y  3 x  10 .
*****Do Exercise #4 now*****
ADVANCED GEOMETRY: A REVIEW OF LINEAR FUNCTIONS
1.
a)
Given the equation y = 0.5x – 5, answer the following questions in the order presented:
What is the slope of this line?
b) Will this line rise or fall? How do you know?
c) Will this line be steeper that y = x or less steep that y = x? How do you know?
d) Will this line cross the y-axis above or below the point (0,0)? How do you know?
e) Fill in the table of values and plot the points for this line to confirm your conclusions above.
x
-2
0
2
6
Y
2. (Read Part Two in the packet before doing this problem and the next)
Given a line that passes through the points (-4, -4) and (0, 8):
a) Plot the points on the grid below and draw a representation of the line.
Before doing any algebra, answer parts b, c and d just by looking at the graph:
b) Will the slope be positive or negative? How do you know?
c)
Will the magnitude of the slope be greater than 1 or less than 1? How do you know?
d) What do you estimate for the b-value in our linear equation? How do you know?
e)
Use the slope-intercept equation (method 1) to find the equation of our line. Show all work (as
I did above).
f)
Use the point-slope equation (method 2) to find the equation of our line. Show all work (as I did
above). In the end, your equation from this part f must match your equation from part e.
g) Which method (slope-intercept OR point-slope) do you prefer to use when finding an equation?
3.
Recalling that parallel lines have the same slope, use any method to find the equation of the
line that passes through the point (1, 3) and is parallel to the line
y  2 x  6 . Show your
work to justify your response.
4. (Read Part Three in the packet before doing this final problem.) Find the equation of a line that
1
is perpendicular to y  
x  4 and pass through the point (4, 3). Use any
4
method. Show your work to justify.
(#4, continued) Now, plot both the original line and the perpendicular line on the grid below.