Graphing Linear Equations
Cartesian Coordinate System
Quadrants – There are four quadrants in the Cartesian coordinate system created by the x-axis (the horizontal axis) and
the y-axis (the vertical axis.) The axes and quadrants are labeled in the above graph.
Ordered pair (x,y) ORDER IS IMPORTANT – The first number listed must be the value of x and the second number
listed must be the value of y. The ordered pair (3,5) is graphed above. Note that the lines from the point connect to the 3
on the x-axis, and the 5 on the y-axis.
Origin – ordered pair of (0,0) located at the intersection of the x and y-axes. See the origin labeled on the above graph.
x-intercept - where the line crosses the x-axis. (x,0) Note that y = 0 On the above graph, the line crosses the x-axis at
2 , so the x-intercept is ( 2 ,0).
y-intercept - where the line crosses the y-axis. (0, y) Note that x = 0 On the above graph, the line crosses the y-axis at
2, so the y-intercept is (0,2).
Slope of a line: For 2 ordered pairs  x1 , y1  and  x2 , y2 
slope of a line =
y  y1
change in y (vertical change or rise)
 2
change in x (horizontal change or run) x2  x1
Ex. For the points: (  3,  6) and (0,  4) the slope is
4  (6) 2

0  (3) 3
Looking at the graph, to get from one point to the next, you need to go up 2 (a rise of 2) and to the right 3(a run of 3).
Note that you need to be working left to right! Thus, again you can see that the slope will be
rise 2

run 3
Slope Summary:
If the slope is positive, the line will rise going from left to right. Notice in the above graph, that the slope is
positive, and the line rises going from left to right.
If the slope is negative, the line will fall going from left to right.
The slope of a horizontal line (a line of the form: y = a number) is 0.
The slope of a vertical line (a line of the form: x = a number) is undefined.
If the slopes of two lines are equal, then they are parallel.
Ex. For the two lines y = 3x+ 6 and y = 3x, both lines have a slope of 3, so they are parallel.
If the slopes of two lines are the negative reciprocals of each other, then they are perpendicular.
Ex. For the two lines y  3x  6 and y  
each other, so the lines are perpendicular.
1
1
x  6 , the slopes, 3 and  are negative reciprocals of
3
3
Graph Linear Equations by:
1. Plotting points
2. Using the x and y intercepts
3. Using one point on the line and the slope
Forms of linear equations
1. Standard form
ax+by=c
2. Slope-Intercept form
y=mx+b where m is the slope, and (0,b) is the y-intercept of the line
Easiest form of the equation to graph from. Plot the y-intercept and then use the slope to plot additional
points.
Easiest form to use to write an equation of the line when you know the slope of the line and the yintercept
Ex. If the slope is 7, and the y-intercept is 9, the graph of the line in slope-intercept form would
be y = 7x + 9.
Can also use when you know a point on the line and the slope of the line. Substitute in the x and y values
from the point for the x and y values in the equation, and then solve for b (the y-intercept). Finally, write
the equation in slope-intercept form using the slope and the y-intercept that you solved for.
Ex. If the point (4,8) was on the line, and the slope of the line was 
1
, substitute the
2
values into the equation y = mx + b,
8= 
1
( 4 ) + b Now solve
2
8=2+b
6=b
Now, substitute back into the original equation:
y= 
1
x+ 6
2
Can also use slope-intercept form when you know 2 points on the line. First calculate the slope using the
2 points, then substitute in the values of one point and the slope to solve for the y-intercept. Finally write
the equation using the values of the slope and y-intercept that you solved for.
3. Point-Slope form
y  y1  m( x  x1 ) where m is the slope, and ( x1 , y1 ) is a point on the line.
Easy choice when you know the slope and 1 point on the line.
Ex. To write the equation of the line with a slope of 7 and going through the point
(0,  6) , substitute the slope, x1 and y1 values into the equation.
y  (6)  7( x  0)
Also easy when you know 2 points on the line. First use the 2 points to determine
the slope. Then use one of the points to substitute in for x1 and y1 in the equation,
and substitute in the slope that you calculated.
Ex. Given the points (  3,7) and (1,9)
First calculate the slope:
97
2 1
 
1  (3) 4 2
Next choose one of the points and substitute it and the slope back into the
equation y  7 
1
( x  (3)) .
2
*Note that you can use algebra to change an equation from one form to another.
Ex. y  18  3(x  2)
y  18  3 x  6
+18
+18
y  3 x  12
 3x  3x
y  3 x  12
point-slope form
slope-intercept form
standard form