Intro. To Graphing Linear Equations
The Coordinate Plane
A. The coordinate plane has 4 quadrants.
B. Each point in the coordinate plain has an x coordinate (the abscissa) and a y coordinate
(the ordinate). The point is stated as an ordered pair (x,y).
C. Horizontal Axis is the X – Axis. (y = 0)
D. Vertical Axis is the Y- Axis (x = 0)
Plot the following points:
a) (3,7)
b) (-4,5)
c) (-6,-1)
d) (6,-7)
e) (5,0)
f) (0,5)
g) (-5,0)
f) (0, -5)
1
Graphing Linear Equations
To graph a line (linear equation), we first want to make sure the equation is in slope intercept form
(y=mx+b). We will then use the slope and the y-intercept to graph the line.
Slope (m): Measures the steepness of a non-vertical line. It is sometimes refereed to as the rise/run
or
. It’s how fast and in what direction y changes compared to x.
y-intercept: The y-intercept is where a line passes through the y axis. It is always stated as an
ordered pair (x,y). The x coordinate is always zero. The y coordinate can be taken from the “b” in
y=mx+b.
Graphing The Linear Equation:
y = 3x - 5
1) Find the slope:
m=3m= 3 . = y .
1
x
2) Find the y-intercept: x = 0 , b = -5  (0, -5)
3) Plot the y-intercept
4) Use slope to find the next point: Start at (0,-5)
m = 3 . = y .  up 3 on the y-axis
1
x
 right 1 on the x-axis
(1,-2) Repeat: (2,1) (3,4) (4,7)
5) To plot to the left side of the y-axis, go to y-int. and
do the opposite. (Down 3 on the y, left 1 on the x)
(-1,-8)
6) Connect the dots.
Do Now:
1) y = 2x + 1
2) y = -4x + 5
3) y = ½ x – 3
4) y= - ⅔x + 2
2
Finding the equation of a line in slope intercept form (y=mx + b)
Example: Find the equation in slope intercept form of the line formed by (3,8) and (-2, -7).
A. Find the slope (m):
m = y2 – y1
x2 – x1
m = (-7) – (8) .
(-2) – (3)
m = -15 .
-5
m= 3
III.
Special Slopes
A. Zero Slope
* No change in Y
* Equation will be Y =
* Horizontal Line
B. Use m and one point to find b:
y = mx + b
m= 3
x= 1
y= 2
2 = 3(1) + b
2=3+b
-3 -3
-1 = b
y = 3x – 1
B. No Slope (undefined slope)
* No change in X
* Equation will be X =
* Vertical Line
3
Slope Intercept Form  y = mx + b
 “y” is by itself
Point-Slope Form
 y – y1 = m(x – x1) 
Standard Form:
 Ax + By = C
 Constant (number) is by itself
Given the slope and 1 point, write the equation of the line in: (a)
slope intercept form, (b) point-slope form, and (c) standard form:
Example: m = ½ ; (-6,-1)
a) Point-Slope Form
b) Slope intercept form
c) Standard Form
1) m = -2; (-3,1)
4
2) m = - ¾ ; (-8, 5)
3) m = ⅔; (-6, -4)
4) m = -1 (5, -1)
5
Find equation in slope intercept form and graph:
1) (3,-2)(-6,-8)
6) m= 4 (-2,-5)
2) (-6,10) (9,-10)
7) m= ⅔ (-6,-7)
3) (3,7) (3,-7)
8) m= -3/2 (8,-1)
4) (7,-6)(-3,4)
9) m = 0 (4,3)
5) (5,-9)(-5,-9)
10) m = undefined (-6, 5)
11) m=-3 (-4,19)
12) 16x -4y =36
13) 8x+24y = 96
14) y+7=2(x+1)
15) y+5=(2/5)(x+10)
16) y-7= ¾ (x-12)
17) y-2=-3(x-2)
Graph both of the lines on the same set of axis:
y = -2x + 6
y = -2x – 5
IV. Parallel and Perpendicular Lines:
A. Parallel Lines
* Do not intersect
* Have same slopes
For the given line, find a line that is parallel and passes through the given point.
Given Line:
Parallel:
Given Line:
Parallel:
7) y = ⅓ x + 4
(6,1)
8) y = 4x – 5
(2,13)
Given Line:
9) y = -⅔ x + 2
Parallel:
(-9,2)
Given Line:
10) –5x + 6
Parallel:
(4,-27)
6
Practice Problems: a) Use the two points to find the equation of the line.
b) For the line found in part a, find a line that is parallel and passes through
the given point.
c) Graph both lines on the same set of axis. (Use your own graph paper.)
Given Line:
1) (-5, 13) (3, -3)
2) (-6,0) (3,6)
3) (2,6)(-3,-19)
4) (-4,3) (-8,6)
5) (2,-5) (-2, -5)
6) (-9,-11)(6,9)
7) (8,-3) (-4,9)
8) (3,6)(3,-6)
9) (4,-3)(-6,-8)
10) (2,4)(-6,-12)
Parallel:
(4,-10)
(6,3)
(5,30)
(-4, 10)
(8,-2)
(-3,-9)
(-2, 1)
(7,-3)
(6,7)
(-3,-5)
11) Find the equation of the line parallel to y = 3x – 2, passing through (-2, 1).
12) Find the equation of the line parallel to y = -½x – 5, passing through (-2, 7)
13) Find the equation of the line parallel to y = -¼ x + 2, passing through (-8, 4)
14) Find the equation of the line parallel to y = (3/2)x + 6, passing through (-6, -11)
15) Find the equation of the line parallel to y = -5, passing through (2,7)
16) Find the equation of the line parallel to x = 5, passing through (6, -4).
7
Solving Systems of Equations
Graphically
A system of equations is a collection of two or more equations with a same set of unknowns. In
solving a system of equations, we try to find values for each of the unknowns that will satisfy every
equation in the system. When solving a system containing two linear equations there will be one
ordered pair (x,y) that will work in both equations.
To solve such a system graphically, we will graph both lines on the same set of axis and look for the
point of intersection. The point of intersection will be the one ordered pair that works in both
equations. We must then CHECK the solution by substituting the x and y coordinates in BOTH
ORIGINAL EQUATIONS.
1) Solve the following system graphically:
y = 2x – 5
y = - ⅓x + 2
8
Solve each of the systems of equations graphically:
2) y + 1 = -3(x – 1)
7x + 7y = 42
3) y – 9 = ¾ (x – 12)
6x + 12y = -60
9
4) 12x – 8y = 48
y – 4 = -2(x – 2)
5) y + 5 = 2(x + 4)
y – 10 = - ½ (x + 4)
10
Solve each system graphically and check: (USE YOUR OWN GRAPH PAPER)
6)
y = -4x -5
10)
y-2= (3/5)(x-10)
y = 2x -7
y+11 =2(x+7)
7)
6x + 3y =21
12x + 16y = -48
11)
6x + 9y = 45
9x +15y = 75
8)
12x – 6y = -6
16x -8y = 40
12)
x=5
y-12 = -3(x+2)
9)
y= -4
x=7
13)
9x – 18y = 126
y = -4
11
Graphing Inequalities
When we solved and graphed inequalities with only one variable (ex: x > 3), we moved on
to compound inequalities (AND/OR). We would graph both inequalities on the same number line
and decide what to keep based on whether it was an AND or an OR problem. When we graphed
linear equations on the coordinate plane we moved on to solving systems of equations graphically.
When we graph inequalities in two variables on the coordinate plane, we do not graph compound
inequalities. We move on to solving systems of inequalities. It takes a little from both inequalities
with one variable and solving systems graphically.
Graph the Inequality:
y>¼x+3
Step 1: Graph the line.
y>¼x+3
m = ¼ = ▲y = up 1
▲x
r4
y-int= (0,3)
Step 2: Test a point one up from the from
the y-int and one down from the y-int):
(0, 2)
2 > ¼ (0)+3
(0, 4)
4 > ¼ (0) + 3
2 > 3
4>3
FALSE
TRUE
Step 3: Shade towards the “true” point (0,4)
When you “test”, you must do it in the original equation!
12
1) 6x - 9y > 36
2) y - 3 > -2(x + 1)
13
3) 12x + 9y < 27
4) y + 4 > -3(x - 3)
14
5) y > 4
6) x < -6
15
Graphing Systems of Inequalities
Solve the system of inequalities graphically:
y>¼x+3
y < 3x – 5
Step 1: Graph the 1st inequality (graph
Step 2: Graph the 2nd inequality (graph the line
and test a point one up from the
the line and test a point one up from the y-int and one down
from the y-int):
y-int and one down from the y-int):
y < 3x – 5
m = 3/1 =
y-int.= (0,-5)
y>¼x+3
m=¼=
y-int= (0,3)
(0, 2)
2 > ¼ (0)+3
TEST
(0, 4)
-3 > ¼ (0) + 3
(0-6)
TEST
-6 < 3(0) - 5
(0,-4)
-4 < 3(0) - 5
2 > 3
4>3
-6 < -5
-4 < -5
FALSE
TRUE
TRUE
FALSE
Step 3: Label the area where the shading intersects with an “S”
16
2) y - 3 < - ⅓(x – 6)
12x – 6y > -12
3) x > 4
y < -5
17
4) 24x + 6x > -6
y>2
5) y – 6 < ⅔(x - 9)
x < -3
18