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) Repeat: (-2,-11) (-3,-14)
6) Connect the dots.
Do Now on GP:
1) y = 2x + 1
2) y = -4x + 5
3) y = ½ x – 3
4) y = - ⅔x + 2
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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):
B. Use m and one point to find b:
m = y2 – y1
y = mx + b
x2 – x1
m= 3
x= 3
y=8
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
-7 = 3(-2) + b
-7 = -6 + b
+6 +6
-1 = b
y = 3x – 1
B. No Slope (undefined slope)
* No change in X
* Equation will be X =
* Vertical Line
Practice Problems: Find the equation in slope intercept form: On some problems , the slope (m) is
given, so you only have to find the y-intercept (b).
1) (-4, -7)(3, 7)
2) m = ¾ , (-8,-9)
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3) (4, -7)(-2, 11)
4) (-2, 9)(2, 9)
5) (4, 6) (-6, -9)
6) (-8, 6) (-8, 11)
7) m = 0, (5, 7)
8) m = undefined, (4, 10)
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Find equation in slope intercept form and graph:
1) (3,-2)(-6,-8)
6) m= 4 (-2,-5)
12) 16x -4y =36
2) (-6,10) (9,-10)
7) m= ⅔ (-6,-7)
13) 8x+24y = 96
3) (3,7) (3,-7)
8) m= -3/2 (8,-1)
14) y-7=2(x+1)
4) (7,-6)(-3,4)
9) m = 0 (4,3)
15) y+5=(2/5)(x-10)
5) (5,-9)(-5,-9)
10) m = undefined (-6, 5)
16) y-7= ¾ (x+12)
11) m=-3 (-4,19)
17) y-2=-3(x-2)
IV. Parallel and Perpendicular Lines:
A. Parallel Lines
* Do not intersect
* Have same slopes
Do in NB:
B. Perpendicular lines
* Intersect to form right angles (90˚)
* Slopes are negative reciprocals.
(Invert fraction and change sign)
(Products of slopes is –1)
For the given line, find a line that is parallel and passes through the given point. Then,
find the equation of a line that is perpendicular and passes through the given point.
Given Line:
7) y = ⅓ x + 4
Parallel:
(6,1)
Perpendicular:
( -2,10)
8) y = 4x – 5
(2,13)
(8.-5)
9) y = -⅔ x + 2
(-9,-11)
(4,-1)
10) –5x + 6
(4,-27)
(-10,6)
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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) Find the equation of a line that is perpendicular and passes through the given
point.
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)
(5,14)
(-4, 10)
(8,-2)
(-3,-9)
(-2, 14)
(11,-3)
(6,7)
(-3,-5)
Perpendicular:
(2,7)
(6,-7)
(5,5)
(-6,-8)
(4,-3)
(-4,10)
(6,-4)
(5,2)
(-5,0)
(-8,4)
11) Find the equation of the line parallel to y = 3x – 2, passing through (-2, 1).
12) Find the equation of the line perpendicular to y = -½x – 5, passing through (-2, -10)
13) Find the equation of the line parallel to y = -¼ x + 2, passing through (-8, 7)
14) Find the equation of the line perpendicular to y = (3/2)x + 6, passing through (-6, 1)
15) Find the equation of the line parallel to y = -5, passing through (2,7)
16) Find the equation of the line perpendicular to y = 5, passing through (6, -4).
17) Find the equation of the line parallel to x= 8, passing through (4, -9)
18) Find the equation of the line perpendicular to x = -3, passing through (6, -7).
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Find the equation for each line and graph together on the same set of coordinate axis:
1) 24x + 8y = 32
y + 7 = ⅓(x + 3)
2) y + 3 = (-3/2)(x + 2)
3x – 12y = -96
3) m = 0, (-3,-6)
m = undefined, (4,-6)
Solve each system graphically:
1) Put both lines into slope intercept form (y = mx + b)
2) Graph both lines on the same set of axis.
3) Find the point of intersection and label it. (This is the solution to the system.)
4) Make sure you label 5 THINGS!!! x-axis, y-axis, 1st line, 2nd line, and point of int.!!!!
Check your solution (point of intersection) in both original
equations!!
5)
19)
y = -4x -5
y = 2x -7
23)
y-2= (3/5)(x-10)
y+11 =2(x+7)
20)
6x + 3y =21
12x + 16y = -48
24)
6x + 9y = 45
9x +15y = 75
21)
12x – 6y = -6
16x -8y = 40
25)
x=5
y-12 = -3(x+2)
22)
y= -4
x=7
26)
9x – 18y = 126
y = -4
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