Name__________________________________________
Geometry Easter Break Assignment
Class___________
Graphing
Directions: This packet is to be completed by you over Easter break. This assignment is due
Monday April 20th. You must hand this in to the Math Office (E114) before 8:30 AM in the
designated “Ms. Lanci” box. This assignment will be graded as a quiz. If you have any questions
please e-mail me.
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. 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 (ON GRAPH PAPER):
Graph each line below on different axes. *For each please write down the slope and y-intercept
before graphing.*
1) y = 2x + 1
2) y = -4x + 5
3) y = ½ x – 3
4) y= - ⅔x + 2
I.
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 (1,2) and (-2, -7).
A. Find the slope (m):
m = y2 – y1
x2 – x1
m = -7 – 2 .
-2 – 1
m = -9 .
-3
m= 3
II.
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
Special Slopes
A. Zero Slope
* No change in Y
* Equation will be Y =
* Horizontal Line
Do Now: Find equation in slope intercept form and graph:
1) (3,-2)(-6,-8)
2) (-6,10) (9,-10)
y = 3x – 1
B. No Slope (undefined slope)
* No change in X
* Equation will be X =
* Vertical Line
3) (3,7) (3,-7)
Example: Find the equation of the line when given the slope and 1 point.
A.) Since we have the slope we must plug the slope and the point into y = mx + b to find
the y-intercept. Then write the equation.
m= 4 (-2,-5)
y = mx + b
-5 = 4(-2) + b
-5 = -8 + b
+8 +8
3=b
y = 4x + 3
Do Now: Find the equation of the line in slope-intercept form.
1.) m= ⅔ (-6,-7)
2.) m = 0 (4,3)
III. Parallel and Perpendicular Lines:
A. Parallel Lines
* Do not intersect
* Have same slopes
3.) m= -3/2 (8,-1)
B. Perpendicular Lines
*Form 90° angles
* Have negative reciprocal slopes
Example: For the given line, find a line that is parallel and passes through the given point.
Given Line:
Parallel:
y=⅓x+4
(6,1)
Parallel lines have the same slope so we need m from the first equation.
m = 1/3
Now we will plug m and the parallel point into y = mx + b. Then write the
equation.
y = mx + b
1 = 1/3(6) + b
1=2+b
-2 -2
-1 = b
y = 1/3x - 1
Do Now:
Given Line:
1) y = 4x – 5
Parallel:
(2,13)
Given Line:
2) y = -⅔ x + 2
Parallel:
(-9,2)
3) Find the equation of the line parallel to y = 3x – 2, passing through (-2, 1).
4) Find the equation of the line parallel to y = -½x – 5, passing through (-2, 7)
Example: For the given line, find a line that is perpendicular and passes through the given point.
Given Line:
Perpendicular:
y=2x-3
(0,10)
Perpendicular lines have negative reciprocal slopes so we need the slope from the
given equation: m = 2 (Now we need to change the sign and flip it) : m = -1/2
Now we will plug m and the perpendicular point into y = mx + b. Then write the
equation.
y = mx + b
10 = -1/2(0) + b
10 = 0+ b
10 = b
y = -1/2x + 10
Do Now:
1.) Find a line that is perpendicular to y = 1/4x -2 and that passes through the point (-8,1).
2.) Find a line that is perpendicular to y = -3/5x - 8/5 and that passes through the point (-6,-5).
3.) Find a line that is perpendicular to 8y – 7x = 16 and that passes through the point (14,6).