Unit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points
Formula for Finding Slope Given Two Points
Using two points: (x1, y1) (x2, y2)
Example 1
Find the slope of the line that passes through
(-3,0) and (3,6)
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Example 2
Find the slope of the line that passes through
(4, -4) and (6,7)
Unit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points
Formula For Finding Slope:
Part 1: Find the slope of a line given two points.
Using Two Points: (x1, y1) & (x2, y2)
y2 – y 1
x2 - x1
1. (2,7) (-5,-3)
2. (10, 3) (9,2)
3. Find the slope of a line that passes
through the origin and the point (-6,2) .
4. Which number represents the slope of the line that contains the following points:
(-2, 8) & (7, -10)
A. -2/3
B. 2
C. -2
D. 2/3
Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your answer
correct? Explain.
Explain Your Answer:
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Unit 2: Graphing Equations
5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points
(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.
6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line
would look like when graphed. Make a sketch of the line.
7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line
would look like when graphed. Make a sketch of the line.
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Unit 2: Graphing Equations
1. Using the slope formula, find the slope of the line
that contains the points: (-3,6) & (5,10) (2 points)
2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points)
A (-5,1)
B (-2, 5)
C (1, -4)
AB_______
AC_______
BC_______
3. Find the missing value using the given slope and points. (1 point)
1. m = 3/2 and (x, 4) (5,7)
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Unit 2: Graphing Equations
Lesson 6: Finding Slope Given Two Points – Answer Key
Part 1: Find the slope of a line given two points.
Formula For Finding Slope:
1. (2, 7) (-5,-3)
x1 y1 x2 y2
Using Two Points: (x1, y1) & (x2, y2)
y2 – y1 = -3 – 7 = -10 = 10
x2 – x1
-5 -2
-7
7
y2 – y 1
x2 - x1
Slope = 10/7
3. Find the slope of the line that passes
2. (10, 3) (9, 2)
x1 y1 x2 y2
through the origin and the point (-6,2)
(0,0) (-6, 2)
x1 y1 x2 y2
y2 – y1 = 2 - 3 = -1= 1
x2 – x1
9 – 10 -1
y2 – y1 = 2 – 0 = 2 = -1
x2 – x1
-6 -0 -6
3
Slope = 1
Slope = -1/3
4. Which number represents the slope of the line that contains the following points:
(-2, 8) & (7, -10)
A. -2/3
y2 – y1 = -10 – 8 = -18 = -2
x2 – x1
7 – (-2)
9
B. 2
C. -2
D. 2/3
b. Graph the line for the two points above (-2, 8) & (7, -10) to check your answer. Was your
answer correct? Explain.
Explain Your Answer: (Your answer may vary based on
your answer above.)
My answer was correct because the slope is -2. From each
point on the line, if I count a rise of 2 and a run of -1 (to the
left), I will find the next point on the line. Therefore, the
slope is -2.
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Unit 2: Graphing Equations
5. Line A passes through the points (2, -7) & ( -3,5). Line B passes through the points
(10, -8) & (22, 9). Which line has a greater slope? Explain your answer.
Line A: y2 – y1 = 5 – (-7) = 12 = -2 &2/5
x2 – x1
-3 – 2
-5
Line B: y2 – y1 = 9 – (-8) = 17 = 1 & 5/12
x2 – x1
22 – 10 12
Line A has the greater slope. Line A’s slope is – 12/5 and line B’s slope is 17/12. 12/5 is greater
than 17/12. When you are determining the steepness of the slope, disregard the negative signs.
Take the absolute value of the number and the larger the number the greater the slope. The
positive and negative signs only determine whether the line rises or falls.
6. Find the slope of the line that passes through the points (2,3) & (2,-5). Describe what this line
would look like when graphed. Make a sketch of the line.
y2 – y1 = -5 – 3 = -8
x2 – x1
2 -2
0
Since the slope has a 0 in the denominator, the slope of the
line is undefined. That means that this is a vertical line.
7. Find the slope of the line that passes through the points (8,4) & (-7,4). Describe what this line
would look like when graphed. Make a sketch of the line.
y2 – y1 = 4-4 = 0
x2 – x1 -7 -8 -15
Since this slope has a 0 in the numerator, the slope
of the line is 0. This is a horizontal line through (0,4).
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Unit 2: Graphing Equations
1. Using the slope formula, find the slope of the line
that contains the points: (-3,6) & (5,10) (2 points)
y2 – y1
x2 - x1
10 – 6 = 4 = 1
5-(-3)
8
2
The slope of the line is 1/2
2. The three points are vertices of a triangle. Find the slope of each side of the triangle. (6 points)
A (-5,1)
B (-2, 5)
C (1, -4)
AB 4/3
Line segment AB
Line segment AC
Line segment BC
A (-5,1)
A (-5,1)
B (-2,5) C (1, -4)
B (-2,5)
C (1, -4)
AC -5/6
BC -3
y2 – y1 5 – 1 = 4
x2 - x1 -2 – (-5) 3
y2 – y1 -4 - 1 = -5
x2 - x1 1 – (-5)
6
y2 – y1 -4 –5 = -9 = -3
x2 - x1 1-(-2) 3
Slope of AB = 4/3
Slope of AC= 5/6
Slope of BC = -3
3. Find the missing value using the given slope and points. (1 point)
1. m = 3/2 and (x, 4) (5,7)
y2 – y1 7 – 4 = 3
x2 - x1 5 – x
2
x = 3 because 5 – 3 = 2. Since the slope is 3/2, we know that the denominator must equal 3 once
the coordinates are substituted into the slope formula.
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