Slope of a Line

Slope of a Line
Consider the equation
π’š = πŸπ’™ βˆ’ πŸ’
We can make an 𝒙, π’š chart to help graph the line represented by this
equation.
𝒙
π’š
𝟏
βˆ’πŸ
𝟐
𝟎
πŸ‘
𝟐
πŸ’
πŸ’
Do you notice a pattern between the change of π’š values and the change
of 𝒙 values?
This pattern is referred to as the slope of the line.
The slope of a line is the ratio of the change in π’š (vertical change) to
the change in 𝒙 (horizontal change). We use the letter π’Ž to represent
slope.
π’Ž=
π’„π’‰π’‚π’π’ˆπ’† 𝒐𝒇 π’š
π’„π’‰π’‚π’π’ˆπ’† 𝒐𝒇 𝒙
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Let’s look at:
π’š = πŸπ’™ βˆ’ πŸ’
+𝟏
The 𝒙 values
increase by 1
+𝟏
+𝟏
π’Ž=
𝒙
𝟏
π’š
βˆ’πŸ
𝟐
𝟎
πŸ‘
𝟐
πŸ’
πŸ’
+𝟐
+𝟐
The π’š values
increase by 2
+𝟐
π’„π’‰π’‚π’π’ˆπ’† 𝒐𝒇 π’š 𝟐
= =𝟐
π’„π’‰π’‚π’π’ˆπ’† 𝒐𝒇 𝒙 𝟏
therefore, π’Ž = 𝟐 for the line π’š = πŸπ’™ βˆ’ πŸ’.
We can use the slope to help graph the line represented by the equation
π’š = πŸπ’™ βˆ’ πŸ’.
+1
+2
+1
+2
2
Looking left to right:
If the line is going up, the
line has a negative slope
If the line is going down,
the line has a positive slope
Slope is
zero
Undefined
slope
All horizontal lines have a
slope equal to zero.
All vertical lines have an
undefined slope.
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Finding the slope of a line given two points on the line:
Let’s call 𝑷1 (point 𝟏) (π’™πŸ , π’šπŸ ) and
𝑷2 (point 𝟐) (𝒙2 , π’š2 )
π’š
π’™πŸ βˆ’ π’™πŸ
π’š2
π‘·πŸ
π’šπŸ βˆ’ π’šπŸ
π’šπŸ
π‘·πŸ
𝒙
𝒙2
π’™πŸ
π’Ž =
π’„π’‰π’‚π’π’ˆπ’† π’Šπ’ π’š
βˆ†π’š
π’šπŸ βˆ’ π’šπŸ
=
=
π’„π’‰π’‚π’π’ˆπ’† π’Šπ’ 𝒙
βˆ†π’™
π’™πŸ βˆ’ π’™πŸ
Example 1
Find the slope of the line that passes through the points (πŸ”, 𝟐) and
(πŸπŸ–, πŸ–).
π’šπŸ βˆ’ π’šπŸ
πŸ–βˆ’πŸ
π’Ž=
=
π’™πŸ βˆ’ π’™πŸ πŸπŸ– βˆ’ πŸ”
=
=
*reduce
4
Parallel Lines:
Two lines are parallel if they
never intersect. Parallel lines
have the same slope.
Perpendicular Lines: Two lines are perpendicular if and only if they
intersect at a πŸ—πŸŽ° angle. The slopes of two perpendicular lines are
negative reciprocals of each other.
π’ŽπŸ
π‘š2
π‘š1
𝟐
πŸ‘
π’ŽπŸ
𝟏
βˆ’
πŸ‘
πŸ‘
𝟐
𝟎
π‘Όπ’π’…π’†π’‡π’Šπ’π’†π’…
πŸ‘
βˆ’
Finding the slope using the equation of a line:
Recall: Equation of a line
π’š = π’Žπ’™ + 𝒃
Step 1: Solve the equation for π’š
Step 2: identify the coefficient of 𝒙 βˆ’ this is the slope of the line.
5
Example 2
Find the slope of the line πŸπ’™ + π’š = πŸ–
Step 1:
πŸπ’™ + π’š = πŸ–
βˆ’πŸπ’™
βˆ’ πŸπ’™
π’š = βˆ’πŸπ’™ + πŸ–
NOTE: we prefer to see the 𝒙 before the constant
Step 2: The coefficient of 𝒙 is – 𝟐
Therefore, the slope of the line 2π‘₯ + 𝑦 = 8 is – 𝟐.
NOTE: Any line parallel to the line πŸπ’™ + π’š = πŸ– has a slope of – 𝟐
𝟏
ands any line perpendicular to the line has a slope of .
𝟐
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Example 3
Identify the lines as being parallel, perpendicular or neither:
𝒙+π’š=πŸ’
πŸ‘π’™ + πŸ‘π’š = πŸ”
First we find the slope of each line.
Step 1: Solve for π’š (for both Line 1 and Line 2)
Line 1:
Line 2:
𝒙+π’š=πŸ’
πŸ‘π’™ + πŸ‘π’š = πŸ”
βˆ’π’™
βˆ’π’™
βˆ’πŸ‘π’™
βˆ’ πŸ‘π’™
π’š = βˆ’π’™ + πŸ’
πŸ‘π’š = βˆ’πŸ‘π’™ + πŸ”
Μ…Μ…Μ…
Μ…Μ…Μ…Μ…
Μ…Μ…Μ…Μ…
πŸ‘
πŸ‘
πŸ‘
Slope of line 1: π’ŽπŸ = βˆ’πŸ
π’š = βˆ’π’™ + 𝟐
Slope of line 2: π’ŽπŸ = βˆ’πŸ
Therefore, these two lines are parallel
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Slope of a Line
Practice Problems
1. Find the slope of the line that passes through the points (πŸ“, πŸ•)
and (– 𝟏, πŸ‘).
2. Match each equation with its corresponding graph:
𝐈:
a) 𝒙 + π’š = 𝟏
𝑦
π‘₯
𝐈𝐈:
b) – 𝒙 + π’š = 𝟏
𝑦
π‘₯
𝐈𝐈𝐈:
𝑦
c) π’š = 𝟏
π‘₯
πˆπ•:
𝑦
π‘₯
d) 𝒙 = 𝟏
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