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DIFFERENTIAL CALCULUS
Study Notes 3

“LINES”

Prepared By: Tutor Win

LINES

07/28/2021

In the last StudyNote 2, we have learned what’s rectangular coordinate
system is all about
...
In this
StudyNote 3, we will also learn the concept associated with lines in Calculus
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Basically, we have already knew how to graph certain lines in a rectangular
system during our previous courses in Math
...
Don’t be intimidated
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SLOPE OF A LINE
It is defined as the steepness of a line in the given graph
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Also, the slope talks about the “rise-over-run” concept in
determining the value
...
Given these two points of the line, the slope m could be defined as,
m=

=
y

L
x2—x1
(run)
B(x2, y2)
y2—y1
(rise)
A(x1, y1)

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Note: If we have the same line L as what we have above, but have replaced
another points that are still part of line L
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Example 1
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or

Step 1
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m =

Step 2
...
m =

Step 2
...
m =

𝟑

Step 4
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If the line is increasing or going upward, the slope is positive
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If the line is decreasing or going downward, the slope is negative
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If the line is a horizontal line, the slope is zero
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If the line is vertical, the slope is undefined
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EQUATIONS OF LINES
Let’s say we have a line L and it passes through a point A(x 1, y1) and
has a slope m
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Thus, for any point (x, y) on the line, the slope m is
m=
If the other point (x, y) is not part of the line, then, it has also its
different slope
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Example 2
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Find the equation of the line using point-slope equation
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m =
Step 2
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Step 3
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Find the equations of the line using point-slope equation that
passes through the points (4, 0) and (3, 4)
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The slope is unknown
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m=
m=
m=
m = −4
Step 2
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There will be two point-slope
equations here in the example
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-4 =


...
-4 =


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In this equation, the slope and the intercept are
required to complete the equation
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m=
y – y1 = m(x – x1)
y – y1 = mx – mx1
y = mx + y1 – mx1
Let b = y1 – mx1
Then, the slope-intercept equation is
y = mx + b
The variable b is actually the y-intercept of the line
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Example 4
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Step 1
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Find the slope first
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Put m=3 to the slope-intercept equation
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Next is to find b
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We can use any of the
points (x1, y1) given to solve for b because the answer will remain the same
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If we use (3,4)

b
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Substitute all of the values to the slope-intercept equation
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The most important thing is they’re parallel to each
other
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Find the slope intercept equation of the line that passes through
(3, -4) and parallel to the line with an equation of 5x – 2y = 4
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The equation of the line is 5x – 2y = 4
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5x – 2y = 4
2y = 5x – 4
y=

−

y= 𝑥− 2

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Step 2
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The other line that passes through
(3, -4) should also have a slope of
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Find the value of b using the formula b = y1 – mx1
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So,
b = -4 –( )(3)
b = -4 –
b=
Step 4
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If Line 1 has a slope of m1 and Line 2 has a
slope of m2, then m1m2 = -1
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Example 6
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Step 1
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Arrange this one in the
form of slope-intercept equation y = mx +b
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The slope of the line is −
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That is, 2 is the slope
of the other line
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Find the value of b using the formula b = y 1 – mx1
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So,
b = 5 –(2)(-2)
b=5+4
b=9
Step 4
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A
...

2
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Find the slope of the line given two points
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Find the equation of the line using point-slope equation
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Find the equation of a line that passes through the point (-2, 6) and
has a slope of
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Find the equations of the line that passes through the points (5, 1)
and (-6, 2)
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Find the equation of the line using slope-intercept equation
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Find the equation of the line that passes through (5, -3) and (2, 8)
7
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D
...

8
...

9
...

10
...


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PRACTICE TEST 3
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1
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m =
Step 2
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m =
Step 4
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Step 1
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m =
Step 3
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m =

3
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m =
Step 2
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m =
B
...
Step 1
...
=

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Step 3
...
Step 1
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We need to solve for it first
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Substitute the value of (x, y)
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a
...


and

b
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C
...
Step 1
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Find the slope first
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Put m = −

11
3

11
to the slope-intercept equation
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Next is to find b
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We can use any of the
points (x1, y1) given to solve for b because the answer will remain the same
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We will use (5,-3)
b = y1 – mx1
b = -3 – (−

11
)(5)
3

b = -3 +
b=
Step 4
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y=−

𝟏𝟏
𝟒𝟔
x+
𝟑
𝟑

7
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m = − and b = -2
Step 2
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y = − 𝑥 – 2
D
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Step 1
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Arrange this one in the
form of slope-intercept equation y = mx +b
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The slope of the line is 2
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Therefore, the equation of the
unknown equation of the other line is
y = 2𝑥 + 𝑏
Step 3
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Remember
that this line passes through (4, 1)
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Therefore, the equation of the line that passes through (4, 1)
and parallel to the line with an equation of 4x – 2y = 5 is y =𝟐𝒙 + 𝟕
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Step 1
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It is already in the form of
y = mx +b
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The slope of the line is 1
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Therefore, the equation of
the unknown equation of the other line is
y=𝑥+𝑏
Step 3
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Remember
that this line passes through (0,0)
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Therefore, the equation of the line that passes through the
origin (0, 0) and parallel to the line with an equation of 4x – 2y = 5 is
y =𝒙
10
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The equation of the line is 3x – 2y = 1
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3x – 2y = 1
2y = 3x – 1
y=

−

y= 𝑥−
Step 2
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The other line that passes through
the origin (0,0) should have a negative reciprocal slope of
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Therefore, the equation of the

unknown equation of the other line is
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y=− 𝑥+𝑏
Step 3
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Remember
that this line passes through (0, 0)
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Therefore, the equation of the line that passes through the
origin (0, 0) and perpendicular to the line with an equation of
3x – 2y = 1 is
𝟐

y =− 𝒙
𝟑

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