PointSlope Form of a Line

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168
3.5
Point-Slope Form of a Line
1. First, plot the point P (−3, 4). Then, because the slope is Δy/Δx = −5/7,
start at the point P (−3, 4), then move 5 units downward and 7 units to the
right, arriving at the point Q(4, −1).
y
6
P (−3, 4)
Δy = −5
x
Q(4, −1)
6
−6
Δx = 7
−6
Next, substitute m = −5/7 and (x0 , y0 ) = (−3, 4) in the point-slope
form of the line:
y − y0 = m(x − x0 )
5
y − 4 = − (x − (−3))
7
Point-slope form.
Substitute: −5/7 for m, −3 for x0 ,
and 4 for y0
Hence, the equation of the line is y − 4 = − 75 (x + 3). Label the line with its
equation.
y
6
P (−3, 4)
Δy = −5
x
Q(4, −1)
6
−6
Δx = 7
−6
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y − 4 = − 75 (x + 3)
3.5. POINT-SLOPE FORM OF A LINE
169
3. First, plot the point P (−2, 4). Then, because the slope is Δy/Δx = −4/5,
start at the point P (−2, 4), then move 4 units downward and 5 units to the
right, arriving at the point Q(3, 0).
y
6
P (−2, 4)
Δy = −4
Q(3, 0)
−6
6
Δx = 5
x
−6
Next, substitute m = −4/5 and (x0 , y0 ) = (−2, 4) in the point-slope
form of the line:
y − y0 = m(x − x0 )
4
y − 4 = − (x − (−2))
5
Point-slope form.
Substitute: −4/5 for m, −2 for x0 ,
and 4 for y0
Hence, the equation of the line is y − 4 = − 45 (x + 2). Label the line with its
equation.
y
6
P (−2, 4)
Δy = −4
Q(3, 0)
−6
Δx = 5
6
x
y − 4 = − 54 (x + 2)
−6
5. First, plot the point P (−1, −4). Then, because the slope is Δy/Δx = 5/3,
start at the point P (−1, −4), then move 5 units upward and 3 units to the
right, arriving at the point Q(2, 1).
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CHAPTER 3. INTRODUCTION TO GRAPHING
170
y
6
Δx = 3
−6
Q(2, 1)
6
x
Δy = 5
P (−1, −4)
−6
Next, substitute m = 5/3 and (x0 , y0 ) = (−1, −4) in the point-slope
form of the line:
y − y0 = m(x − x0 )
5
y − (−4) = (x − (−1))
3
Point-slope form.
Substitute: 5/3 for m, −1 for x0 ,
and −4 for y0
Hence, the equation of the line is y + 4 =
equation.
5
3 (x
+ 1). Label the line with its
y
y + 4 = 53 (x + 1)
6
Δx = 3
−6
Q(2, 1)
6
x
Δy = 5
P (−1, −4)
−6
7. Plot the points P (−4, 0) and Q(4, 3) and draw a line through them.
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3.5. POINT-SLOPE FORM OF A LINE
171
y
6
Subtract the coordinates of poing
P (−4, 0) from the point Q(4, 3).
Q(4, 3)
Δy
Δx
3 − (0)
=
4 − (−4)
3
=
8
m=
−6
P (−4, 0)
6
x
−6
Next, substitute m = 3/8, then substitute either point P (−4, 0) or point
Q(4, 3) for (x0 , y0 ) in the point-slope form of the line. We’ll substitute P (−4, 0)
for (x0 , y0 ).
y − y0 = m(x − x0 )
3
y − (0) = (x − (−4))
8
Point-slope form.
Substitute: 3/8 for m, −4 for x0 ,
and 0 for y0
Hence, the equation of the line is y = 38 (x+4). Label the line with its equation.
y
6
y = 38 (x + 4)
Q(4, 3)
−6
P (−4, 0)
6
x
−6
Alternately, if you substitute the point Q(4, 3) for (x0 , y0 ) in the point-slope
form y − y0 = m(x − x0 ), you get an equivalent equation y − 3 = 38 (x − 4).
9. Plot the points P (−3, −3) and Q(2, 0) and draw a line through them.
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CHAPTER 3. INTRODUCTION TO GRAPHING
172
y
6
Subtract the coordinates of poing
P (−3, −3) from the point Q(2, 0).
Δy
Δx
0 − (−3)
=
2 − (−3)
3
=
5
m=
−6
Q(2, 0)
6
x
P (−3, −3)
−6
Next, substitute m = 3/5, then substitute either point P (−3, −3) or
point Q(2, 0) for (x0 , y0 ) in the point-slope form of the line. We’ll substitute
P (−3, −3) for (x0 , y0 ).
y − y0 = m(x − x0 )
3
y − (−3) = (x − (−3))
5
Point-slope form.
Substitute: 3/5 for m, −3 for x0 ,
and −3 for y0
Hence, the equation of the line is y + 3 =
equation.
3
5 (x
+ 3). Label the line with its
y
6
y + 3 = 35 (x + 3)
−6
Q(2, 0)
6
x
P (−3, −3)
−6
Alternately, if you substitute the point Q(2, 0) for (x0 , y0 ) in the point-slope
form y − y0 = m(x − x0 ), you get an equivalent equation y = 35 (x − 2).
11. Plot the points P (−3, 1) and Q(4, −4) and draw a line through them.
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3.5. POINT-SLOPE FORM OF A LINE
173
y
6
Subtract the coordinates of poing
P (−3, 1) from the point Q(4, −4).
P (−3, 1)
Δy
Δx
−4 − 1
=
4 − (−3)
5
=−
7
m=
−6
6
x
Q(4, −4)
−6
Next, substitute m = −5/7, then substitute either point P (−3, 1) or
point Q(4, −4) for (x0 , y0 ) in the point-slope form of the line. We’ll substitute
P (−3, 1) for (x0 , y0 ).
y − y0 = m(x − x0 )
5
y − 1 = − (x − (−3))
7
Point-slope form.
Substitute: −5/7 for m, −3 for x0 ,
and 1 for y0
Hence, the equation of the line is y − 1 = − 57 (x + 3). Label the line with its
equation.
y
6
P (−3, 1)
−6
6
x
Q(4, −4)
−6
y − 1 = − 57 (x + 3)
Alternately, if you substitute the point Q(4, −4) for (x0 , y0 ) in the point-slope
form y − y0 = m(x − x0 ), you get an equivalent equation y + 4 = − 57 (x − 4).
13. Start by plotting the line y = − 45 x + 1. Comparing this equation with the
slope-intercept form y = mx + b, we note that m = −5/4 and b = 1. Thus,
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CHAPTER 3. INTRODUCTION TO GRAPHING
174
the slope is −5/4 and the y-intercept is (0, 1). First plot the y-intercept (0, 1).
Because the slope is Δy/Δx = −5/4, start at the y-intercept, then move −5
units downward and 4 units to the right, arriving at the point (4, −4). Draw
the line y = − 45 x + 1 through these two points.
Next, plot the point P (−3, 2). Because the line through P must be
parallel to the first line, it must have the same slope −5/4. Start at the point
P (−3, 2), then move −5 units downward and 4 units to the right, arriving at
the point Q(1, −3). Draw the line through these two points.
y = − 54 x + 1
y
y = − 54 x + 1
6
y
6
P (−3, 2)
(0, 1)
−6
6
Δy = −5
x
Δy
−6 = −5
(4, −4)
6
Q(1, −3)
Δx = 4
Δx = 4
−6
−6
To find the equation of the second line, we’ll substitute the point (x0 , y0 ) =
(−3, 2) and slope m = −5/4 into the point-slope form of the line.
y − y0 = m(x − x0 )
5
y − 2 = − (x − (−3))
4
Point-slope form.
Substitute: −5/4 for m, −3 for x0 ,
and 2 for y0 .
Hence, the equation of the line is y − 2 = − 45 (x + 3). Label the line with its
equation.
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x
3.5. POINT-SLOPE FORM OF A LINE
175
y
y = − 54 x + 1
6
P (−3, 2)
−6
6
−6
x
y − 2 = − 54 (x + 3)
15. Start by plotting the line y = 35 x. Comparing this equation with the
slope-intercept form y = mx + b, we note that m = 3/5 and b = 0. Thus,
the slope is 3/5 and the y-intercept is (0, 0). First plot the y-intercept (0, 0).
Because the slope is Δy/Δx = 3/5, start at the y-intercept, then move 3 units
upward and 5 units to the right, arriving at the point (5, 3). Draw the line
y = 35 x through these two points.
Next, plot the point P (−4, 0). Because the line through P must be
parallel to the first line, it must have the same slope 3/5. Start at the point
P (−4, 0), then move 3 units upward and 5 units to the right, arriving at the
point Q(1, 3). Draw the line through these two points.
y
y
6
6
Δx = 5
Δx = 5
Q(1, 3)
(5, 3)
Δy = 3
−6
Δy = 3
(0, 0)
y = 35 x
6
x
−6
P (−4, 0)
6
y = 35 x
−6
−6
To find the equation of the second line, we’ll substitute the point (x0 , y0 ) =
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x
CHAPTER 3. INTRODUCTION TO GRAPHING
176
(−4, 0) and slope m = 3/5 into the point-slope form of the line.
y − y0 = m(x − x0 )
3
y − 0 = (x − (−4))
5
Point-slope form.
Substitute: 3/5 for m, −4 for x0 ,
and 0 for y0 .
Hence, the equation of the line is y = 35 (x+4). Label the line with its equation.
y
y = 35 (x + 4)
6
−6
P (−4, 0)
6
x
y = 35 x
−6
17. Start by plotting the line y = 34 x + 1. Comparing this equation with the
slope-intercept form y = mx + b, we note that m = 3/4 and b = 1. Thus,
the slope is 3/4 and the y-intercept is (0, 1). First plot the y-intercept (0, 1).
Because the slope is Δy/Δx = 3/4, start at the y-intercept, then move 3 units
upward and 4 units to the right, arriving at the point (4, 4). Draw the line
y = 34 x + 1 through these two points.
Next, plot the point P (−3, 0). Because the line through P must be
parallel to the first line, it must have the same slope 3/4. Start at the point
P (−3, 0), then move 3 units upward and 4 units to the right, arriving at the
point Q(1, 3). Draw the line through these two points.
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3.5. POINT-SLOPE FORM OF A LINE
177
y
y
6
6
Δx = 4
Δx = 4
(4, 4)
Δy = 3
Q(1, 3)
Δy = 3
(0, 1)
−6
6
x
y = 34 x + 1
−6
P (−3, 0)
6
x
y = 34 x + 1
−6
−6
To find the equation of the second line, we’ll substitute the point (x0 , y0 ) =
(−3, 0) and slope m = 3/4 into the point-slope form of the line.
y − y0 = m(x − x0 )
3
y − 0 = (x − (−3))
4
Point-slope form.
Substitute: 3/4 for m, −3 for x0 ,
and 0 for y0 .
Hence, the equation of the line is y = 34 (x+3). Label the line with its equation.
y
6
−6
P (−3, 0)
y = 34 (x + 3)
6
x
y = 34 x + 1
−6
19. Start by plotting the line passing through the points P (−2, 0) and Q(2, −3)
and then calculating its slope.
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CHAPTER 3. INTRODUCTION TO GRAPHING
178
y
6
Subtract coordinates of point
P (−2, 0) from the coordinates of
the point Q(2, −3).
P (−2, 0)
−6
6
x
Q(2, −3)
Δy
Δx
−3 − (0)
=
2 − (−2)
3
=−
4
m=
−6
Thus, the slope of the line through points P and Q is −3/4. The slope of any
line perpendicular to this line is the negative reciprocal of this number, that is
4/3.
The next step is to plot the point R(−1, 0). To draw a line through R
with slope 4/3, start at the point R, then move upward 4 units and right 3
units, arriving at the point X(2, 4).
y
6
Δx = 3
To find the equation of the perpendicular line, substitute 4/3 for m
and (−1, 0) for (x0 , y0 ) in the pointslope form.
X(2, 4)
Δy = 4
−6
R(−1, 0)
6
x
y − y0 = m(x − x0 )
4
y − (0) = (x − (−1))
3
−6
Hence, the equation line through the point R perpendicular to the line through
points P and Q is y = 43 (x + 1). Label the line with its equation.
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3.5. POINT-SLOPE FORM OF A LINE
179
y
y = 43 (x + 1)
6
−6
R(−1, 0)
6
x
−6
21. Start by plotting the line passing through the points P (−2, −4) and Q(1, 4)
and then calculating its slope.
y
6
Subtract coordinates of point
P (−2, −4) from the coordinates of
the point Q(1, 4).
Q(1, 4)
−6
6
P (−2, −4)
x
Δy
Δx
4 − (−4)
=
1 − (−2)
8
=
3
m=
−6
Thus, the slope of the line through points P and Q is 8/3. The slope of any
line perpendicular to this line is the negative reciprocal of this number, that is
−3/8.
The next step is to plot the point R(−4, −1). To draw a line through R
with slope −3/8, start at the point R, then move downward 3 units and right
8 units, arriving at the point X(4, −4).
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CHAPTER 3. INTRODUCTION TO GRAPHING
180
y
6
To find the equation of the perpendicular line, substitute −3/8 for
m and (−4, −1) for (x0 , y0 ) in the
point-slope form.
−6
R(−4, −1)
6
Δy = −3
x
y − y0 = m(x − x0 )
3
y − (−1) = − (x − (−4))
8
X(4, −4)
Δx = 8
−6
Hence, the equation line through the point R perpendicular to the line through
points P and Q is y + 1 = − 38 (x + 4). Label the line with its equation.
y
6
−6
R(−4, −1)
−6
6
x
y + 1 = − 38 (x + 4)
23. Start by plotting the line passing through the points P (−2, 3) and Q(1, −1)
and then calculating its slope.
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3.5. POINT-SLOPE FORM OF A LINE
181
y
6
Subtract coordinates of point
P (−2, 3) from the coordinates of
the point Q(1, −1).
P (−2, 3)
−6
6
x
Q(1, −1)
Δy
Δx
−1 − 3
=
1 − (−2)
4
=−
3
m=
−6
Thus, the slope of the line through points P and Q is −4/3. The slope of any
line perpendicular to this line is the negative reciprocal of this number, that is
3/4.
The next step is to plot the point R(−3, −4). To draw a line through
R with slope 3/4, start at the point R, then move upward 3 units and right 4
units, arriving at the point X(1, −1).
y
6
To find the equation of the perpendicular line, substitute 3/4 for
m and (−3, −4) for (x0 , y0 ) in the
point-slope form.
−6
Δx = 4
6
X(1, −1)
Δy = 3
x
y − y0 = m(x − x0 )
3
y − (−4) = (x − (−3))
4
R(−3, −4)
−6
Hence, the equation line through the point R perpendicular to the line through
points P and Q is y + 4 = 34 (x + 3). Label the line with its equation.
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CHAPTER 3. INTRODUCTION TO GRAPHING
182
y
6
y + 4 = 34 (x + 3)
−6
6
x
R(−3, −4)
−6
25. Set up a coordinate system, placing the time t on the horizontal axis and
velocity v on the vertical axis. Label and scale each axis, including units in
your labels. Plot the points (3, 50) and (14, 30) and draw a line through the
points.
v (ft/s)
100
80
60
(3, 50)
40
(14, 30)
20
t (s)
0
0
2
4
6
Compute the slope.
30 − 50
14 − 3
−20
=
11
Slope =
Second Edition: 2012-2013
8
10
12
14
16
18
20
Substitute −20/11 for m and
(3, 50) for (x0 , y0 ) in the
point-slope form.
y − y0 = m(x − x0 )
20
y − 50 = − (x − 3)
11