Chabot Mathematics
§2.5 Line Eqn
Point-Slope
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Review §
2.4
MTH 55
 Any QUESTIONS About
• §’s2.4 → Slope-Intercept Eqn,
Modeling
 Any QUESTIONS About HomeWork
• §’s2.4 → HW-06
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
The Point-Slope Equation
 The equation y−y1 = m(x−x1) is
called the point-slope equation for
the line with slope m that contains
the point (x1,y1).
 Note that (x1,y1) is a KNOWN point
• e.g.; (x1,y1) = (−7,11)
• Sometimes (x1,y1) is called the
ANCHOR Point
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Point-Slope Derivation
 Suppose that a line through
(x1, y1) has slope m. Every
other point (x, y) on the line
must satisfy the equation
y  y1
m
x  x1
 Because any two points can be used to
find the slope. Multiply both sides by
(x − x1) yielding: y  y  m x  x
1

1

• which is the point-slope form of the
equation of the line.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Point-Slope Eqn
 Find m & b for the
line through
(−1, −1) and (3, 4).
 Find m by y-chg
divided by x-chg
 Use Pt-Slope Eqn
and Solve for y to
reveal b
 Last Line to shows
both m & b
Chabot College Mathematics
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y2  y1 4  ( 1) 5
m


x2  x1 3  ( 1) 4
y  y1  m( x  x1 )
5
y  4  ( x  3)
4
5
15
y4  x
4
4
5
1
y  x
4
4
m
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Point-Slope
 Write a point-slope equation for the
line with slope 2/3 that contains the
point (4, 9)
 SOLUTION: Substitute 2/3 for m, and 4
for x1, and 9 for y1 in the Pt-Slope Eqn:
y  y1  mx  x1 
Chabot College Mathematics
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2
y  9  x  4
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Pt-Slope → Slp-Inter
 Write the slope-intercept equation for
the line with slope 3 and point (4, 3)
 SOLUTION: There are two parts to this
solution. First, write an equation in
point-slope form:
y  y1  mx  x1 
y  3  3x  4
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Pt-Slope → Pt-Inter
 slope 3 & point (4, 3) → y = mx + b
 SOLUTION: Next, we find an equivalent
equation of the form y = mx + b:
y  3  3x  4
y  3  3x  12
y  3x  9
Chabot College Mathematics
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By Distributive Law
Add 3 to Both Sides to
yield Slope-Intercept Line:
y = mx + b = 3x + (−9)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Graphing and Point-Slope Form
When we know a
line’s slope and
a point that is on
the line (i.e., an
ANCHOR Point),
we can draw the
graph.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Graph y − 3 = 2(x − 1)
 SOLUTION: Since
y − 3 = 2(x − 1) is in
point-slope form,
we know that the
line has slope 2 and
passes through the
point (1, 3).
 We plot (1, 3) and
then find a second
point by moving up
2 units and to the
right 1 unit.
Chabot College Mathematics
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right 1
up 3
(1, 3)
Anchor
Point
 Connect the Dots
to Draw the Line
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Graph y+3 = (−4/3)(x+2)
10
9
8
7
6
5
4
3
2
1
0
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
-1
-2
-3
-4
-5
-6
-7
-8
-9
file =XY_Plot_0211.xls
file =XY_Plot_0211.xls
Chabot College Mathematics
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-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Graph y+3 = (−4/3)(x+2)
 SOLUTION: Find an
equivalent equation:
4
y  3   x  2
3
4
y   3   x   2
3
 The line passes
through Anchor-Pt
(−2, −3) and has
slope of −4/3
Chabot College Mathematics
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(2, 3)
down 4
right 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Parallel and Perpendicular Lines
 Two lines are parallel (||) if they lie in
the same plane and do not intersect
no matter how far they are extended.
 Two lines are perpendicular (┴) if
they intersect at a right angle
(i.e., 90°). E.g., if one line is vertical
and another is horizontal, then they
are perpendicular.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Para & Perp Lines Described
 Let L1 and L2 be two distinct lines with
slopes m1 and m2, respectively. Then
• L1 is parallel to L2 if and only if
m1 = m2 and b1 ≠ b2
– If m1 = m2. and b1 = b2 then the Lines are CoIncident
• L1 is perpendicular L2 to if and only if
m1•m2 = −1.
• Any two Vertical or Horizontal lines are parallel
• ANY horizontal line is perpendicular to
ANY vertical line
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Parallel Lines by Slope-Intercept
 Slope-intercept form allows us to quickly
determine the slope of a line by simply
inspecting, or looking at, its equation.
 This can be especially helpful when
attempting to decide whether two
lines are parallel
 These Lines All
Have the SAME
Slope
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Parallel Lines
 Determine whether the graphs of the
lines y = −2x − 3 and 8x + 4y = −6
are parallel.
8 x  4 y  6
 SOLUTION
• Solve General
Equation for y
• Thus the Eqns are
–
y = −2x − 3
–
y = −2x − 3/2
Chabot College Mathematics
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4 y  8 x  6
1
y   8 x  6 
4
3
y  2 x 
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Parallel Lines
 The Eqns y = −2x − 3 & y = −2x − 3/2
show that
• m1 = m2 = −2
• −3 = b1 ≠ b2 = −3/2
 Thus the Lines
ARE Parallel
• The Graph confirms
the Parallelism
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  ║& ┴ Lines
 Find equations in general form for the
lines that pass through the point (4, 5)
and are (a) parallel to & (b)
perpendicular to the line 2x − 3y + 4 = 0
 SOLUTION
2x  3y  4  0
• Find the Slope by
ReStating the
Line Eqn in
Slope-Intercept
Form
Chabot College Mathematics
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3y  2x  4
2
4
y x
3
3
m  2 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  ║& ┴ Lines
 SOLUTION cont.
• Thus Any line parallel
to the given line must
have a slope of 2/3
• Now use the Given
Point, (4,5) in the
Pt-Slope Line Eqn
 Thus ║- Line Eqn
2 x  3 y  7
Chabot College Mathematics
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y  y1  m x  x1 
2
y  5  x  4 
3
3y  5   2 x  4 
3y  15  2x  8
3y  2x  7  0
2x  3y  7  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  ║& ┴ Lines
 SOLUTION cont.
• Any line perpendicular
to the given line must
have a slope of −3/2
• Now use the Given
Point, (4,5) in the
Pt-Slope Line Eqn
 Thus ┴ Line Eqn
y  y1  m x  x1 
3
y  5   x  4 
2
2 y  5   3x  4 
2y  10  3x  12
3x  2y  22  0
3x  2 y  22
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  ║& ┴ Lines
 SOLUTION Graphically
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Estimates & Predictions
 It is possible to use line graphs to estimate
real-life quantities that are not already
known. To do so, we calculate the
coordinates of an unknown point by using
two points with known coordinates.
• When the unknown point is located
BETWEEN the two points, this process
is called interpolation.
• Sometimes a graph passing through the
known points is EXTENDED to predict
future values. Making predictions in
this manner is called extrapolation
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Aerobic Exercise

A person’s target heart rate is the
number of beats per minute that bring
the most aerobic benefit to his or her
heart. The target heart rate for a
20-year-old is 150 beats per minute
(bpm) and for a 60-year-old, 120 bpm
a) Graph the given data and calculate the
target heart rate for a 46-year-old
b) Calculate the target heart rate for
a 70-year-old
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Aerobic Exercise
 Solution a) We draw
the axes and label,
using a scale that
will permit us to view
both the given and
the desired data.
The given
information allows
us to then plot
(20, 150) and
(60, 120)
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Aerobic Exercise
 Find the Slope of the Line
Chg in y 150  120 bpm
m

20  60 years
Chg in x
30 bpm
3
m
  bpm per year
- 40 years
4
 Use One Point
to Write Line
Equation
Chabot College Mathematics
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3
y  150   x  20
4
3
y  150   x  15
4
3
y   x  165
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Aerobic Exercise
 Solution a) To
calculate the target
heart rate for a 46year-old, we sub 46
for x in the slopeintercept equation:
3
y   46   165
4
y  34.5  165  130.5
 The graph confirms
the target heart rate
Chabot College Mathematics
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130
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Example  Aerobic Exercise
 Solution b) To
calculate the target
heart rate for a
70-year-old, we
substitute 70 for x in
the slope-intercept
equation:
3
y   70   165
4
y  52.5  165  112.5
 The graph confirms
the target heart rate
Chabot College Mathematics
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112
70
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
WhiteBoard Work
 Problems From §2.5 Exercise Set
• PPT-example, 14, 22, 26, 40, 52, 62

The Line
Equations
Chabot College Mathematics
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General  Ax  By  C
Slp  Inter  y  mx  b
Pt  Slp  y  y1  mx  x1 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
DropOut Rates  Scatter Plot
 Given Column Chart
 Read Chart to
Construct T-table
Year x = Yr-1970
1970
0
1980
10
1990
20
1996
26
1997
27
2000
30
2001
31
Chabot College Mathematics
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y=%
15%
14.1%
12.1%
11.1%
11.0%
10.9%
10.7%
 Use T-table to Make
Scatter Plot on the
next Slide
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
SCATTER PLOT: % of USA High School Students Dropping Out
16%
y (% USA HiSchool DropOuts)
14%
12%
10%
8%
6%
 Zoom-in to more
accurately calc the
Slope
4%
2%
0%
0
2
4
6
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
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8
10
12
14
16
18
20
22
24
26
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
28
30
32
SCATTER PLOT: % of USA High School Students Dropping Out
16%
Intercept  15.2%
y (% USA HiSchool DropOuts)
15%
(x1,y1) = (8yr, 14%)
“Best” Line
(EyeBalled)
14%
13%
Rise  3%
12%
11%
Run  20 yrs
10%
0
4
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
31
8
12
16
20
24
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
28
32
DropOut Rates  Scatter Plot
 Calc Slope from
Scatter Plot
Measurements
rise  3%
m

run 20 yrs
 m  0.15 % yr
 Read Intercept from
Measurement
b  yx  0  15.2%
Chabot College Mathematics
32
 Thus the Linear
Model for the Data in
SLOPE-INTER Form
 0.15% 
y  
x  15.2%

 yr 
 To Find Pt-Slp Form
use Known-Pt from
Scatter Plot
• (x1,y1) = (8yr, 14%)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
DropOut Rates  Scatter Plot
 Thus the Linear
Model for the Data
in PT-SLOPE Form
 X for 2010 →
x = 2010 − 1970 = 40
y  y1  mx  x1  
 0.15% 
40 yr   15.2%
y2010   

 yr 
y2010  6%  15.2%
 0.15% 
x  8 yr 
y  14%   

 yr 
 Now use Slp-Inter
Eqn to Extrapolate
to DropOut-% in
2010
Chabot College Mathematics
33
 In Equation
y2010  9.2%
 The model Predicts
a DropOut Rate of
9.2% in 2010
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
SCATTER PLOT: % of USA High School Students Dropping Out
16%
y (% USA HiSchool DropOuts)
15%
14%
13%
12%
11%
10%
9%
9.2%
8%
0
5
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
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10
15
20
25
30
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
35
40
All Done for Today
Community
College
Enrollment
Rates
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt