W A T K I N S - J O H N S O N C O M P A N Y Semiconductor Equipment Group

1
Chabot Mathematics
2.5 Line EqnPoint-Slope
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu
2
Review

• Any QUESTIONS About
• s2.4 ? Slope-Intercept Eqn, Modeling
• Any QUESTIONS About HomeWork
• s2.4 ? HW-06

3
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

4
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

• Because any two points can be used to find the
  slope. Multiply both sides by (x - x1)
  yielding

• which is the point-slope form of the equation of
  the line.

5
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

m
b
6
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

7
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

8
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

By Distributive Law
Add 3 to Both Sides to yield Slope-Intercept
Liney mx b 3x (-9)
9
Graphing and Point-Slope Form

• When we know a lines slope and a point that is
  on the line (i.e., an ANCHOR Point), we can
  draw the graph.

10
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.

right 1
up 3
(1, 3)AnchorPoint

• Connect the Dots to Draw the Line

11
Example ? Graph y3 (-4/3)(x2)
12
Example ? Graph y3 (-4/3)(x2)

• SOLUTION Find an equivalent equation

(?2, ?3)
down 4

• The line passes through Anchor-Pt (-2, -3) and
  has slope of -4/3

right 3
13
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.

14
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 m1m2
  -1.
• Any two Vertical or Horizontal lines are parallel
  
• ANY horizontal line is perpendicular to ANY
  vertical line

15
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

16
Example ? Parallel Lines

• Determine whether the graphs of the lines y
  -2x - 3 and 8x 4y -6 are parallel.
• SOLUTION
• Solve General Equation for y

• Thus the Eqns are
• y -2x - 3
• y -2x - 3/2

17
Example ? Parallel Lines

• The Eqns y -2x - 3 y -2x - 3/2 show that
• m1 m2 -2
• -3 b1 ? b2 -3/2
• Thus the LinesARE Parallel
• The Graph confirmsthe Parallelism

18
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
• Find the Slope by ReStating the Line Eqn in
  Slope-Intercept Form

19
Example ? - Lines

• SOLUTION cont.
• Thus Any line parallel to the given line must
  have a slope of 2/3
• Now use the GivenPoint, (4,5) in thePt-Slope
  Line Eqn

• Thus - Line Eqn

20
Example ? - Lines

• SOLUTION cont.
• Any line perpendicular to the given line must
  have a slope of -3/2
• Now use the GivenPoint, (4,5) in thePt-Slope
  Line Eqn

• Thus - Line Eqn

21
Example ? - Lines

• SOLUTION Graphically

22
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

23
Example ? Aerobic Exercise

• A persons 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
• Graph the given data and calculate the target
  heart rate for a 46-year-old
• Calculate the target heart rate for a 70-year-old

24
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)

25
Example ? Aerobic Exercise

• Find the Slope of the Line

• Use One Point to Write Line Equation

26
Example ? Aerobic Exercise

• Solution a) To calculate the target heart rate
  for a 46-year-old, we sub 46 for x in the
  slope-intercept equation

130
46

• The graph confirms the target heart rate

INterpolation
27
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

112
70

• The graph confirms the target heart rate

EXtrapolation
28
WhiteBoard Work

• Problems From 2.4 Exercise Set
• PPT-example 14, 22, 26, 40, 52, 62

• The LineEquations

29
DropOut Rates ? Scatter Plot

• Given Column Chart

• Read Chart to Construct T-table

• Use T-table to Make Scatter Plot on the next Slide

30

• Zoom-in to more accurately calc the Slope

31
Intercept ? 15.2
(x1,y1) (8yr, 14)
Best Line(EyeBalled)
32
DropOut Rates ? Scatter Plot

• Thus the Linear Model for the Data in SLOPE-INTER
  Form

• Calc Slope from Scatter Plot Measurements

• To Find Pt-Slp Form use Known-Pt from Scatter
  Plot
• (x1,y1) (8yr, 14)

• Read Intercept from Measurement

33
DropOut Rates ? Scatter Plot

• X for 2010 ? x 2010 - 1970 40
• In Equation

• Thus the Linear Model for the Data in PT-SLOPE
  Form

• Now use Slp-Inter Eqn to Extrapolate to DropOut-
  in 2010

• The model Predicts a DropOut Rate of 9.2 in 2010

34
?9.2
35
All Done for Today
CommunityCollegeEnrollmentRates
36
Chabot Mathematics
Appendix
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu