bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • MTH55_Lec-09_sec_2-4_Pt-Slp_Line-Eqn.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§2.5 Line §2.5 Line EqnEqn

Point-SlopePoint-Slope



Review §Review §

Any QUESTIONS About• §’s2.4 → Slope-Intercept Eqn,

Modeling

Any QUESTIONS About HomeWork• §’s2.4 → HW-06

2.4 MTH 55



The Point-Slope EquationThe Point-Slope Equation

The equation yy−−yy11 = = m(xm(x−−xx11)) is called the point-slopepoint-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



Point-Slope DerivationPoint-Slope Derivation

Suppose that a line through (x1, y1) has slope m. Every other point (x, y) on the line must satisfy the equation

mxxyy

1

1

Because any two points can be used to find the slope. Multiply both sides by (x − x1) yielding: 11 xxmyy • which is the point-slope form of the

equation of the line.



Example Example Point-Slope Eqn 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

4

1

4

54

15

4

54

34

54

4

5

13

14

11

12

12

xy

xy

xy

xxmyy

xx

yym

)(

)(

)(

)(

Last Line to shows both m & b

m b



Example Example Point-Slope 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:

11 xxmyy

43

29 xy



Example Example Pt-Slope → Slp-Inter 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:

11 xxmyy

433 xy



Example Example Pt-Slope → Pt-Inter 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:

433 xy

1233 xy By Distributive Law

93 xy Add 3 to Both Sides to yield Slope-Intercept Line:y = mx + b = 3x + (−9)



Graphing and Point-Slope FormGraphing 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.



Example Example Graph Graph yy −− 3 = 2( 3 = 2(xx −− 1) 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.

up 3

right 1

(1, 3)AnchorPoint

Connect the Dots to Draw the Line



Example Example Graph Graph yy+3 = (+3 = (−−4/3)(4/3)(xx+2)+2)

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

file =XY_Plot_0211.xlsfile =XY_Plot_0211.xls



Example Example Graph Graph yy+3 = (+3 = (−−4/3)(4/3)(xx+2)+2) SOLUTION: Find an

equivalent equation:

23

43 xy

23

43 xy

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

down 4

right 3

(2, 3)



Parallel and Perpendicular LinesParallel 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.



Para & Perp Lines DescribedPara & 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



Parallel Lines by Slope-InterceptParallel 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



Example Example Parallel Lines Parallel Lines

Determine whether the graphs of the lines y = −2x − 3 and 8x + 4y = −6 are parallel.

SOLUTION• Solve General

Equation for y

8 4 6x y

4 8 6y x

18 6

4y x

32

2y x

• Thus the Eqns are– y = −2x − 3

– y = −2x − 3/2



Example Example Parallel Lines 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 confirms

the Parallelism



Example Example ║& ┴ Lines║& ┴ 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

2x 3y 4 0

3y 2x 4

y2

3x

4

3

32m




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

y y1 m x x1 y 5

2

3x 4

3 y 5 2 x 4 3y 15 2x 8

3y 2x 7 0

2x 3y 7 0 Thus ║- Line Eqn

732 yx




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

y y1 m x x1 y 5

3

2x 4

2 y 5 3 x 4 2y 10 3x 12

3x 2y 22 0 Thus ┴ Line Eqn

2223 yx



Example Example ║& ┴ Lines║& ┴ Lines SOLUTION Graphically



Estimates & PredictionsEstimates & 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 interpolationinterpolation.

• Sometimes a graph passing through the known points is EXTENDED to predict future values. Making predictions in this manner is called extrapolationextrapolation



Example Example Aerobic Exercise 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



Example Example Aerobic Exercise 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)



Example Example Aerobic Exercise Aerobic Exercise

Find the Slope of the Line years 6020

bpm 120150

in x Chg

yin Chg

m

yearper bpm 4

3

years 40-

bpm 30m

Use One Point to Write Line Equation

204

3150 xy

154

3150 xy

1654

3y x



Example Example Aerobic Exercise 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:

165464

3y

5.1301655.34y The graph confirms

the target heart rate

46

130



Example Example Aerobic Exercise 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:

165704

3y

5.1121655.52y The graph confirms

the target heart rate

70

112



WhiteBoard WorkWhiteBoard Work

Problems From §2.5 Exercise Set• PPT-example, 14, 22, 26, 40, 52, 62

The LineEquations

11 xxmyySlpPt

bmxyInterSlp

CByAxGeneral



DropOut Rates DropOut Rates Scatter Plot Scatter Plot Given Column Chart

Read Chart to Construct T-table

Year x = Yr-1970 y = %

1970 0 15%1980 10 14.1%1990 20 12.1%1996 26 11.1%1997 27 11.0%2000 30 10.9%2001 31 10.7%

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



SCATTER PLOT: % of USA High School Students Dropping Out

0%

2%

4%

6%

8%

10%

12%

14%

16%

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

x (years since 1970)

y (

% U

SA

HiS

cho

ol

Dro

pO

uts

)

M55_§JBerland_Graphs_0806.xls

Zoom-in to more accurately calc the Slope




10%

11%

12%

13%

14%

15%

16%

0 4 8 12 16 20 24 28 32


y (

% U

SA

HiS

cho

ol

Dro

pO

uts

)


%3Rise

yrs 20Run

“Best” Line(EyeBalled)

Intercept 15.2%

(x1,y1) = (8yr, 14%)



DropOut Rates DropOut Rates Scatter Plot Scatter Plot Calc Slope from

Scatter Plot Measurements

yr% 15.0

20

%3

run

rise

m

yrsm

Read Intercept from Measurement

%.2150 xyb

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

%.%.

215150

x

yry

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



DropOut Rates DropOut Rates Scatter Plot Scatter Plot Thus the Linear

Model for the Data in PT-SLOPE Form

yrxyr

y

xxmyy

8150

14

11

%.%

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

X for 2010 → x = 2010 − 1970 = 40

In Equation

%.

%.%

%.%.

29

2156

21540150

2010

2010

2010

y

y

yryr

y

The model Predicts a DropOut Rate of 9.2% in 2010




8%

9%

10%

11%

12%

13%

14%

15%

16%

0 5 10 15 20 25 30 35 40


y (

% U

SA

HiS

cho

ol

Dro

pO

uts

)


9.2%



All Done for TodayAll Done for Today

CommunityCollege

EnrollmentRates



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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