The equation of a line can be written in many different for - PowerPoint Presentation

Slide1

The equation of a line can be written in many different forms. The

point-slope

and

slope-intercept forms

of a line are equivalent. Because the slope of a vertical line is undefined, these forms cannot be used to write the equation of a vertical line.Slide2
Slide3

A line with

y

-intercept

b contains the point (0, b).A line with x-intercept a contains the point (a, 0).

Remember!Slide4

Example 1A: Writing Equations In Lines

Write the equation of each line in the given form.

the line with slope 6 through (3, –4) in point-slope form

y

–

y

1

=

m(x – x1)

y – (–4) = 6(x – 3)

Point-slope form

Substitute 6 for m, 3 for x

1

, and -4 for y

1

.Slide5

Example 1B: Writing Equations In Lines

Write the equation of each line in the given form.

the line through (–1, 0) and (1, 2) in slope-intercept form

y

=

mx

+

b

0 = 1(-1) + b

1 = b

y = x + 1

Slope-intercept form

Find the slope.

Substitute 1 for m, -1 for x, and 0 for y.

Write in slope-intercept form using m = 1 and b = 1.Slide6

Example 1C: Writing Equations In Lines

Write the equation of each line in the given form.

the line with the

x-intercept 3 and y

-intercept –5 in point slope form

y

–

y

1 = m(x –

x1)Point-slope form

Use the point (3,-5) to find the slope.

Simplify.

Substitute for m, 3 for x

1

, and 0 for y

1

.

5

3

y

= (

x

- 3)

5

3

y

– 0 = (

x

– 3)

5

3Slide7

Check It Out!

Example 1a

Write the equation of each line in the given form.

the line with slope 0 through (4, 6) in slope-intercept form

y

= 6

y

–

y1

= m(x – x1)

y – 6 = 0(

x

– 4)

Point-slope form

Substitute 0 for m, 4 for x

1

, and 6 for y

1

.Slide8

Check It Out!

Example 1b

Write the equation of each line in the given form.

the line through (–3, 2) and (1, 2) in point-slope form

y

- 2 = 0

Find the slope.

y

–

y

1 = m(x – x1

)

Point-slope form

Simplify.

Substitute 0 for m, 1 for x

1

, and 2 for y

1

.

y

– 2 = 0(

x

– 1)Slide9

A system of two linear equations in two variables represents two lines. The lines can be parallel, intersecting, or coinciding. Lines that coincide are the same line, but the equations may be written in different forms.Slide10
Slide11

Determine whether the lines are parallel, intersect, or coincide.

Example 3A: Classifying Pairs of Lines

y

= 3

x

+ 7,

y

= –3

x – 4The lines have different slopes, so they intersect. Slide12

Determine whether the lines are parallel, intersect, or coincide.

Example 3B: Classifying Pairs of Lines

Solve the second equation for y to find the slope-intercept form.

6

y

= –2

x

+ 12

Both lines have a slope of , and the

y

-intercepts are different. So the lines are parallel. Slide13

Determine whether the lines are parallel, intersect, or coincide.

Example 3C: Classifying Pairs of Lines

2

y

– 4

x

= 16,

y

– 10 = 2(x - 1)Solve both equations for

y to find the slope-intercept form. 2y

– 4x = 16

Both lines have a slope of 2 and a

y

-intercept of 8, so they coincide.

2

y

= 4

x

+ 16

y

= 2

x

+ 8

y

– 10 = 2(

x

– 1)

y

– 10 = 2

x

- 2

y

= 2

x

+ 8Slide14

Check It Out!

Example 3

Determine whether the lines 3

x + 5y = 2 and 3x + 6 = -5

y

are parallel, intersect, or coincide.

Both lines have the same slopes but different

y

-intercepts, so the lines are parallel.

Solve both equations for y to find the slope-intercept form.

3x + 5y = 2

5

y

= –3

x

+ 2

3

x

+ 6 = –5

ySlide15

Erica is trying to decide between two car rental plans. For how many miles will the plans cost the same?

Example 4: Problem-Solving ApplicationSlide16

1

Understand the Problem

The answer is the number of miles for which the costs of the two plans would be the same. Plan A costs $100.00 for the initial fee and $0.35 per mile. Plan B costs $85.00 for the initial fee and $0.50 per mile.Slide17

2

Make a Plan

Write an equation for each plan, and then graph the equations. The solution is the intersection of the two lines. Find the intersection by solving the system of equations.Slide18

Solve

3

Plan A:

y

= 0.35

x

+ 100

Plan B:

y

= 0.50x + 85

0 = –0.15

x

+ 15

x

= 100

y

= 0.50

(100)

+ 85 = 135

Subtract the second equation from the first.

Solve for x.

Substitute 100 for x in the first equation.Slide19

Check your answer for each plan in the original problem.

For 100 miles, Plan A costs

$100.00 + $0.35(100) = $100 + $35 = $135.00.

Plan B costs $85.00 + $0.50(100) = $85 + $50 = $135, so the plans cost the same.Look Back

4Slide20

Check It Out!

Example 4

What if…?

Suppose the rate for Plan B was also $35 per month. What would be true about the lines that represent the cost of each plan?

The lines would be parallel.