Mrs. Rivas Find the slope of the line passing through the given points.

Post on 05-Jan-2016

219 views 1 download

Tags:

Transcript of Mrs. Rivas Find the slope of the line passing through the given points.

HomeworkMrs. RivasFind the slope of the line passing through the given points.

1.

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

ΒΏπŸ–βˆ’πŸŽβˆ’πŸ”βˆ’πŸ

ΒΏπŸ–βˆ’πŸ–

ΒΏβˆ’πŸ

HomeworkMrs. RivasFind the slope of the line passing through the given points.

2.

ΒΏβˆ’πŸ‘βˆ’πŸβˆ’πŸ—βˆ’πŸ—

ΒΏβˆ’πŸ’βˆ’πŸπŸ–

ΒΏπŸπŸ—

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

HomeworkMrs. RivasFind the slope of the line passing through the given points.

3.

ΒΏπŸ–βˆ’(βˆ’πŸ)πŸβˆ’(βˆ’πŸ‘)

ΒΏπŸ–+𝟏𝟐+πŸ‘

ΒΏπŸ—πŸ“

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

HomeworkMrs. RivasFind the slope of the line passing through the given points.

4.

ΒΏπŸ•πŸ”

HomeworkMrs. RivasFind the slope of the line passing through the given points.

5.

ΒΏβˆ’πŸ’πŸ‘

HomeworkMrs. RivasGraph each line.

6.

Starting point(0,-4)

π’š=π’Žπ’™+𝒃

𝟏𝟏

HomeworkMrs. RivasGraph each line.

7.

Starting point(0,3)

π’š=π’Žπ’™+𝒃

𝟐𝟏

HomeworkMrs. RivasGraph each line.

8.

Starting point(0,0)

π’š=π’Žπ’™+𝒃

πŸπŸ’

HomeworkMrs. RivasGraph each line.

9.

Starting point(0,-1)

π’š=π’Žπ’™+𝒃

βˆ’πŸ‘πŸ’

HomeworkMrs. RivasUse the given information to write an equation of each line.

10. slope -intercept

π’š=π’Žπ’™+𝒃

π’š=πŸπŸ‘π’™+πŸ”

HomeworkMrs. RivasUse the given information to write an equation of each line.

11. slope -intercept

π’š=π’Žπ’™+𝒃

π’š=βˆ’πŸπŸŽπ’™βˆ’πŸ‘

HomeworkMrs. RivasUse the given information to write an equation of each line.

12. slope 5, passes through

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’ (βˆ’πŸ‘ )=βˆ’πŸ“ (π’™βˆ’πŸ)

π’š+πŸ‘=βˆ’πŸ“ (π’™βˆ’πŸ)

π’š+πŸ‘=βˆ’πŸ“ 𝒙+πŸπŸŽπ’š=βˆ’πŸ“ 𝒙+πŸ•

HomeworkMrs. RivasUse the given information to write an equation of each line.

13. Slope , passes through

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=πŸ‘πŸ’

(π’™βˆ’(βˆ’πŸ–))

π’š βˆ’πŸ=πŸ‘πŸ’π’™+πŸ”

π’š=πŸ‘πŸ’π’™+πŸ–

HomeworkMrs. RivasUse the given information to write an equation of each line.

14. passes through and

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

ΒΏβˆ’πŸβˆ’πŸ”πŸ’βˆ’πŸŽ

ΒΏβˆ’πŸ–πŸ’

ΒΏβˆ’πŸ

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ”=βˆ’πŸ(π’™βˆ’πŸŽ)

π’š βˆ’πŸ”=βˆ’πŸ 𝒙+𝟎

π’š=βˆ’πŸ 𝒙+πŸ”

HomeworkMrs. RivasUse the given information to write an equation of each line.

15. passes through and

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

ΒΏβˆ’πŸ’βˆ’πŸ–πŸ“βˆ’(βˆ’πŸ)

ΒΏβˆ’πŸπŸπŸ”

ΒΏβˆ’πŸ

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ–=βˆ’πŸ(π’™βˆ’(βˆ’πŸ))

π’š βˆ’πŸ–=βˆ’πŸ(𝒙+𝟏)

π’š βˆ’πŸ–=βˆ’πŸ π’™βˆ’πŸπ’š=βˆ’πŸ 𝒙+πŸ”

HomeworkMrs. RivasWrite the equations of the horizontal and vertical lines through the given

point.16.

π‘―π’π’“π’Šπ’›π’π’π’•π’‚π’ π‘³π’Šπ’π’† :π’š=πŸ”

π‘½π’†π’“π’•π’Šπ’„π’‚π’ π‘³π’Šπ’π’† :𝒙=πŸ“

HomeworkMrs. RivasWrite the equations of the horizontal and vertical lines through the given

point.17.

π‘―π’π’“π’Šπ’›π’π’π’•π’‚π’ π‘³π’Šπ’π’† :π’š=βˆ’πŸ‘

π‘½π’†π’“π’•π’Šπ’„π’‚π’ π‘³π’Šπ’π’† :𝒙=βˆ’πŸ

HomeworkMrs. RivasWrite the equations of the horizontal and vertical lines through the given

point.18.

π‘―π’π’“π’Šπ’›π’π’π’•π’‚π’ π‘³π’Šπ’π’† :π’š=βˆ’πŸ

π‘½π’†π’“π’•π’Šπ’„π’‚π’ π‘³π’Šπ’π’† :𝒙=πŸ–

HomeworkMrs. RivasWrite the equations of the horizontal and vertical lines through the given

point.19.

π‘―π’π’“π’Šπ’›π’π’π’•π’‚π’ π‘³π’Šπ’π’† :π’š=𝟎

π‘½π’†π’“π’•π’Šπ’„π’‚π’ π‘³π’Šπ’π’† :𝒙=𝟏𝟎

HomeworkMrs. RivasWrite each equation in slope-intercept form.

20.

π’š βˆ’πŸ“=πŸ‘(𝒙 βˆ’πŸ’)

π’š βˆ’πŸ“=πŸ‘ π’™βˆ’πŸπŸπ’š=πŸ‘ π’™βˆ’πŸ•

π’š=π’Žπ’™+𝒃

HomeworkMrs. RivasWrite each equation in slope-intercept form.

21.

π’š+𝟐=βˆ’πŸ“ (π’™βˆ’πŸ)

π’š+𝟐=βˆ’πŸ“ 𝒙+πŸ“π’š=βˆ’πŸ“ 𝒙+πŸ‘

π’š=π’Žπ’™+𝒃

HomeworkMrs. RivasWrite each equation in slope-intercept form.

22.

𝟐 𝒙+πŸ’ π’š=πŸ–

π’š=π’Žπ’™+𝒃

βˆ’πŸ 𝒙 βˆ’πŸ π’™πŸ’ π’š=βˆ’πŸ 𝒙+πŸ–πŸ’ πŸ’ πŸ’

π’š=βˆ’πŸπŸπ’™+𝟐

HomeworkMrs. RivasWrite each equation in slope-intercept form.

23.

πŸπŸŽπ’š+πŸπŸ”π’™+πŸ’=𝟐 π’š

π’š=π’Žπ’™+𝒃

βˆ’πŸπŸŽπ’š βˆ’πŸπŸŽπ’šπŸπŸ”π’™+πŸ’=βˆ’πŸ– π’šβˆ’πŸ– βˆ’πŸ– βˆ’πŸ–

π’š=βˆ’πŸ π’™βˆ’πŸπŸ

HomeworkMrs. Rivas24.Coordinate Geometry The vertices of a quadrilateral are , ,

, and .a. Write an equation for the line through A and B.

ΒΏπŸ’βˆ’πŸ

πŸβˆ’(βˆ’πŸ)

ΒΏπŸ’βˆ’πŸπŸ+𝟏

¿33=𝟏

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=𝟏(𝒙 βˆ’(βˆ’πŸ))

π’š βˆ’πŸ=𝟏(𝒙+𝟏)

π’š βˆ’πŸ=𝒙+πŸπ’š=𝒙+𝟐

HomeworkMrs. Rivas24.Coordinate Geometry The vertices of a quadrilateral are , ,

, and .

b. Write an equation for the line through C and D.

ΒΏβˆ’πŸβˆ’(βˆ’πŸ’)πŸŽβˆ’πŸ

ΒΏπŸβˆ’πŸ

ΒΏβˆ’πŸ

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’ (βˆ’πŸ’ )=βˆ’πŸ(𝒙 βˆ’πŸ)

π’š+πŸ’=βˆ’(𝒙 βˆ’πŸ)

π’š+πŸ’=βˆ’ 𝒙+πŸπ’š=βˆ’π’™βˆ’πŸ

HomeworkMrs. Rivas24.Coordinate Geometry The vertices of a quadrilateral are , ,

, and .c. Without graphing the lines, what can you tell about the lines from their

slopes?

π’š=𝒙+𝟐 π’š=βˆ’π’™βˆ’πŸ

One line has a positive slope and the other has a negative slope.

We can also say that they are perpendicular since their slopes are opposite reciprocal.

HomeworkMrs. RivasFor Exercises 25 and 26, are lines and parallel? Explain.

25.

β„“πŸ  =πŸβˆ’πŸŽ

πŸ‘βˆ’(βˆ’πŸ‘)

ΒΏπŸπŸ”

ΒΏπŸπŸ‘

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

β„“πŸ  =βˆ’πŸβˆ’(βˆ’πŸ‘)πŸ“βˆ’(βˆ’πŸ)

ΒΏπŸπŸ”

ΒΏπŸπŸ‘

Yes, the lines are parallel because the have the same slopes.

HomeworkMrs. RivasFor Exercises 25 and 26, are lines and parallel? Explain.

26.

β„“πŸ  =βˆ’πŸ‘βˆ’πŸ”βˆ’πŸβˆ’(βˆ’πŸ‘)

ΒΏβˆ’πŸ—πŸ

ΒΏβˆ’πŸ—πŸ

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

β„“πŸ  =πŸŽβˆ’πŸ–πŸ”βˆ’πŸ’

ΒΏβˆ’πŸ–πŸ

ΒΏβˆ’πŸ’

No, the lines are NOT parallel because the don’t have the same slopes.

HomeworkMrs. RivasWrite an equation of the line parallel to the given line that contains.

27. β€œSame Slope”

π’Ž=βˆ’πŸ“ π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)π’š βˆ’(βˆ’πŸ)=βˆ’πŸ“(π’™βˆ’πŸ“)

π’š+𝟐=βˆ’πŸ“ (π’™βˆ’πŸ“)Use the distributive property

π’š+𝟐=βˆ’πŸ“ 𝒙+πŸπŸ“Solve for y:

π’š=βˆ’πŸ“ 𝒙+πŸπŸ‘

HomeworkMrs. RivasWrite an equation of the line parallel to the given line that contains.

28. β€œSame Slope”

π’Ž=𝟐 π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=𝟐(𝒙 βˆ’πŸ–)Use the distributive property

π’š βˆ’πŸ=𝟐 π’™βˆ’πŸπŸ”Solve for y:

π’š=𝟐 π’™βˆ’πŸπŸ“

HomeworkMrs. RivasWrite an equation of the line parallel to the given line that contains.

29. β€œSame Slope”

π’Ž=πŸπŸ‘

π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ”=πŸπŸ‘

(π’™βˆ’πŸŽ)Use the distributive property

π’š βˆ’πŸ”=πŸπŸ‘π’™βˆ’πŸŽSolve for y:

π’š=πŸπŸ‘π’™+πŸ”

HomeworkMrs. RivasRewrite each equation in slope-intercept form, if necessary. Then

determine whether the lines are parallel. Explain.

30.

π’š=π’Žπ’™+𝒃

𝟐 π’š+πŸ” 𝒙=πŸπŸ–βˆ’πŸ” 𝒙 βˆ’πŸ” π’™πŸ π’š=βˆ’πŸ” 𝒙+πŸπŸ–πŸ 𝟐 𝟐

π’š=βˆ’πŸ‘ 𝒙+πŸ—

πŸ’ π’š+πŸπŸπ’™=πŸπŸ’βˆ’πŸπŸπ’™ βˆ’πŸπŸπ’™πŸ’ π’š=βˆ’πŸπŸπ’™+πŸπŸ–πŸ’ πŸ’ πŸ’

π’š=βˆ’πŸ‘ 𝒙+πŸ—

Yes, the lines are parallel because the have the same slopes.

HomeworkMrs. RivasRewrite each equation in slope-intercept form, if necessary. Then

determine whether the lines are parallel. Explain.

31.

π’š=π’Žπ’™+𝒃

π’š=𝒙+πŸ– π’™βˆ’πŸπ’š=πŸ’βˆ’π’™ βˆ’π’™

βˆ’πŸ π’š=βˆ’π’™+πŸ’βˆ’πŸ βˆ’πŸ βˆ’πŸ

π’š=βˆ’πŸπŸπ’™βˆ’πŸ

No, the lines are NOT parallel because the don’t have the same slopes.

HomeworkMrs. RivasRewrite each equation in slope-intercept form, if necessary. Then

determine whether the lines are parallel. Explain.

32.

πŸ’ π’š βˆ’πŸ‘ 𝒙=𝟐𝟎+πŸ‘ 𝒙 +πŸ‘ π’™πŸ’ π’š=πŸ‘ 𝒙+πŸπŸŽπŸ’ πŸ’ πŸ’

π’š=πŸ‘πŸ’π’™+πŸ“

𝟐 π’š=πŸ‘πŸπ’™+πŸ’

𝟐𝟐

𝟐

Yes, the lines are parallel because the have the same slopes.

32Γ·21ΒΏ32Γ—12ΒΏ34

π’š=πŸ‘πŸ’π’™+𝟐

HomeworkMrs. RivasUse slopes to determine whether the opposite sides of

quadrilateral WXYZ are parallel.

33.

𝑾 𝑿

𝒀𝒁

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

𝑾𝑿=βˆ’πŸβˆ’(βˆ’πŸ)βˆ’πŸ‘βˆ’(βˆ’πŸ)

ΒΏβˆ’πŸ+πŸβˆ’πŸ‘+𝟏

ΒΏπŸŽβˆ’πŸ

¿𝟎

𝒁𝒀=πŸ‘βˆ’πŸ’

πŸβˆ’(βˆ’πŸ)

ΒΏπŸ‘βˆ’πŸ’πŸ+𝟐

ΒΏβˆ’πŸπŸ’

ΒΏβˆ’πŸπŸ’

HomeworkMrs. RivasUse slopes to determine whether the opposite sides of

quadrilateral WXYZ are parallel.

33.

𝑾 𝑿

𝒀𝒁

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

𝑾 𝒁=βˆ’πŸβˆ’πŸ‘βˆ’πŸβˆ’πŸ

ΒΏβˆ’πŸ’βˆ’πŸ‘

ΒΏπŸ’πŸ‘

𝟎

𝑿𝒀=πŸ’βˆ’(βˆ’πŸ)βˆ’πŸβˆ’(βˆ’πŸ‘)

ΒΏπŸ’+πŸβˆ’πŸ+πŸ‘

ΒΏπŸ“βˆ’πŸ

βˆ’πŸπŸ’

No, the lines are NOT parallel because the don’t have the same slopes.

ΒΏβˆ’πŸ“

HomeworkMrs. RivasUse slopes to determine whether the opposite sides of

quadrilateral WXYZ are parallel.

34.

𝑾 𝑿

π’€π’π’Ž=

π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

𝑾𝑿=πŸ’βˆ’πŸ

πŸβˆ’(βˆ’πŸ)

ΒΏπŸ’βˆ’πŸπŸ+𝟏

ΒΏπŸ‘πŸ‘

¿𝟏

𝒁𝒀=βˆ’πŸβˆ’πŸπŸβˆ’πŸ’

ΒΏβˆ’πŸ‘βˆ’πŸ‘

¿𝟏

Yes, the lines are parallel because the have the same slopes.

HomeworkMrs. RivasFor Exercises 35 and 36, are lines and perpendiular? Explain.

35.

β„“πŸ  =βˆ’πŸ’βˆ’(βˆ’πŸ)πŸβˆ’(βˆ’πŸ)

ΒΏβˆ’πŸ’+𝟏𝟏+𝟐

ΒΏβˆ’πŸ“πŸ‘

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

β„“πŸ  =βˆ’πŸ–βˆ’(βˆ’πŸ‘)βˆ’πŸβˆ’πŸ“

ΒΏβˆ’πŸ–+πŸ‘βˆ’πŸ•

ΒΏπŸ“πŸ•

No, the lines are NOT Perpendicular because the don’t have opposite reciprocal slopes.

HomeworkMrs. RivasFor Exercises 35 and 36, are lines and perpendiular? Explain.

36.

β„“πŸ  =βˆ’πŸβˆ’πŸ”βˆ’πŸβˆ’(βˆ’πŸ“)

ΒΏβˆ’πŸβˆ’πŸ”βˆ’πŸ+πŸ“

ΒΏβˆ’πŸ–πŸ’

π’Ž=π’šπŸβˆ’π’šπŸ

π’™πŸβˆ’π’™πŸ

β„“πŸ  =πŸ‘βˆ’πŸŽ

πŸβˆ’(βˆ’πŸ“)

ΒΏπŸ‘βˆ’πŸŽπŸ+πŸ“

¿𝟏𝟐

ΒΏβˆ’πŸ

ΒΏπŸ‘πŸ”

Yes, the lines are Perpendicular because the have opposite reciprocal slopes.

HomeworkMrs. RivasWrite an equation of the line perpendicular to the given line that

contains D.37. β€œOpposite Reciprocal slopeβ€π’Ž=

πŸπŸ‘ π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=πŸπŸ‘

(π’™βˆ’πŸ”)Use the distributive property

π’š βˆ’πŸ=πŸπŸ‘π’™βˆ’πŸSolve for y:

π’š=πŸπŸ‘π’™

HomeworkMrs. RivasWrite an equation of the line perpendicular to the given line that

contains D.38. β€œOpposite Reciprocal slope”

π’Ž=βˆ’πŸπŸ

=βˆ’πŸ π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)π’š βˆ’(βˆ’πŸ‘)=βˆ’πŸ(π’™βˆ’πŸŽ)

Use the distributive property

π’š+πŸ‘=βˆ’πŸ (π’™βˆ’πŸŽ)

π’š+πŸ‘=βˆ’πŸ π’™βˆ’πŸŽSolve for y:

π’š=βˆ’πŸ π’™βˆ’πŸ‘

HomeworkMrs. RivasWrite an equation of the line perpendicular to the given line that

contains D.39. β€œOpposite Reciprocal slopeβ€π’Ž=

πŸ‘πŸ π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=πŸ‘πŸ

(π’™βˆ’(βˆ’πŸ–))

π’š βˆ’πŸ=πŸ‘πŸ

(𝒙+πŸ–)Use the distributive property

π’š βˆ’πŸ=πŸ‘πŸπ’™+𝟏𝟐

π’š=πŸ‘πŸπ’™+πŸπŸ‘

Solve for y:

HomeworkMrs. RivasWrite an equation of the line perpendicular to the given line that

contains D.40. β€œOpposite Reciprocal slopeβ€π’Ž=βˆ’

πŸπŸ“ π’š βˆ’π’šπŸ=π’Ž(π’™βˆ’π’™πŸ)

π’š βˆ’πŸ=βˆ’πŸπŸ“

(π’™βˆ’πŸ)Use the distributive property

π’š βˆ’πŸ=βˆ’πŸπŸ“π’™+

πŸπŸ“Solve for y:

25+21ΒΏ2+105

ΒΏ125 π’š=βˆ’

πŸπŸ“π’™+

πŸπŸπŸ“