mrs. rivas

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H om ew ork Mrs. Rivas Find the slope of the line passing through the given points. 1. = βˆ’ βˆ’ ΒΏ βˆ’ βˆ’ βˆ’ ΒΏ βˆ’ ΒΏ βˆ’

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Mrs. Rivas. Find the slope of the line passing through the given points. 1. Mrs. Rivas. Find the slope of the line passing through the given points. 2. Mrs. Rivas. Find the slope of the line passing through the given points. 3. Mrs. Rivas. - PowerPoint PPT Presentation

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Page 1: Mrs. Rivas

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

Page 2: Mrs. Rivas

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

Page 3: Mrs. Rivas

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

Page 4: Mrs. Rivas

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

4.

ΒΏπŸ•πŸ”

Page 5: Mrs. Rivas

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

5.

ΒΏβˆ’πŸ’πŸ‘

Page 6: Mrs. Rivas

HomeworkMrs. RivasGraph each line.

6.

Starting point

π’š=π’Žπ’™+𝒃

Page 7: Mrs. Rivas

HomeworkMrs. RivasGraph each line.

7.

Starting point

π’š=π’Žπ’™+𝒃

Page 8: Mrs. Rivas

HomeworkMrs. RivasGraph each line.

8.

Starting point

π’š=π’Žπ’™+𝒃

Page 9: Mrs. Rivas

HomeworkMrs. RivasGraph each line.

9.

Starting point

π’š=π’Žπ’™+𝒃

Page 10: Mrs. Rivas

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”

Page 11: Mrs. Rivas

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‘

Page 12: Mrs. Rivas

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•

Page 13: Mrs. Rivas

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–

Page 14: Mrs. Rivas

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”

Page 15: Mrs. Rivas

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”

Page 16: Mrs. Rivas

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“

Page 17: Mrs. Rivas

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

Page 18: Mrs. Rivas

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–

Page 19: Mrs. Rivas

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

Page 20: Mrs. Rivas

HomeworkMrs. RivasWrite each equation in slope-intercept form.

20.

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

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

Page 21: Mrs. Rivas

HomeworkMrs. RivasWrite each equation in slope-intercept form.

21.

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

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

Page 22: Mrs. Rivas

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

Page 23: Mrs. Rivas

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

Page 24: Mrs. Rivas

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

Page 25: Mrs. Rivas

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

Page 26: Mrs. Rivas

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.

Page 27: Mrs. Rivas

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.

Page 28: Mrs. Rivas

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.

Page 29: Mrs. Rivas

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‘

Page 30: Mrs. Rivas

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“

Page 31: Mrs. Rivas

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”

Page 32: Mrs. Rivas

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.

Page 33: Mrs. Rivas

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.

Page 34: Mrs. Rivas

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

Page 35: Mrs. Rivas

HomeworkMrs. RivasUse slopes to determine whether the opposite sides of

quadrilateral WXYZ are parallel.

33.

𝑾 𝑿

π’€π’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’

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

Page 36: Mrs. Rivas

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.

Page 37: Mrs. Rivas

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.

Page 38: Mrs. Rivas

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.

Page 39: Mrs. Rivas

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’™

Page 40: Mrs. Rivas

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‘

Page 41: Mrs. Rivas

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:

Page 42: Mrs. Rivas

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“