Mrs. Rivas 𝟐(𝟐𝒙 + 𝟏) = 𝟏𝟖 𝟒𝒙 + 𝟐 = 𝟏𝟖 𝟒𝒙 = 𝟏𝟔 𝒙 = 𝟒 (5-1) Algebra Find the value of x. 1. 𝟐(𝟐𝒙 + 𝟏) = 𝟏𝟖 𝟒𝒙 + 𝟐 = 𝟏𝟖 𝟒𝒙 = 𝟏𝟔 𝒙 = 𝟒

Mrs. Rivas 𝟐(𝟑𝒙) = 𝟑𝟎 𝟔𝒙 = 𝟑𝟎 𝒙 = 𝟓 (5-1) Algebra Find the value of x. 2. 𝟐(𝟑𝒙) = 𝟑𝟎 𝟔𝒙 = 𝟑𝟎 𝒙 = 𝟓

Mrs. Rivas (5-1) Algebra Find the value of x. 3. 𝟐(𝟑𝒙) = 𝟐𝟏 𝟔𝒙 = 𝟐𝟏 𝒙 = 𝟑.𝟓

Mrs. Rivas 𝟗 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 . 4. Find XZ. 𝟗

Mrs. Rivas 𝟐𝟎 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 . 5. If XY = 10, find MO. 𝟐𝟎

Mrs. Rivas 𝟔𝟒 𝟔𝟒 X is the midpoint of 𝑴𝑵 . Y is the midpoint of 𝑶𝑵 . 6. If 𝑚𝑀 is 64, find 𝑚𝑌. 𝟔𝟒 𝟔𝟒

Mrs. Rivas 𝟓.𝟓 Use the diagram at the right for Exercises 7 and 8. 7. What is the distance across the lake? 𝟓.𝟓

Mrs. Rivas 𝟒 𝟓.𝟓 BC is shorter. BC is half od 8 and AB is half od 11. Use the diagram at the right for Exercises 7 and 8. 8. Is it a shorter distance from A to B or from B to C? Explain. 𝟒 BC is shorter. BC is half od 8 and AB is half od 11. 𝟓.𝟓

Mrs. Rivas 𝟓𝒙 + 𝟑 = 𝟕𝒙 − 𝟏 𝟓𝒙 𝟓𝒙 𝟑 = 𝟐𝒙 − 𝟏 + 𝟏 + 𝟏 𝟒 = 𝟐𝒙 𝟐 = 𝒙 (5-2) Algebra Find the indicated variables and measures. 10. x, EH, EF 𝟓𝒙 + 𝟑 = 𝟕𝒙 − 𝟏 𝟓𝒙 𝟓𝒙 𝟑 = 𝟐𝒙 − 𝟏 + 𝟏 + 𝟏 𝟒 = 𝟐𝒙 𝟐 = 𝒙 𝑬𝑯=𝟓𝒙+𝟑=𝟓 𝟐 +𝟑=𝟏𝟑 𝑬𝑭=𝟕𝒙−𝟏=𝟕 𝟐 −𝟏=𝟏𝟑

Mrs. Rivas 𝟐𝒙 – 𝟔 = 𝟑𝒙 – 𝟐𝟓 𝟐𝒙 𝟐𝒙 – 𝟔 = 𝒙 – 𝟐𝟓 + 𝟐𝟓 + 𝟐𝟓 𝟏𝟗 = 𝒙 (5-2) Algebra Find the indicated variables and measures. 11. x, mTPS, mRPS 𝟐𝒙 – 𝟔 = 𝟑𝒙 – 𝟐𝟓 𝟐𝒙 𝟐𝒙 – 𝟔 = 𝒙 – 𝟐𝟓 + 𝟐𝟓 + 𝟐𝟓 𝟏𝟗 = 𝒙 𝒎∠𝑻𝑷𝑺=𝟐𝒙−𝟔=𝟐 𝟏𝟗 −𝟔=𝟑𝟐 𝒎∠𝑹𝑷𝑺=𝟑𝒙−𝟐𝟓=𝟑 𝟏𝟗 −𝟐𝟓=𝟑𝟐

Mrs. Rivas 𝟑𝒂 – 𝟐 = 𝒂 + 𝟏𝟎 𝟐𝒂 – 𝟐 = 𝟏𝟎 𝟐𝒂 = 𝟏𝟐 𝒂 = 𝟔 𝟑𝒃 – 𝟏𝟓 = 𝟐𝒃 + 𝟓 (5-2) Algebra Find the indicated variables and measures. 12. a, b 𝟑𝒂 – 𝟐 = 𝒂 + 𝟏𝟎 𝟐𝒂 – 𝟐 = 𝟏𝟎 𝟐𝒂 = 𝟏𝟐 𝒂 = 𝟔 𝟑𝒃 – 𝟏𝟓 = 𝟐𝒃 + 𝟓 𝒃 – 𝟏𝟓 = 𝟓 𝒃 =𝟐𝟎

Mrs. Rivas 1.

Mrs. Rivas 2.

Mrs. Rivas 3.

Mrs. Rivas 4.

Mrs. Rivas 5.

Mrs. Rivas 6.

Mrs. Rivas 7.

Mrs. Rivas 𝒙 + 𝟓 = 𝟑𝒙 + 𝟕 8. −𝟐𝒙 + 𝟓 = 𝟕 −𝟐𝒙 =𝟐 𝒙=−𝟏

Mrs. Rivas 𝒙 + 𝟐 = 𝟐𝒙 – 𝟑 9. −𝒙 + 𝟐 = – 𝟑 −𝒙 = –𝟓 𝒙 = 𝟓

Mrs. Rivas 10. 𝟓𝒙 + 𝟕 = 𝒙 + 𝟖 𝟒𝒙 + 𝟕 = 𝟖 𝟒𝒙 =𝟏 𝒙 = 𝟏 𝟒

Mrs. Rivas 𝑪𝑿= 𝟐 𝟑 𝑪𝑾 𝑪𝑿= 𝟐 𝟑 (𝟏𝟓) 𝑿𝑾=𝑪𝑾−𝑪𝑿 𝑪𝑿= 𝟑𝟎 𝟑 𝑿𝑾=𝟏𝟓−𝟏𝟎 𝑿𝑾=𝟓 (5-4) In ∆ABC, X is the centroid. 11. If CW = 15, find CX and XW. 𝑪𝑿= 𝟐 𝟑 𝑪𝑾 𝟏𝟓 𝑪𝑿= 𝟐 𝟑 (𝟏𝟓) 𝑿𝑾=𝑪𝑾−𝑪𝑿 𝑪𝑿= 𝟑𝟎 𝟑 𝑿𝑾=𝟏𝟓−𝟏𝟎 𝑿𝑾=𝟓 𝑪𝑿=𝟏𝟎

Mrs. Rivas 𝑩𝑿= 𝟐 𝟑 𝑩𝒀 𝟖= 𝟐 𝟑 𝑩𝒀 𝑿𝒀=𝑩𝒀−𝑩𝑿 𝟑 𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝑿𝒀=𝟏𝟐−𝟖 (5-4) In ∆ABC, X is the centroid. 12. If BX = 8, find BY and XY. 𝑩𝑿= 𝟐 𝟑 𝑩𝒀 𝟖 𝟏𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝟖= 𝟐 𝟑 𝑩𝒀 𝟑 𝟐 𝑿𝒀=𝑩𝒀−𝑩𝑿 𝑿𝒀=𝟏𝟐−𝟖 𝟐𝟒 𝟐 =𝑩𝒀 𝑿𝑾=𝟒 𝟏𝟐=𝑩𝒀

Mrs. Rivas 𝑿𝒁= 𝟏 𝟑 𝑨𝒁 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝑨𝑿=𝑨𝒁−𝑿𝒁 𝑨𝑿=𝟗−𝟑 𝟗=𝑨𝒁 (5-4) In ∆ABC, X is the centroid. 13. If XZ = 3, find AX and AZ. 𝑿𝒁= 𝟏 𝟑 𝑨𝒁 𝟗 𝟑 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝟑= 𝟏 𝟑 𝑨𝒁 𝟑 𝟏 𝑨𝑿=𝑨𝒁−𝑿𝒁 𝑨𝑿=𝟗−𝟑 𝟗=𝑨𝒁 𝑨𝑿=𝟔

Mrs. Rivas Is 𝑨𝑩 a median, an altitude, or neither? Explain. 15. 14. Altitude; 𝑨𝑩 is perpendicular to the opposite side. Median; 𝑨𝑩 bisects the opposite side.

Mrs. Rivas Is 𝑨𝑩 a median, an altitude, or neither? Explain. 17. 16. Altitude; 𝑨𝑩 is perpendicular to the opposite side. Neither; 𝑨𝑩 is not perpendicular to nor does it bisect the opposite side.

Mrs. Rivas 𝑪𝑱 In Exercises 18–22, name each segment. 18. a median in ∆ABC 𝑪𝑱

Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment. 19. an altitude for ∆ABC 𝑨𝑯

Mrs. Rivas 𝑰𝑯 In Exercises 18–22, name each segment. 20. a median in ∆AHC 𝑰𝑯

Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment. 21. an altitude for ∆AHB 𝑨𝑯

Mrs. Rivas 𝑨𝑯 In Exercises 18–22, name each segment. 22. an altitude for ∆AHG. 𝑨𝑯

Mrs. Rivas 23. A(0, 0), B(0, 2), C(3, 0). Find the orthocenter of ∆ABC.

Mrs. Rivas 24. In which kind of triangle is the centroid at the same point as the orthocenter? equilateral

Mrs. Rivas 𝑴𝒆𝒅𝒊𝒂𝒏 𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒄𝒐𝒏𝒄𝒖𝒓𝒓𝒆𝒏𝒄𝒚 𝑪𝒆𝒏𝒕𝒓𝒐𝒊𝒅 𝑰𝒏𝒄𝒆𝒏𝒕𝒆𝒓 𝑪𝒐𝒏𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒍𝒊𝒏𝒆𝒔 𝑪𝒊𝒓𝒄𝒖𝒎𝒄𝒆𝒏𝒕𝒆𝒓 𝑨𝒍𝒕𝒊𝒕𝒖𝒕𝒆 𝑶𝒓𝒕𝒉𝒐𝒄𝒆𝒏𝒕𝒆𝒓 𝑽𝒆𝒓𝒕𝒆𝒙