10th Grade Coordinates: Quadrants, Midpoints and Distance Worksheet — Level 3 — Practice Set 2 — Answers
Answer sheet with worked solutions for the 10th Grade geometry worksheet.
10th Grade Coordinates: Quadrants, Midpoints and Distance Worksheet — Level 3 — Practice Set 2 — Answers
Midpoint = (average of the x-values, average of the y-values). Distance: use Pythagoras with the horizontal and vertical changes.
1.Find the distance between (1, 1) and (10, −11).
Answer: 15
- Horizontal change = 9, vertical change = 12
- d = √(92 + 122) = √225
- = 15
2.Find the midpoint of the segment joining (6, −1) and (1, 7).
Answer: (3.5, 3)
- x: (6 + 1) ÷ 2 = 3.5
- y: (−1 + 7) ÷ 2 = 3
3.Find the midpoint of the segment joining (6, −5) and (8, 0).
Answer: (7, −2.5)
- x: (6 + 8) ÷ 2 = 7
- y: (−5 + 0) ÷ 2 = −2.5
4.Find the distance between (3, 6) and (−6, −6).
Answer: 15
- Horizontal change = 9, vertical change = 12
- d = √(92 + 122) = √225
- = 15
5.Find the midpoint of the segment joining (9, −5) and (−8, 5).
Answer: (0.5, 0)
- x: (9 + (−8)) ÷ 2 = 0.5
- y: (−5 + 5) ÷ 2 = 0
6.Find the midpoint of the segment joining (4, −3) and (−6, −2).
Answer: (−1, −2.5)
- x: (4 + (−6)) ÷ 2 = −1
- y: (−3 + (−2)) ÷ 2 = −2.5
7.Find the distance between (−3, 2) and (−12, 14).
Answer: 15
- Horizontal change = 9, vertical change = 12
- d = √(92 + 122) = √225
- = 15
8.Find the midpoint of the segment joining (−1, −4) and (3, −3).
Answer: (1, −3.5)
- x: (−1 + 3) ÷ 2 = 1
- y: (−4 + (−3)) ÷ 2 = −3.5
9.Find the distance between (1, 6) and (7, −2).
Answer: 10
- Horizontal change = 6, vertical change = 8
- d = √(62 + 82) = √100
- = 10
10.Find the midpoint of the segment joining (−4, 6) and (2, −3).
Answer: (−1, 1.5)
- x: (−4 + 2) ÷ 2 = −1
- y: (6 + (−3)) ÷ 2 = 1.5
11.Find the distance between (4, −5) and (9, 7).
Answer: 13
- Horizontal change = 5, vertical change = 12
- d = √(52 + 122) = √169
- = 13
12.Find the midpoint of the segment joining (8, 4) and (0, −5).
Answer: (4, −0.5)
- x: (8 + 0) ÷ 2 = 4
- y: (4 + (−5)) ÷ 2 = −0.5
Marking tips
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