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7th Grade Coordinates: Quadrants, Midpoints and Distance Worksheet — Level 3 — Practice Set 4 — Answers

Answer sheet with worked solutions for the 7th Grade geometry worksheet.

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7th Grade Coordinates: Quadrants, Midpoints and Distance Worksheet — Level 3 — Practice Set 4 — 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 (−6, 3) and (−15, −9).

Answer: 15

  1. Horizontal change = 9, vertical change = 12
  2. d = √(92 + 122) = √225
  3. = 15
2.Find the midpoint of the segment joining (5, 5) and (2, 5).

Answer: (3.5, 5)

  1. x: (5 + 2) ÷ 2 = 3.5
  2. y: (5 + 5) ÷ 2 = 5
3.Find the midpoint of the segment joining (−8, −5) and (7, 2).

Answer: (−0.5, −1.5)

  1. x: (−8 + 7) ÷ 2 = −0.5
  2. y: (−5 + 2) ÷ 2 = −1.5
4.Find the distance between (−6, −3) and (−11, 3), correct to 1 decimal place.

Answer: 7.8

  1. Horizontal change = 5, vertical change = 6
  2. d = √(52 + 62) = √61
  3. = 7.8
5.Find the midpoint of the segment joining (−3, 2) and (1, 1).

Answer: (−1, 1.5)

  1. x: (−3 + 1) ÷ 2 = −1
  2. y: (2 + 1) ÷ 2 = 1.5
6.Find the midpoint of the segment joining (−9, 8) and (5, −5).

Answer: (−2, 1.5)

  1. x: (−9 + 5) ÷ 2 = −2
  2. y: (8 + (−5)) ÷ 2 = 1.5
7.Find the distance between (−2, −3) and (7, −15).

Answer: 15

  1. Horizontal change = 9, vertical change = 12
  2. d = √(92 + 122) = √225
  3. = 15
8.Find the midpoint of the segment joining (0, 5) and (−8, 6).

Answer: (−4, 5.5)

  1. x: (0 + (−8)) ÷ 2 = −4
  2. y: (5 + 6) ÷ 2 = 5.5
9.Find the distance between (1, −3) and (9, 12).

Answer: 17

  1. Horizontal change = 8, vertical change = 15
  2. d = √(82 + 152) = √289
  3. = 17
10.Find the midpoint of the segment joining (0, −5) and (−8, 9).

Answer: (−4, 2)

  1. x: (0 + (−8)) ÷ 2 = −4
  2. y: (−5 + 9) ÷ 2 = 2
11.Find the distance between (4, 2) and (0, 7), correct to 1 decimal place.

Answer: 6.4

  1. Horizontal change = 4, vertical change = 5
  2. d = √(42 + 52) = √41
  3. = 6.4
12.Find the distance between (2, −5) and (−1, −1).

Answer: 5

  1. Horizontal change = 3, vertical change = 4
  2. d = √(32 + 42) = √25
  3. = 5
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