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7th Grade Coordinates: Quadrants, Midpoints and Distance Worksheet — Level 3 — Practice Set 2 — 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 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 midpoint of the segment joining (6, −2) and (9, 6).

Answer: (7.5, 2)

  1. x: (6 + 9) ÷ 2 = 7.5
  2. y: (−2 + 6) ÷ 2 = 2
2.Find the midpoint of the segment joining (−5, 3) and (−2, 7).

Answer: (−3.5, 5)

  1. x: (−5 + (−2)) ÷ 2 = −3.5
  2. y: (3 + 7) ÷ 2 = 5
3.Find the distance between (−4, −4) and (−12, −5), correct to 1 decimal place.

Answer: 8.1

  1. Horizontal change = 8, vertical change = 1
  2. d = √(82 + 12) = √65
  3. = 8.1
4.Find the midpoint of the segment joining (−8, −7) and (1, 6).

Answer: (−3.5, −0.5)

  1. x: (−8 + 1) ÷ 2 = −3.5
  2. y: (−7 + 6) ÷ 2 = −0.5
5.Find the midpoint of the segment joining (0, −4) and (−7, −4).

Answer: (−3.5, −4)

  1. x: (0 + (−7)) ÷ 2 = −3.5
  2. y: (−4 + (−4)) ÷ 2 = −4
6.Find the midpoint of the segment joining (7, −2) and (8, −3).

Answer: (7.5, −2.5)

  1. x: (7 + 8) ÷ 2 = 7.5
  2. y: (−2 + (−3)) ÷ 2 = −2.5
7.Find the distance between (5, −5) and (−2, −9), correct to 1 decimal place.

Answer: 8.1

  1. Horizontal change = 7, vertical change = 4
  2. d = √(72 + 42) = √65
  3. = 8.1
8.Find the midpoint of the segment joining (7, 0) and (1, −3).

Answer: (4, −1.5)

  1. x: (7 + 1) ÷ 2 = 4
  2. y: (0 + (−3)) ÷ 2 = −1.5
9.Find the midpoint of the segment joining (7, 4) and (−5, −7).

Answer: (1, −1.5)

  1. x: (7 + (−5)) ÷ 2 = 1
  2. y: (4 + (−7)) ÷ 2 = −1.5
10.Find the distance between (2, 2) and (10, 17).

Answer: 17

  1. Horizontal change = 8, vertical change = 15
  2. d = √(82 + 152) = √289
  3. = 17
11.Find the distance between (−1, −1) and (8, −13).

Answer: 15

  1. Horizontal change = 9, vertical change = 12
  2. d = √(92 + 122) = √225
  3. = 15
12.Find the distance between (2, −2) and (−1, −6).

Answer: 5

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