Distance and midpoint in the coordinate plane

About this lesson

Grade 8 › Pythagorean theorem and irrational numbers

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Distance and midpoint in the coordinate plane

Warm-up

Three quick questions from earlier lessons.

Discover

From (1,1)(1, 1) to (4,5)(4, 5) you move 3 right and 4 up. How long is the straight segment?

  1. 5
  2. 7

Put A at (0, 0) and B at (4, 6). Where is M?

  1. (2, 3)
  2. (4, 6)

Key idea

Pythagoras on the grid, and averaging for the middle.

Distance: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Midpoint: (x1+x22,y1+y22)\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right).

Worked example

A(−2,3)A(-2, 3), B(4,−5)B(4, -5).

d=36+64=10d = \sqrt{36 + 64} = 10
Δx = 6, Δy = −8.
Midpoint (1,−1)(1, -1)
Average the coordinates.

Your turn, guided

Your turn: the distance from (0,0)(0, 0) to (5,12)(5, 12).

1313
√(25 + 144).

Practice ★

  1. Find the distance between (−1,−3)(-1, -3) and (2,1)(2, 1).
  2. Find the distance between (−3,3)(-3, 3) and (0,7)(0, 7).
  1. Tap the midpoint of the segment from (−6,−5)(-6, -5) to (−4,−1)(-4, -1).
  2. Tap the midpoint of the segment from (−7,−8)(-7, -8) to (1,−4)(1, -4).

Watch out!

Practice ★★

  1. Find the exact distance between (0,−2)(0, -2) and (1,−4)(1, -4).
  2. Find the exact distance between (2,−2)(2, -2) and (3,−6)(3, -6).

Exit ticket

Three questions to check you have got it.

  1. Find the distance between (−1,−3)(-1, -3) and (2,1)(2, 1).
  2. Find the distance between (−3,3)(-3, 3) and (0,7)(0, 7).
  3. Find the distance between (2,5)(2, 5) and (11,17)(11, 17).

Summary card

Distance and midpoint

d=Δx2+Δy2d = \sqrt{\Delta x^2 + \Delta y^2}; midpoint = average of the coordinates.

Watch out: average, not half the difference.

End of the lesson

Back to the chapter

Answers: Distance and midpoint in the coordinate plane

A

  1. 55
  2. 55

B

  1. (−5,−3)(-5, -3)
  2. (−3,−6)(-3, -6)

C

  1. 5\sqrt{5}
  2. 17\sqrt{17}

D

  1. 55
  2. 55
  3. 1515

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