Distance and midpoint

Any two points on a grid are the ends of the long side of a right triangle. The horizontal and vertical legs are easy to count, and the Pythagorean theorem does the rest.

From

A (- 2, - 1)

B (2, 2)

, the legs are 4 and 3, so the distance is 5.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

and

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Work through it

Find the distance from $(1, 2)$ to $(7, 10)$ .

Find the legs. $6$ and $8$

Use Pythagoras. $36 + 64 = 10$

Find the midpoint of $(2, - 3)$ and $(8, 5)$ .

Average the $x$ 's. $\frac{2 + 8}{2} = 5$

Average the $y$ 's. $\frac{- 3 + 5}{2} = 1$

So it's here. $(5, 1)$

Watch out for this

Subtracting in the midpoint formula.

Midpoint adds and halves. It's distance that subtracts.

The short version
Distance: draw a right triangle and use Pythagoras. Midpoint: average the

x

's and average the

y

's.

Your turn

Prefer paper? Print the Coordinate geometry and transformations worksheet, with an answer key.

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Work through it

Find the distance from (1,2) to (7,10).

Find the midpoint of (2,−3) and (8,5).

Watch out for this

Your turn

Find the distance from $(1, 2)$ to $(7, 10)$ .

Find the midpoint of $(2, - 3)$ and $(8, 5)$ .