Connected ideas

Pythagoras and coordinates

Connect coordinate changes to the distance between two points.

Predict, test and explain

Step 1 of 3

Predict

What is the distance from A(−1, 1) to B(2, 5)?

A(−1,  −2),B(2,  2)A(-1,\;-2),\quad B(2,\;2)
−6-6
−6-6
−4-4
−4-4
−2-2
−2-2
22
22
44
44
66
66
A(−1,  −2)A(-1,\;-2)
B(2,  2)B(2,\;2)

Scroll the diagram sideways to see every label.

Horizontal changeΔx=3\Delta x=3Vertical changeΔy=4\Delta y=4

The dashed sides are perpendicular. Their lengths are the absolute coordinate changes; the direct segment is the hypotenuse.

Check the working
QuantityValue
Horizontal side length3
Vertical side length4
Squared distance32+42=253^2+4^2=25
Exact distanceAB=5AB=5
Approximate distance5 units

InvestigateSwap the endpoints. Does the distance change?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

Find the distance between A(−1, 1) and B(2, 5).

A(−1,1),B(2,5)A(-1,1),\quad B(2,5)

Hint

Find the horizontal and vertical changes, then apply Pythagoras.

Reveal answer

The perpendicular side lengths are three and four, giving a direct distance of five units.

AB=32+42=5AB=\sqrt{3^2+4^2}=5

Build an explanation

Hint 1

Subtract the x-coordinates and the y-coordinates to find the signed changes.

Hint 2

Use the absolute changes as the lengths of two perpendicular sides.

Hint 3

Square those lengths, add them, and take the square root to find the direct distance.

Worked example

This example uses fixed values, separate from the diagram controls.

Find the distance between A(−1, −2) and B(2, 2).

  1. The horizontal change is 2 − (−1) = 3.
  2. The vertical change is 2 − (−2) = 4.
  3. The squared distance is 3² + 4² = 25.
  4. The distance is √25 = 5 units. Swapping the points gives the same distance.

Answer: 5 units.

Connect this idea

Watch out: A negative coordinate change is not a negative side length. If one change is zero, the distance is the other side length; coincident points have distance zero.