9.7 Distances on a Grid

One of the most useful spin-offs of the theorem is a way to measure the straight-line distance between two points drawn on squared paper.

Suppose two dots sit at grid positions P and Q. Walk from P to Q along the gridlines: first across by some horizontal amount, then up (or down) by some vertical amount. Those two legs meet at a right angle, and the straight segment PQ is the hypotenuse. So the distance is just \sqrt{(\text{horizontal step})^2 + (\text{vertical step})^2}.

Worked Example

Find the distance between P at (3,4) and Q at (15,9).

Horizontal step = 15 - 3 = 12; vertical step = 9 - 4 = 5. These are the legs of a right triangle, so PQ = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13. The two points are exactly 13 units apart.

Try This Find the distance between (1,2) and (7,10). (Steps 6 and 8, do you recognise the triple?) What is the distance between any point and itself, by this formula?

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