Distance Formula Questions with Worked Answers

Question one: cross both coordinate axes

Find the distance from A = (−4,−1) to B = (2,7). The coordinate differences are 6 and 8, so d = √(6² + 8²) = √100 = 10. Subtracting a negative coordinate requires parentheses: 2 − (−4), not 2 − 4. The result is positive because distance measures separation, regardless of which endpoint is written first. Reversing A and B gives differences −6 and −8, whose squares produce the same result.

Question two: leave an exact irrational answer

For P = (1,−2) and Q = (4,3), the changes are 3 and 5. Distance is √34, approximately 5.8310. If the exercise requests exact form, √34 is the answer; replacing it with 5.83 changes an exact expression into a rounded estimate. The horizontal change alone, 3, is too small, while adding the two absolute changes, 8, describes an axis-aligned two-leg path rather than the direct segment. The direct distance lies between those useful comparison values.

Question three: solve for a missing coordinate

Point A = (1,2) is 5 units from B = (4,k). Squaring the distance equation gives 25 = (4 − 1)² + (k − 2)², so (k − 2)² = 16. Both k − 2 = 4 and k − 2 = −4 are possible, giving k = 6 or k = −2. There are two answers because the second point can lie above or below A at the same separation. Test both complete pairs in the calculator rather than silently discarding the negative branch.

Question four: spot an impossible condition

If A = (1,2) and B = (7,k), the horizontal separation is already 6. Their direct distance cannot be 5 for any real k, because 25 = 36 + (k − 2)² would require a negative square. This is a useful pre-calculation check: distance cannot be less than either absolute coordinate change. A problem can contain inconsistent measurements or a transcription mistake; reporting no real solution is more accurate than taking the square root of a negative value as a real length. For every exercise, substitute the final candidate back into the original distance equation; a squared intermediate equation alone may obscure the requested interpretation.

Keep coordinate pairs and units together

An ordered pair is written (x,y), with the horizontal coordinate first. Label the points before substituting numbers, and use the same point order in every difference. Negative coordinates are positions relative to the chosen origin; they do not make a geometric length negative. If coordinates represent measured lengths, put both axes in the same length unit before using the ordinary distance formula. A graph whose horizontal axis is time and vertical axis is temperature can have a meaningful rate of change, but the square root of their squared differences is not an ordinary physical distance.

Check the answer against the original points

Plotting a small sketch helps distinguish a plausible calculation from an input-order mistake. A segment’s midpoint must lie between its endpoints, and its distance must be at least the absolute horizontal change and at least the absolute vertical change. Reversing the endpoints leaves distance and midpoint unchanged. It also leaves a defined slope unchanged because both differences reverse sign. Keep exact fractions or square roots until the requested final rounding step; rounding each coordinate before subtracting can noticeably change a short segment between nearby points.

Use a plane-coordinate tool for plane-coordinate data

The linked calculator accepts two numeric coordinate pairs. It does not read coordinates from a photograph, identify a street route, or transform latitude and longitude into a local map projection. Record how the coordinates were obtained, the axis scale and the requested output before entering them. For a drawing measured in centimetres, the resulting distance is in centimetres; for a unitless exercise, report coordinate units. The input fields contain numbers only, so retain the unit labels in your working rather than appending them to a numeric field.

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