Draw the right triangle hiding in the grid.
across = 5 − (−3) = 8
the x-difference
up = 8 − 2 = 6
the y-difference
The two points are the ends of a hypotenuse. The legs run along the grid, so their lengths are just subtractions.
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The Distance Formula
Pythagoras wearing coordinates — nothing more.
The problem
How far apart are (−3, 2) and (5, 8)?
The steps are dx = 5 − (−3) = 8 and dy = 8 − 2 = 6, so d = √8² + 6² = √100 = 10. B forgets the square root, and D just adds the legs. Watch the double negative in the x-step — it is 8, not 2.
The formula is Pythagoras, relabelled.
d = √ (x₂ − x₁)² + (y₂ − y₁)²
a² + b² = c²
the theorem
the same thing
a and b are the two differences
If you ever forget it, sketch the triangle and use Pythagoras directly.
Substitute, carefully.
d = √ (5 − (−3))² + (8 − 2)²
brackets round each difference
= √ 8² + 6²
minus a minus adds
= √ 64 + 36
square them
= √100 = 10
the 3-4-5 triple, doubled
Squares before adding
√64 + 36 is √100 = 10. It is not √64 + √36 = 14. A square root never splits across a plus.
The order of the points does not matter.
(−3 − 5)² = (−8)² = 64
reversed
(5 − (−3))² = 8² = 64
the original
Squaring destroys the sign, so distance is the same in either direction.
That is the difference from slope, where the order does matter — a reversed slope changes sign, but a reversed distance does not.
Backwards: the distance is given, a coordinate is missing.
The distance from (2, k) to (−3, 1) is 13. Find k.
(2 − (−3))² + (k − 1)² = 13²
square both sides — no root to carry
25 + (k − 1)² = 169
the known leg is 5
(k − 1)² = 144
the missing leg squared
k − 1 = 12 or k − 1 = −12
a square root has two signs
k = 13 or k = −11
one point above, one below
Two answers, then read the condition
The point can sit 12 above or 12 below y = 1, so there are always two values. The question then picks one with a condition such as "k > 0" or "in the second quadrant". Taking only +12 and stopping is the usual slip.
Equally far from two points. A point P on the y-axis is the same distance from (3, 1) as from (5, 3). Find P.
P = (0, k)
on the y-axis, x is 0 (on the x-axis, y is 0)
(0 − 3)² + (k − 1)² = (0 − 5)² + (k − 3)²
set the two squared distances equal
9 + k² − 2k + 1 = 25 + k² − 6k + 9
expand both brackets
4k = 24, so k = 6
the k² cancels — it is linear
P = (0, 6)
check: 9 + 25 = 34 and 25 + 9 = 34 ✓
Variations: a root that does not come out, a level pair, fractions.
Not a perfect square. How far is (−1, 4) from (3, −2)?
d = √ (3 − (−1))² + (−2 − 4)²
brackets round each difference
= √ 4² + (−6)² = √16 + 36
the minus dies when squared
= √52 = √4 × 13
pull out the biggest square factor
= 2√13
about 7.2
What is different: 52 is not a square number, so the answer stays a simplified root — the choices will show 2√13, not a decimal.
Only the square factor comes out
√52 = √4 × √13 = 2√13. Writing 4√13 brings out the 4 itself instead of its root, and 2√26 splits 52 as 2 × 26, where 2 is not a square.
Two points at the same height. How far is (−7, 3) from (2, 3)?
up = 3 − 3 = 0
same y: the segment is level
d = √ 9² + 0² = √81
the formula still works
= 9
the same as 2 − (−7)
What is different: one difference is 0, so there is no triangle — the distance is just the other difference, made positive.
Fractions. How far is (½, −1) from (2, 1)?
across = 2 − ½ = 32, up = 1 − (−1) = 2
the two legs
d = √ (32)² + 2² = √ 94 + 164
write 4 as 164 to add
= √254
one fraction under the root
= 52 = 2½
root the top and the bottom
What is different: square the fraction whole — (32)² is 94, not 92 — and put both squares over the same denominator before adding.
What it gets used for.
| Question | What to do |
|---|---|
| perimeter of a triangle from its vertices | three distances, added |
| is the triangle isosceles? | compare the three lengths |
| is the triangle right-angled? | test a² + b² = c² on the lengths |
| radius of a circle | distance from centre to a point on it |
| which point is nearer? | compare distances — no need to root |
For that last one, comparing the values under the root is enough. The bigger square is the bigger distance.
Worked: classify the triangle with vertices A(5, 2), B(1, 1) and C(0, 5).
AB² = 4² + 1² = 17
keep the squares
BC² = 1² + 4² = 17
AB = BC, so isosceles
AC² = 5² + 3² = 34
the longest side
17 + 17 = 34 ✓
Pythagoras holds, so right-angled at B
On the exam
Distance questions rarely stop at one length: they ask for a missing coordinate, the farthest point, or what kind of triangle three points make. Work with the squared distances throughout — compare them, test a² + b² = c² on them — and only take a root if the answer needs a length.
Check: count the squares.
on the grid
8 right, 6 up
matches the subtractions ✓
Pythagoras
64 + 36 = 100
√100 = 10 ✓
sensible?
10 > 8 and 10 > 6
the hypotenuse is longest ✓
Now you try
Practice problem
Find the distance
Find the distance between (1, −4) and (−5, 4).
dx = −5 − 1 = −6 and dy = 4 − (−4) = 8, so d = √36 + 64 = √100 = 10. The negative dx is harmless because squaring removes the sign — which is why you never need to worry about which point you start from.