Pythagoras and Coordinate Distances
Pythagoras links the three side lengths of a right-angled triangle. You will learn when to add, when to subtract, how to check a right angle and how to find the distance between coordinates.
Foundation & Higher. Three-dimensional applications follow in a separate Higher lesson.
Identify the hypotenuse
The hypotenuse is opposite the right angle and is the longest side. Call it c. The two shorter sides are a and b. For a right-angled triangle, a²+b²=c². The letters may change in a question; the position opposite 90° decides which side is the hypotenuse.
Finding the longest side
Square both shorter lengths, add the squares, then take the positive square root. The sum is c², not c.
Worked example — two shorter sides
A right triangle has shorter sides 6 cm and 8 cm. c²=6²+8²=36+64=100. Therefore c=√100=10 cm. The answer is longer than either given side.
Worked example — an exact answer
With shorter sides 3 cm and 5 cm, c=√(9+25)=√34 cm. This is exact. If asked for one decimal place, c≈5.8 cm. Keep the square root on your calculator until the final step.
Finding a shorter side
When the hypotenuse is known, subtract the square of the known shorter side from the square of the hypotenuse. Rearrange first: a²=c²−b².
Worked example — a ladder
A 13 m ladder reaches 12 m vertically up a wall on level ground. The wall and ground meet at 90°. The ladder is the hypotenuse. If the foot is x metres from the wall, x²+12²=13², so x²=169−144=25 and x=5 m. Adding here would give a distance longer than the ladder, which is impossible.
Coordinate distances
Join two coordinate points and draw horizontal and vertical lines to make a right triangle. Subtract x-coordinates for its horizontal length and y-coordinates for its vertical length. Use positive lengths; squaring removes a negative difference anyway.
Worked example — across quadrants
A(−2,1) and B(4,9) differ by 4−(−2)=6 horizontally and 9−1=8 vertically. AB=√(6²+8²)=10 units. The midpoint is a different question: averaging coordinates gives (1,5).
In general, distance=√[(x₂−x₁)²+(y₂−y₁)²]. For a horizontal or vertical segment, one difference is zero.
Testing for a right angle and explaining the theorem
To test three side lengths, first identify the longest. A triangle with sides 5,12,13 is right-angled because 5²+12²=13². Sides 5,6,8 do not give a right angle: 25+36≠64. Do not use the rule on a triangle just because it looks right-angled.
Why a²+b²=c²
Arrange four identical right triangles, with shorter sides a and b, inside a square of side a+b. Their hypotenuses bound a central square of side c: adjacent acute angles sum to 90°, so its corners are right angles. The large square has area (a+b)². The four triangles have total area 4×½ab=2ab. Therefore c²=(a+b)²−2ab=a²+b².
Think Like a Mathematician
Which side must be longest before you calculate? That prediction tells you whether to add or subtract and provides a quick check afterwards.
Common Mistakes
Do not square a sum: 6²+8² is 100, whereas (6+8)² is 196. Do not stop at c². Use consistent units and give a length answer, not square units. Establish a right angle before using the theorem.
Practice — Pythagoras and distances
1. A right triangle has shorter sides 9 cm and 12 cm. Find its hypotenuse.
c=√(81+144)=√225=15 cm.
2. A right triangle has hypotenuse 17 m and one shorter side 8 m. Find the other.
x=√(289−64)=√225=15 m.
3. Find the exact hypotenuse when the shorter sides are 2 cm and 5 cm.
√(4+25)=√29 cm.
4. Find the distance from (−3,−1) to (5,5).
Differences are 8 and 6. Distance=√(64+36)=10 units.
5. Are the sides 7,24,25 those of a right triangle? Explain.
Yes: 7²+24²=49+576=625=25². The right angle is opposite the side 25.
6. A rectangle is 1.2 m by 50 cm. Find its diagonal in metres.
50 cm=0.5 m. Diagonal=√(1.44+0.25)=1.3 m.
Frequently Asked Questions
Can I use this for every triangle?
No. The theorem requires a right angle. Non-right-angled triangles need another method, such as the cosine rule at Higher.
Why take only the positive square root?
A geometric side length is positive. The negative algebraic root does not represent a length.
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