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A short class with Sarah and her students. The other ways are under More.
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Plot A(2, 3) and B(6, 7). How could Pythagoras help us find the distance between them?
Draw a horizontal line from A and a vertical line from B. They make a right triangle with AB as the hypotenuse.
How long are the two legs?
The horizontal one is 6 − 2 = 4 and the vertical one is 7 − 3 = 4.
So AB² is 4² + 4². What is AB?
√32, which is 4√2, about 5.66.
So in general it's the square root of the x-difference squared plus the y-difference squared.
What if we subtract the other way, 2 − 6?
We get −4, but squaring gives 16 again, so the answer is the same.
Now find the distance from P(−3, 2) to Q(5, −4).
5 − (−3) is 8, and −4 − 2 is −6.
8² + (−6)² = 64 + 36 = 100, so the distance is exactly 10.
Good. Brackets around negative numbers stop most sign mistakes.
The distance from (1, k) to (4, 6) is 5. Find k.
3² + (6 − k)² = 25, so (6 − k)² = 16.
Then 6 − k is 4 or −4, so k = 2 or k = 10. Two answers!
Now, what kind of triangle is P(1, 2), Q(4, 6), R(8, 3)?
PQ and QR are both 5 and PR is √50. It's isosceles.
And 25 + 25 = 50, so it's also right-angled at Q.
d = √((x₂ − x₁)² + (y₂ − y₁)²) for points (x₁, y₁) and (x₂, y₂).
the coordinate differences are the legs of a right triangle and the distance is its hypotenuse.
the order of the points does not matter, because squaring removes negative signs.
simplify surds such as √32 = 4√2 before giving a decimal.
find unknown coordinates, classify triangles, find perimeters and test whether points lie on a circle.
A(2, 3) to B(6, 7): d = √(4² + 4²) = √32 = 4√2 ≈ 5.66 units.
P(−3, 2) to Q(5, −4): d = √(8² + (−6)²) = √100 = 10 units.
If the distance from (1, k) to (4, 6) is 5, then 9 + (6 − k)² = 25, so k = 2 or k = 10.
P(1, 2), Q(4, 6), R(8, 3): PQ = QR = 5 and PR = √50, and 5² + 5² = 50, so the triangle is right-angled and isosceles.
Subtract the x's, subtract the y's, square them both and add; the root of that is the distance, and that's not bad.
A crow and a taxi leave the same corner for a café 3 blocks east and 4 blocks north. The taxi drives 7 blocks, but the crow flies straight and covers only 5, the distance the formula gives.
"Difference, Square, Add, Root" - the four steps in order.
D.S.A.R. - Difference, Square, Add, Root.
Term
State the distance formula.
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Term
Which theorem is the distance formula based on?
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Term
Find the distance between (2, 3) and (6, 7).
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Term
Does it matter which point you call (x₁, y₁)?
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Find the distance from the origin to (−5, 12).
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d = √((x₂ − x₁)² + (y₂ − y₁)²), the straight-line distance between two points.
The ordered pair (x, y) that gives a point's position on the Cartesian plane.
In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
The longest side of a right triangle, opposite the right angle.
An irrational square root left in exact form, such as 4√2.
A triangle with two sides of equal length.
If a² + b² = c² for the sides of a triangle, then the triangle has a right angle opposite side c.
The distance between two points A(x₁, y₁) and B(x₂, y₂) is d = √((x₂ − x₁)² + (y₂ − y₁)²), a formula that comes straight from Pythagoras' theorem. For A(2, 3) and B(6, 7) it gives √32 = 4√2 ≈ 5.66 units.
This lesson develops the distance formula and practises using it. Joining two points with a horizontal and a vertical line creates a right triangle whose legs are the differences in the x- and y-coordinates, so Pythagoras' theorem gives the straight-line distance: d = √((x₂ − x₁)² + (y₂ − y₁)²). The lesson works through the example A(2, 3) to B(6, 7), shows that the order of the points does not matter, handles negative coordinates and horizontal or vertical segments, and simplifies surd answers. It then uses the formula to find an unknown coordinate, to classify a triangle as isosceles or right-angled, and to test whether a point lies on a circle.
Where the formula comes from. Plot A(x₁, y₁) and B(x₂, y₂). Draw a horizontal line from A and a vertical line from B; they meet at a right angle at C(x₂, y₁). The horizontal leg AC has length |x₂ − x₁| and the vertical leg CB has length |y₂ − y₁|. The segment AB is the hypotenuse, so by Pythagoras' theorem:
AB² = (x₂ − x₁)² + (y₂ − y₁)², so d = √((x₂ − x₁)² + (y₂ − y₁)²)
Squaring removes any negative signs, so the absolute value bars are not needed in the formula.
Example: A(2, 3) and B(6, 7).
Useful facts.
Using the formula to solve problems.
Common mistakes. Adding the coordinates instead of subtracting them, mixing an x-value with a y-value, forgetting to square a negative difference correctly ((−6)² = 36, not −36), and forgetting the final square root.