Distance on the Coordinate Plane — Print

Draw the right triangle and the distance formula stops being something to memorise.

What the practice questions cover

These questions use 11 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

You are at the corner of 1st and Oak. Your friend is four blocks east and three blocks north.

Walking the streets is 4 + 3 = 7 blocks. Easy.

four blocks east, three blocks north — or cut across the park the park 4 blocks 3 blocks the short cut walking the streets is 4 + 3 = 7 blocks

But there is a park between you, and you can cut straight across it. How far is the short cut?

It is not 7 — the short cut is obviously shorter. It is not 3.5 either.

The two streets meet at a right angle, so you are standing on a right triangle whose legs are 4 and 3.

And you already know what to do with those: 42 + 32 = 25, so the short cut is √25 = 5 blocks. The distance formula is Pythagoras wearing coordinates.

Distance between two points as a right triangle the distance formula is the Pythagorean theorem in disguise A(1, 1) B(5, 4) 4 across 3 up d^2 = 4^2 + 3^2 d^2 = 16 + 9 = 25 d = 5 subtract the x values for the run, the y values for the rise, then use a^2 + b^2 = c^2

Why it is just Pythagoras

Plot the two points. Draw a horizontal segment and a vertical segment to make a right triangle with your two points at the ends of the slanted side.

The horizontal leg is the difference in the x values. The vertical leg is the difference in the y values.

Then a2 + b2 = c2 gives the distance. That is the entire derivation of the formula.

Points A(−2, −1) and B(3, 2) at the ends of a right triangle's slanted side, with legs Δx = 5 and Δy = 3 draw a horizontal and a vertical leg: the distance is the hypotenuse xy 0 A(−2, −1)B(3, 2) (3, −1) Δx = 5Δy = 3 d Δx = x₂ − x₁ = 3 − (−2) = 5 Δy = y₂ − y₁ = 2 − (−1) = 3 d² = Δx² + Δy² d² = 5² + 3² = 34 d = √34 ≈ 5.83 the corner (3, −1) takes itsx from B and its y from A

Draw it every time

Even once you know the formula, sketching the triangle catches sign errors and tells you roughly what the answer should be.

Using the formula

d = √(x2 - x1)2 + (y2 - y1)2. Subtract, square, add, then square root.

The squaring removes any minus signs, so subtracting in the wrong order does no harm here.

Leave the answer as an exact root, like √29, unless you are asked to round.

Square before adding

√32 + 42 = √25 = 5, not 3 + 4 = 7. The root of a sum is not the sum of the roots.

Midpoint

The midpoint is the average of the two points: (x1 + x22, y1 + y22).

Add and halve each coordinate separately.

Sanity check: the midpoint must sit between the two points in both directions.

The midpoint of A(1, 1) and B(7, 5) is M(4, 3): add and halve the x values, then add and halve the y values the midpoint is the average: add and halve each coordinate separately xy A(1, 1)B(7, 5) M(4, 3) 174 153 x:1 + 72= 4 y:1 + 52= 3 M = (4, 3) check: 4 is between 1 and 7,and 3 is between 1 and 5

Add for midpoint, subtract for distance

Midpoint adds then halves. Distance subtracts then squares. Mixing them up is common, so say which one you need before you start.

Perimeter and shape

To find the perimeter of a shape on a grid, compute each side separately and add.

Equal side lengths can prove a shape is a rhombus; equal diagonals can help prove a rectangle.

Two sides with the same slope are parallel — that plus lengths settles most "what shape is it?" questions.

Worked examples

Distance between two points

Find the distance from (1, 2) to (7, 10).

  1. Horizontal leg: 7 - 1 = 6 The run.
  2. Vertical leg: 10 - 2 = 8 The rise.
  3. d = √62 + 82 = √36 + 64 = √100 Square each leg, then add.
  4. d = 10 A 6-8-10 triangle, which is a 3-4-5 doubled.

d = 10

An exact answer

Find the distance from (-3, 4) to (2, -1).

  1. Run: 2 - (-3) = 5 Subtracting a negative adds.
  2. Rise: -1 - 4 = -5 Negative is fine; squaring will fix it.
  3. d = √52 + (-5)2 = √50 25 + 25 = 50.
  4. √50 = 5√2 ≈ 7.07 50 = 25 × 2, so pull out the 5.

5√2 ≈ 7.07

Midpoint

Find the midpoint of the segment from (-4, 3) to (10, -7).

  1. x: -4 + 102 = 62 = 3 Average the x values.
  2. y: 3 + (-7)2 = -42 = -2 Average the y values.
  3. Midpoint is (3, -2). Between -4 and 10, and between 3 and -7 ✓

(3, -2)

Words to know

Leg
A side of a right triangle next to the right angle.
Hypotenuse
The longest side, opposite the right angle.
Distance formula
d = √(x2 - x1)2 + (y2 - y1)2.
Midpoint
The point halfway between two others: average the x values and average the y values.

Mistakes to watch for

Adding the legs instead of squaring them. √32 + 42 = 5, but 3 + 4 = 7 — that is the street route, not the short cut. The root of a sum is never the sum of the roots.

Subtracting for the midpoint. Midpoint adds then halves; distance subtracts then squares. Say which one you need before you start, because the two formulas look alike and do opposite things.

Forgetting the square root at the end. 16 + 9 = 25 is d2, not d. Stopping there gives an answer five times too big, and it looks plausible enough to survive a glance.

Mishandling a negative coordinate. The run from -2 to 4 is 4 - (-2) = 6, not 2. Subtracting a negative adds — and the squaring afterwards will hide the error rather than reveal it.

A rope with 12 evenly spaced knots, pulled into a 3-4-5 triangle, gives a perfect right angle. Egyptian surveyors used exactly this to re-mark field boundaries after the Nile flooded.

They were doing it a thousand years before Pythagoras was born, and the Babylonian tablet Plimpton 322 lists such triples from around 1800 BC.

What the Greeks added was not the fact but the proof — the argument that it must hold for every right triangle, not just the ones anyone had tried.

Problem 1

Find the distance between (0, 0) and (3, 4).

Worked answer

5

32 + 42 = 25

√25 = 5

Problem 2

Find the distance between (1, 1) and (7, 9).

Worked answer

10

Run: 6. Rise: 8

36 + 64 = 100

√100 = 10

Problem 3

Find the distance between (2, 5) and (2, 12).

Worked answer

7

It is a vertical segment

12 - 5 = 7

No square root needed

Problem 4

Find the midpoint of (2, 4) and (8, 10).

Worked answer

(5, 7)

x: 2 + 82 = 5

y: 4 + 102 = 7

Problem 5

Find the midpoint of (-3, 6) and (5, -2).

Worked answer

(1, 2)

x: -3 + 52 = 1

y: 6 - 22 = 2

Problem 6

Find the distance between (-2, 3) and (4, 11).

Worked answer

10

Run: 4 - (-2) = 6

Rise: 11 - 3 = 8

√36 + 64 = 10

Problem 7

Find the exact distance between (1, 2) and (4, 6)... then between (0, 0) and (2, 3).

Worked answer

sqrt(13)

22 + 32 = 13

d = √13

Leave it exact

Problem 8

Find the exact distance between (0, 0) and (5, 5), in simplest radical form.

Worked answer

5sqrt(2)

25 + 25 = 50

√50 = √25 · 2

= 5√2

Problem 9

The midpoint of AB is (4, 5) and A is (1, 2). Find B.

Worked answer

(7, 8)

1 + x2 = 4, so x = 7

2 + y2 = 5, so y = 8

Problem 10

A triangle has vertices (0, 0), (6, 0) and (6, 8). Find its perimeter.

Worked answer

24

Base: 6. Height: 8

Hypotenuse: √36 + 64 = 10

6 + 8 + 10 = 24

Problem 11

Which point is closer to the origin, (3, 4) or (1, 5)?

  1. (3, 4)
  2. (1, 5)
  3. They are the same distance
  4. Cannot be determined
Worked answer

(3, 4)

(3,4): 9 + 16 = 25

(1,5): 1 + 25 = 26

25 < 26, so (3, 4) is closer

Problem 12

Find the distance between (-5, -2) and (-1, 1).

Worked answer

5

Run: -1 - (-5) = 4

Rise: 1 - (-2) = 3

√16 + 9 = 5

Problem 13

The distance from (2, 3) to (2, k) is 9, with k > 3. Find k.

Worked answer

12

|k - 3| = 9

k - 3 = 9 since k > 3

k = 12

Problem 14

A quadrilateral has vertices (0,0), (4,0), (4,4), (0,4). Find the length of a diagonal in simplest radical form.

Worked answer

4sqrt(2)

From (0,0) to (4,4)

16 + 16 = 32

√32 = 4√2

Problem 15

The midpoint of a segment is (0, 0) and one endpoint is (-6, 8). Find the other endpoint.

Worked answer

(6, -8)

-6 + x2 = 0, so x = 6

8 + y2 = 0, so y = -8

Problem 16

Find the perimeter of the triangle with vertices (1, 1), (4, 5) and (1, 5).

Worked answer

12

(1,1) to (1,5): 4

(1,5) to (4,5): 3

(1,1) to (4,5): √9+16 = 5

4 + 3 + 5 = 12

Browse all courses