Finding the distance between two points in the coordinate plane. Includes deriving the formula from the Pythagorean theorem, applying it to pairs of points, why the order of subtraction does not matter, using it with the converse of the theorem to test three points for a right angle, and measuring a real distance across a coordinate grid.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
The Distance Formula
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-735 — the lesson these objectives are drawn from
Warm-up
Lesson 12.6 related the sides of a right triangle. This lesson puts that triangle into the coordinate plane, where its legs can be read off directly.
Discussion prompt
Two points are four units apart horizontally and three units apart vertically. How far apart are they in a straight line?
Hint: Draw the right triangle.
Answer:
\[ d^2 = 4^2 + 3^2 = 25 \;\Longrightarrow\; d = 5 \]
The horizontal and vertical gaps are the legs of a right triangle and the straight-line distance is its hypotenuse. The distance formula is nothing more than that observation written with coordinates.
Concept
The distance between two points is the hypotenuse of a right triangle whose legs are the horizontal and vertical gaps between them. Those gaps are the differences of the coordinates.
distance formula — The distance between the points with coordinates x one, y one and x two, y two is the square root of the sum of the squares of the differences of the coordinates.
It is the Pythagorean theorem in coordinates.
Figure (svg): The distance formula stated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730
Section
Section 1
Concept
Given two points, the third vertex of a right triangle sits at the corner where a horizontal line through one meets a vertical line through the other. Its legs are the coordinate differences.
\[ (x_2 - x_1)^2 + (y_2 - y_1)^2 = d^2 \]
Solving for d gives the distance formula.
Figure (svg): A right triangle drawn between two points in the plane
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730 — the derivation of the distance formula from the Pythagorean theorem
Picture it
The corner is free.
Figure (svg): A right triangle drawn between two points in the plane
The right angle is guaranteed because one leg is horizontal and the other vertical. Nothing has to be assumed about the two original points.
Worked example
The textbook's own derivation.
\[ \text{Derive } d \text{ for points } (x_1, y_1) \text{ and } (x_2, y_2). \]
Place the third vertex
Why: Same x as one, same y as the other.
\[ (x _{2}, y _{1}) \]
Find the horizontal leg
Why: Difference of the x values.
\[ x _{2} - x _{1} \]
Find the vertical leg
Why: Difference of the y values.
\[ y _{2} - y _{1} \]
Apply the theorem
Why: And take the positive root.
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]
Figure (svg): A right triangle drawn between two points in the plane
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Verify: test on a familiar triangle
Why: For the points at the origin and at four, three, the legs are four and three and the formula gives the root of twenty-five, which is five. That is the three-four-five triangle, recovered from coordinates.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730
Faded example
Differences, not sums.
Fill in the blanks
\texty_1 x_2 - x_1 \textd^2 y_2 - y_1, \text___ d^2 = (x_2 - x_1)^2 + (y_2 - ___)^2 \text___ d = \sqrt___}
Why: Each leg is the difference of one pair of coordinates. The formula is then the theorem with those differences substituted in.
Worked example
The step that guarantees the right angle.
\[ \text{Why is the angle at } (x_2, y_1) \text{ a right angle?} \]
Look at one side
Why: It shares a y value with one point.
Look at the other
Why: It shares an x value with the other.
Recall the axes
Why: Horizontal and vertical are perpendicular.
\[ 90 ^\circ \]
Conclude
Why: The theorem applies.
Figure (svg): A right triangle drawn between two points in the plane
\[ \text{horizontal} \perp \text{vertical} \]
Verify: check it holds for any two points
Why: Whatever the two points are, the corner can always be placed this way and the two legs will always be horizontal and vertical. The construction never fails, which is why the formula is general.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730
Trap
The distance formula is a separate rule to be memorised.
Learn it as a string of symbols
Why: It looks unlike anything before it.
It is the Pythagorean theorem with the legs written as coordinate differences. Someone who sees that can rebuild it from a sketch; someone who does not must recall it exactly or not at all.
Sketch the right triangle and read the legs off the coordinates.
Rebuild the formula rather than recalling it
Why: It takes about ten seconds.
The same is true of the quadratic formula from Lesson 12.5.
Elimination
In the derivation.
Eliminate the wrong options
Why is the constructed triangle right-angled?
Survives elimination: A
Why: The corner point shares a coordinate with each of the originals, making one leg horizontal and the other vertical. Those directions are perpendicular by the construction of the plane itself.
Prediction
Both on the same horizontal line, say.
Predict first
What does the formula give?
Correct: The horizontal gap alone, since the vertical difference is zero.
\[ (x_2 - x_1)^2 + 0^2 = (x_2-x_1)^2 \]
Why: If the y values match, their difference is nought and its square contributes nothing, leaving the root of the horizontal difference squared — which is that difference itself. The formula degenerates gracefully into simple subtraction, which is what it should do when the two points lie on a horizontal line. The same happens vertically.
Socratic
The formula is short enough to learn.
Discussion prompt
Say what deriving the distance formula gives you that memorising does not. Then name another formula in this course with the same property.
Hint: What happens if you half-remember it?
Answer:
A half-remembered formula is useless and can be worse than useless, since a plausible wrong version gives plausible wrong answers. A derivation can be redone in ten seconds from a sketch, so it survives being partly forgotten — and it also explains why the differences are squared, which is otherwise arbitrary.
The quadratic formula from Lesson 12.5 has exactly this property: it can be rebuilt by completing the square on the general equation. In both cases the derivation is short and the formula is long, which is the usual sign that understanding the derivation is the better investment.
Section
Section 2
Concept
Substituting two points into the formula gives the distance between them. The answer is usually a radical, which may then be approximated.
Most distances between lattice points are irrational.
Figure (svg): The distance between two specific points
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730 — Example 1, Find the Distance Between Two Points
Picture it
Four operations in order.
Figure (svg): The distance between two specific points
The exact answer is the root of ten and the decimal is a reading of it. Which to report depends on whether the question asks to simplify or to measure.
Worked example
This is Example 1 from the textbook.
\[ \text{Find the distance between } (-1, 4) \text{ and } (2, 3). \]
Subtract the x values
Why: Two minus negative one.
\[ 3 \]
Subtract the y values
Why: Three minus four.
\[ -1 \]
Square and add
Why: Nine plus one.
\[ 10 \]
Take the root
Why: And approximate.
\[ \sqrt{10} \approx 3.16 \]
Figure (svg): The distance between two specific points
\[ d = \sqrt{10} \approx 3.16 \]
Verify: check against a sketch
Why: The two points are three apart horizontally and one apart vertically, so the direct distance should be a little over three. The answer of about 3.16 fits, and it is less than the four you would travel going along the two legs.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730
Faded example
Both squares are positive.
Fill in the blanks
d = \sqrt1 = \sqrt10}} = \sqrt___}
Why: The second difference is negative and its square is positive, so the two are added. The formula never subtracts the squared terms from one another.
Worked example
Guided Practice items of the same kind.
\[ \text{Find the distance between } (2, 5) \text{ and } (0, 4); \; (8, 0) \text{ and } (0, 6). \]
Take the first pair
Why: Gaps of two and one.
\[ \sqrt{4 + 1} = \sqrt{5} \]
Approximate it
Why: To two decimal places.
\[ \approx 2.24 \]
Take the second pair
Why: Gaps of eight and six.
\[ \sqrt{64 + 36} \]
Simplify
Why: A perfect square.
\[ \sqrt{100} = 10 \]
Figure (svg): The distance between two specific points
\[ \sqrt{5} \approx 2.24, \qquad 10 \]
Verify: recognise the second triangle
Why: Gaps of eight and six give the root of a hundred, which is exactly ten — that is twice the three-four-five triangle. When a distance comes out whole, a familiar triangle is usually the reason.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 731-731
Error analysis
The student found the distance between two points.
Annotate
On: \( \begin{aligned} d &= \sqrt{(2 - (-1))^2 + (3 - 4)^2} \\ &= \sqrt{3^2 - 1^2} \\ &= \sqrt{8} \approx 2.83 \end{aligned} \)
The formula always adds the two squares, whatever signs the differences have, because squaring makes both non-negative. Writing each squared difference as a positive number before adding removes the ambiguity entirely.
Matching
Gaps, squared and added.
Match the pairs
Why: Two of these come out whole because the gaps form familiar right triangles. The other two give radicals, which is the more usual outcome for arbitrary points.
Sorting
Which form does the question want?
Sort into buckets
Sort each request by the form of answer it calls for.
Many questions ask for both, an exact form first and a decimal after. Giving only one when both were asked for is a common way to lose half the marks.
Socratic
The points have whole-number coordinates.
Discussion prompt
Explain why the distance between two lattice points is usually a radical. Then say when it comes out whole.
Hint: What has to be true of the sum of two squares?
Answer:
The distance is the root of a sum of two squares, and most such sums are not themselves perfect squares — nine plus one is ten, which has no whole root. Since perfect squares are sparse among the integers, the sum only rarely lands on one.
It comes out whole exactly when the two gaps form a Pythagorean triple, such as three and four giving five, or six and eight giving ten. Those are uncommon enough that a whole-number answer is a signal you have met a familiar triangle rather than a coincidence.
Section
Section 3
Concept
Subtracting the coordinates in the other order changes the sign of each difference, and squaring removes it. Either point may therefore be taken as the first.
\[ (x_2 - x_1)^2 = (x_1 - x_2)^2 \]
Consistency within each bracket is what matters.
Figure (svg): Subtracting the coordinates in either order
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-730 — the distance formula's structure, in which both differences are squared
Picture it
Both give the same square.
Figure (svg): Subtracting the coordinates in either order
This is a real convenience: there is no rule about which point is first, so no rule to misremember. The only requirement is not to mix the orders within one calculation.
Worked example
Confirming the order is free.
\[ \text{Find the distance from } (-1, 4) \text{ to } (2, 3) \text{ with each point taken first.} \]
Take the first point first
Why: Differences of three and negative one.
\[ 9 + 1 = 10 \]
Take the second first
Why: Differences of negative three and one.
\[ 9 + 1 = 10 \]
Compare
Why: The squares are identical.
Conclude
Why: Either order works.
\[ d = \sqrt{10} \]
Figure (svg): Subtracting the coordinates in either order
\[ \sqrt{10} \text{ either way} \]
Verify: say why this must happen
Why: Reversing a subtraction changes only its sign, and squaring makes both signs positive. So the two calculations differ only at a stage whose result is then discarded.
Faded example
The square is unchanged.
Fill in the blanks
(2 - (-1))^2 = 9, \quad ((-1) - 2)^2 = (-3)^2 = 9
Why: The two differences are opposites and their squares are equal. Squaring is what makes the order irrelevant.
Worked example
The one thing the freedom does not extend to.
\[ \text{Why is } (x_2 - x_1)^2 + (y_1 - y_2)^2 \text{ still correct, but mixing within a bracket is not?} \]
Check the mixed version
Why: Different orders in the two brackets.
Explain why
Why: Each bracket is squared separately.
Consider a genuine error
Why: Mixing coordinates within one bracket.
\[ x _{2} - y _{1} \]
Reject it
Why: That is not a coordinate difference.
Figure (svg): Subtracting the coordinates in either order
\[ (x_2-x_1)^2 + (y_1-y_2)^2 = d^2 \]
Verify: test the mixed-order version numerically
Why: For the points above it gives nine plus one again, which is right. Subtracting an x from a y, by contrast, would compare two unrelated quantities and give a number with no meaning.
Trap
\[ d = \sqrt{(2 - 4)^2 + (3 - (-1))^2} \]
Pair the numbers as they appear in the coordinates
Why: Two and four are both second-listed, so they were subtracted.
The two is an x coordinate and the four a y coordinate, so their difference measures nothing. The x values must be subtracted from each other and the y values from each other.
\[ d = \sqrt{(2 - (-1))^2 + (3 - 4)^2} \]
Subtract like coordinates from like
Why: x with x and y with y.
Writing the two points one above the other, with their x values aligned, prevents this entirely.
Elimination
For the points (-1, 4) and (2, 3).
Eliminate the wrong options
Which expression does not give the distance?
Survives elimination: A
Why: The first bracket subtracts an x from a y, comparing quantities that measure different directions. The other three are all legitimate, differing only in which point is taken first in each bracket.
Hypothesis
Order usually matters in subtraction.
Predict first
What makes the choice of first point irrelevant?
Correct: Each difference is squared, which removes its sign.
\[ (a - b)^2 = (b - a)^2 \text{ for all } a, b \]
Why: Reversing a subtraction gives the opposite number, and opposites have equal squares — so the two orders produce identical terms. That the distance is positive is a consequence rather than the cause: the formula would still be sign-free even if it were not. Without the squaring, the order would matter a great deal, which is exactly the situation in the next lesson's midpoint formula where the coordinates are added instead.
Socratic
Not every formula is sign-free.
Discussion prompt
Name a situation in coordinate geometry where the order of subtraction does matter, and say why. Then say what distinguishes the two cases.
Hint: Think about slope.
Answer:
Slope is the difference of the y values over the difference of the x values, and there the order matters within each difference — though reversing both gives the same slope, since two sign changes cancel. Reversing only one gives the negative of the correct slope, which is a genuinely different line.
The difference is that slope divides the two differences rather than squaring them, so the signs survive into the answer. Any formula that squares its differences is sign-free and any that keeps them linear is not, which is a useful thing to notice when meeting a new formula.
Section
Section 4
Concept
To decide whether three points form a right triangle, find the three side lengths with the distance formula and then apply the converse of the Pythagorean theorem.
Working with the squares avoids the radicals entirely.
Figure (svg): Three points tested for a right angle using distances
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 731-731 — Example 2, Check a Right Triangle
Picture it
Then one comparison.
Figure (svg): Three points tested for a right angle using distances
The squares are what the test compares, so the radicals never need to be evaluated. Keeping them as squares is both exact and quicker.
Worked example
This is Example 2 from the textbook.
\[ \text{Are } (3, 2), \; (2, 0) \text{ and } (-1, 4) \text{ vertices of a right triangle?} \]
Find the first distance
Why: Gaps of one and two.
\[ \sqrt{5} \]
Find the second
Why: Gaps of four and two.
\[ \sqrt{20} \]
Find the third
Why: Gaps of three and four.
\[ \sqrt{25} = 5 \]
Apply the converse
Why: Five plus twenty is twenty-five.
Figure (svg): Three points tested for a right angle using distances
\[ 5 + 20 = 25 \;\Longrightarrow\; \text{right-angled} \]
Verify: check the longest side was used
Why: Five is the largest of the three lengths, since the root of twenty is about 4.47. Only the longest can be the hypotenuse, and testing with a shorter one would have given a false negative.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 731-731
Faded example
The largest stands alone.
Fill in the blanks
5 + 20 = 25, \text25 ___
Why: The two smaller squares are added and compared with the largest. Getting that arrangement right is what the test depends on.
Worked example
Avoiding the radicals altogether.
\[ \text{Why is it easier to compare } 5, 20 \text{ and } 25 \text{ than the three lengths?} \]
Note what the formula produces
Why: A square, before the root.
\[ d ^{2} \]
Note what the test needs
Why: Squares of the sides.
\[ \text{also } d ^{2} \]
Skip the root
Why: It would only be squared again.
Compare directly
Why: Whole numbers.
\[ 5 + 20 = 25 \]
Figure (svg): Three points tested for a right angle using distances
\[ d_1^2 + d_2^2 = d_3^2 \]
Verify: see what taking roots would have cost
Why: The lengths are about 2.24, 4.47 and 5, whose squares would then have to be recomputed with rounding error. Stopping at the squares is both exact and shorter, which is worth doing every time this test is applied.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 731-731
Trap
\[ 5 + 25 = 30 \ne 20 \;\Longrightarrow\; \text{not a right triangle} \]
Add two of the squares and compare with the third
Why: Any two were chosen.
Twenty-five is the largest square, so it must be the one on its own. Adding it to another and comparing with a smaller one will fail even when the triangle is right-angled.
\[ 5 + 20 = 25 \;\checkmark \]
Identify the largest square first, then add the other two
Why: Only the longest side can be the hypotenuse.
This is the same requirement as when using the converse with plain side lengths.
Sorting
The largest is the hypotenuse's.
Sort into buckets
For squares of 5, 20 and 25, sort each by its role in the test.
Only the largest square is ever alone on one side of the comparison. Sorting the three values before testing takes a moment and prevents the commonest error.
Prediction
Say the two smaller squares total 61 and the largest is 64.
Predict first
What follows?
Correct: The triangle is not right-angled.
A triangle with sides 5, 6 and 8 has an angle of about 97 degrees, which is close to 90 and not equal to it.
Why: The converse requires exact equality, so sixty-one and sixty-four give a definite no. The triangle is a perfectly ordinary one with no right angle, which is a legitimate answer rather than a failure. Being close is not a mathematical category here: three degrees off a right angle is simply not a right angle, and the test is designed to detect exactly that.
Socratic
Neither answers the question alone.
Discussion prompt
Explain why this problem needs both the distance formula and the converse of the theorem. Then say what each contributes.
Hint: What information does each supply?
Answer:
The points are given as coordinates and the question is about an angle, so something must connect the two. The distance formula converts coordinates into side lengths, and the converse converts side lengths into a statement about the angle — neither bridge alone spans the gap.
It is worth noticing how often problems work this way: one tool turns the given information into a form a second tool can use. Recognising which conversion is needed is frequently harder than either tool, and it is the skill that makes a collection of formulas into a method.
Section
Section 5
Concept
Placing coordinate axes over a field, a map or a screen turns any straight-line distance into a hypotenuse. The formula then measures it from the coordinates alone.
Where the origin is placed does not affect the answer.
Figure (svg): A kick measured across a coordinate grid
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 731-735 — Example 3, on the distance a soccer ball was kicked
Picture it
Two gaps, one path.
Figure (svg): A kick measured across a coordinate grid
The dashed lines are the right triangle that the formula uses. Nothing about the real space needs to be square for the method to work.
Worked example
The lesson opener's situation, with coordinates supplied.
\[ \text{A ball travels from } (10, 20) \text{ to } (34, 27) \text{ on a grid marked in yards. How far?} \]
Find the horizontal gap
Why: Thirty-four minus ten.
\[ 24 \]
Find the vertical gap
Why: Twenty-seven minus twenty.
\[ 7 \]
Square and add
Why: 576 plus 49.
\[ 625 \]
Take the root
Why: A perfect square.
\[ 25 \text{ yards} \]
Figure (svg): A kick measured across a coordinate grid
\[ d = \sqrt{625} = 25 \text{ yards} \]
Verify: check the size against the gaps
Why: The answer of twenty-five is a little more than the larger gap of twenty-four and much less than their sum of thirty-one, which is where a hypotenuse must lie. Seven and twenty-four is a Pythagorean triple, which is why it came out whole.
Faded example
Gaps of 24 and 7.
Fill in the blanks
d = \sqrt49 = \sqrt25}} = \sqrt___ = ___
Why: Seven and twenty-four is one of the Pythagorean triples, so the answer is a whole number of yards. That is unusual and is why the numbers were chosen this way.
Worked example
Testing whether the choice of axes affects the answer.
\[ \text{Shift both points by the same amount. Does the distance change?} \]
Shift both points
Why: Add ten to every coordinate.
\[ (20, 30) \text{ and } (44, 37) \]
Find the gaps again
Why: The shifts cancel.
\[ 24 \text{ and } 7 \]
Compute the distance
Why: The same squares.
\[ 25 \]
Conclude
Why: The answer is unchanged.
Figure (svg): A kick measured across a coordinate grid
\[ d = 25 \text{ either way} \]
Verify: say why this must be so
Why: The formula uses only differences of coordinates, and adding the same amount to both values leaves their difference alone. So the origin may be placed wherever is convenient, which is what makes the method practical on a real field.
Trap
\[ d = 24 + 7 = 31 \text{ yards} \]
Add the horizontal and vertical distances
Why: That is how far you would walk.
Thirty-one yards is the distance along the two legs, which is what a player running the two sides would cover. The ball travelled the hypotenuse, which is twenty-five.
\[ d = \sqrt{24^2 + 7^2} = 25 \text{ yards} \]
Use the formula, since the path is direct
Why: A straight line is a hypotenuse.
The direct route is always shorter than going round two sides, which is the same saving the diamond's diagonal gave.
Hypothesis
Two people place their axes differently.
Predict first
Will they compute the same distance?
Correct: Yes, since the formula uses only differences.
\[ (x_2 + k) - (x_1 + k) = x_2 - x_1 \]
Why: Moving the origin changes every coordinate by the same amount, and that amount cancels when the differences are taken — so the two gaps, and hence the distance, are unchanged. Units do matter, since a distance in yards is a different number from the same distance in feet, but that is about scale rather than about position. The freedom to place the origin anywhere convenient is what makes coordinate methods usable on real spaces.
Sorting
Direct or along the grid.
Sort into buckets
Sort each description by which distance it asks for.
Both are legitimate distances and they answer different questions. A thrown or kicked object travels the hypotenuse; anything confined to a grid of streets or edges travels the sum.
Socratic
It comes from a geometry theorem.
Discussion prompt
Say why the distance formula appears far more often than the theorem it came from. Then name two settings where it is used routinely.
Hint: What form is data usually in?
Answer:
Data about positions almost always arrives as coordinates rather than as a drawn triangle, so the formula is the version that fits the input. It also needs no diagram, which makes it usable by a computer as readily as by a person.
Navigation systems compute distances between latitude and longitude pairs, and computer graphics compute distances between pixel positions constantly — to decide what is nearest, what collides, or how bright a light should be. Both are the same formula applied millions of times a second, which is a fair claim to being the most-used result in this book.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Straight line | Along the grid | |
|---|---|---|
| How to compute | the distance formula | add the two gaps |
| Always the | shorter of the two | longer of the two |
| Suits | a kick, a throw, a flight | walking streets or following edges |
The two answer different questions and both are useful. Reading which one a problem wants is a matter of asking what is actually travelling.
Pattern
To find the distance between two points, these five moves cover it.
Step one prevents the only structural error available here, which is subtracting an x from a y. Everything after it is arithmetic.
OpenStax Intermediate Algebra 2e, §11.1 Distance and Midpoint Formulas; Circles §11.1
Check
Subtract, square, add, root.
Check your understanding
What is the distance between (-1, 4) and (2, 3)?
Answer: A
Why: The gaps are three and negative one, whose squares are nine and one, totalling ten.
Check
Squaring removes the sign.
Check your understanding
Does it matter which point you take first?
Answer: A
Why: Reversing a subtraction changes only its sign, and squaring makes both orders give the same value.
Check
Compare the squares.
Check your understanding
Three points give side lengths root 5, root 20 and 5. Is the triangle right-angled?
Answer: A
Why: The converse compares the squares of the sides, and five plus twenty equals twenty-five, which is five squared.
Real world
This is the soccer question from the lesson opener. With a coordinate grid laid over the pitch, a kick from one point to another can be measured from the coordinates alone.
Discussion prompt
A ball travels from the point at 10, 20 to the point at 34, 27, with units in yards. How far was it kicked, and how does that compare with running along the two grid directions?
Hint: The kick is the hypotenuse of the two gaps.
Answer:
\[ d = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ yards} \]
Running twenty-four yards across and then seven up would cover thirty-one yards, so the direct kick saves six.
Note also that the answer does not depend on where the origin was placed: shifting both points by the same amount leaves both gaps unchanged, so any convenient corner of the pitch can be used as the origin. That independence is what makes coordinate methods practical outside a textbook, where nobody has marked an origin on the grass.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is the distance between (-1, 4) and (2, 3)?
Correct: root 10, about 3.16.
\[ d = \sqrt{3^2 + (-1)^2} = \sqrt{10} \]
Why: The horizontal gap is two minus negative one, which is three, and the vertical gap is three minus four, which is negative one — and both are squared, giving nine and one. The first option subtracts the squares instead of adding them, which is what happens when the minus sign inside the second bracket is mistaken for an operation between the two squared terms; the formula always adds, because squaring makes both terms non-negative whatever the differences were. The third option adds the gaps, giving the distance you would walk along two sides rather than the direct line. A sketch settles it quickly: the points are three across and one up, so the direct distance must be a little over three.
Explain it
They subtracted the two squared terms because one difference was negative.
Discussion prompt
In no more than four sentences, explain why the formula always adds. Then remind them where the formula comes from.
Hint: What is the sign of a square?
Answer:
A usable answer: each difference is squared before anything is added, and squaring makes every result positive. So the minus sign inside a bracket disappears at that point and never reaches the addition between the two terms.
It helps to remember that the formula is just the Pythagorean theorem: the two differences are the legs of a right triangle, and the theorem adds the squares of the legs. Nobody would subtract there, and this is the same calculation written with coordinates.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Pairing coordinates is fixed by writing the two points one above the other with their x values aligned. The addition is fixed by remembering the formula is the Pythagorean theorem, which adds. Combining tools is fixed by asking what each one converts. Exact against rounded is fixed by reading whether the question names a precision. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page plot two points on a set of axes, draw the right triangle whose legs are the horizontal and vertical gaps, label those legs with coordinate differences, and derive the distance formula from the Pythagorean theorem beneath the picture. Underneath, find the distance between three pairs of points, keeping the exact radical and writing the rounded value beside it, and mark which pairs came out whole and why. In the middle, compute one of those distances twice with the points taken in opposite orders, and write one sentence saying why the two agree. Beneath that, take three points, compute all three squared distances without ever taking a root, identify the largest, and test the converse of the theorem on them. In the lower corner, sketch a real space with a grid over it, mark two positions, compute the direct distance and the along-the-grid distance, and note how much the direct route saves. Finally, in the margin, write the one structural error the formula allows and how to prevent it.
Every distance on your page should be shorter than the sum of its two gaps and longer than the larger of them. Those two bounds catch any arithmetic slip without rechecking the working.
Recap
Five things, and the first makes the formula something you can rebuild.
| If the question says | Your first move is |
|---|---|
| Find the distance | Subtract like coordinates, then square and add |
| Which point comes first? | Either; the squaring removes the sign |
| Are these a right triangle? | Find all three squared distances |
| Round to the nearest hundredth | Keep the radical until the last step |
| How far did it travel? | Direct line means the formula; along a grid means adding |
Lesson 12.8 asks a related question about the same two points: not how far apart they are, but where the point exactly between them lies. The answer turns out to be an average rather than a root.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.7 The Distance Formula §12.7, pp. 730-735 — everything on these slides traces back here
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