Finding the midpoint of a line segment in the coordinate plane. Includes the midpoint as an average of the coordinates, computing each average separately, checking a midpoint with the distance formula, recovering an endpoint from a midpoint and the other endpoint, and using midpoints to place objects on a screen.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
The Midpoint 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 12.8 The Midpoint Formula §12.8, pp. 736-739 — the lesson these objectives are drawn from
Warm-up
Lesson 12.7 found how far apart two points are. This lesson asks a different question about the same two points.
Discussion prompt
What number lies halfway between negative two and four on a number line? How did you find it?
Hint: Average the two.
Answer:
\[ \tfrac{-2 + 4}{2} = 1 \]
Averaging the two endpoints gives the point halfway between them, and a quick check confirms it: one is three from negative two and three from four. In the plane the same thing is done twice, once for each coordinate.
Concept
The midpoint of a line segment is the point on it that is the same distance from both ends. Its coordinates are the averages of the endpoints' coordinates.
midpoint formula — The midpoint between the points with coordinates x one, y one and x two, y two is the point whose coordinates are the average of the x values and the average of the y values.
A midpoint can be thought of as an average.
Figure (svg): The midpoint formula stated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-736
Section
Section 1
Concept
The midpoint's x coordinate is the average of the two x values and its y coordinate is the average of the two y values. Each average is computed independently.
\[ \left(\dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2}\right) \]
The Study Tip says a midpoint can be thought of as an average.
Figure (svg): The midpoint formula stated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-736 — the Midpoint Formula and its Study Tip on averages
Picture it
Averaging is the whole idea.
Figure (svg): A midpoint on a number line, as the one-dimensional case
On a line the halfway point is obviously an average. The plane adds nothing conceptually — it just requires the same calculation in a second direction.
Worked example
This is Example 1 from the textbook.
\[ \text{Find the midpoint of the segment joining } (-2, 3) \text{ and } (4, 2). \]
Average the x values
Why: Negative two plus four, over two.
\[ 1 \]
Average the y values
Why: Three plus two, over two.
\[ \tfrac{5}{2} \]
Write the point
Why: The two averages together.
\[ \left(1, \tfrac{5}{2}\right) \]
Check on a sketch
Why: It looks halfway.
Figure (svg): Averaging the coordinates separately
\[ \left(1, \tfrac{5}{2}\right) \]
Verify: check each coordinate lies between
Why: One lies between negative two and four, and two and a half lies between two and three. A midpoint's coordinates must always fall between the corresponding endpoint values, which is a quick sanity check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-736
Faded example
Add, then halve.
Fill in the blanks
\left(\tfrac15/2, \tfrac______\right) = \left(___, ___\right)
Why: The two averages are computed separately and written together as one point. A fractional coordinate is perfectly ordinary and simply means the midpoint falls between gridlines.
Worked example
Guided Practice items of the same kind.
\[ \text{Find the midpoints for } (2,3) \text{ and } (4,1); \; (0,0) \text{ and } (4,6). \]
Average the first pair's x values
Why: Two plus four.
\[ 3 \]
Average its y values
Why: Three plus one.
\[ 2 \]
Take the second pair
Why: Nought plus four, nought plus six.
\[ (2, 3) \]
Note the special case
Why: One endpoint at the origin.
Figure (svg): Averaging the coordinates separately
\[ (3, 2), \qquad (2, 3) \]
Verify: notice the origin case
Why: When one endpoint is the origin, the midpoint is simply half of the other point, since averaging with nought halves. That is worth recognising, since it makes those cases instant.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-736
Trap
\[ \left(\tfrac{4 - (-2)}{2}, \tfrac{2 - 3}{2}\right) = (3, -\tfrac{1}{2}) \]
Take half the differences
Why: The distance formula subtracted, so this one did too.
That gives half the gaps rather than the middle position. The point three, negative a half is not even on the segment, which runs from y equal to two to y equal to three.
\[ \left(\tfrac{-2 + 4}{2}, \tfrac{3 + 2}{2}\right) = \left(1, \tfrac{5}{2}\right) \]
Add the coordinates and halve, rather than subtracting
Why: A midpoint is an average.
Checking that each coordinate lies between the two endpoint values catches this immediately.
Matching
Average each pair.
Match the pairs
Why: The last pair share a y value, so the midpoint shares it too — averaging two equal numbers returns that number. Only one average actually needed computing there.
Prediction
Compared with the endpoints.
Predict first
What must be true of the midpoint's x coordinate?
Correct: It lies between the two endpoint x values.
\[ -2 < 1 < 4 \quad \text{and} \quad 2 < \tfrac{5}{2} < 3 \]
Why: An average of two numbers always falls between them, so the midpoint's coordinates are each trapped between the corresponding endpoint values. That gives a free check on any answer: a coordinate outside the range means the arithmetic went wrong, most likely by subtracting instead of adding. The same applies in the vertical direction.
Socratic
The Study Tip asserts it.
Discussion prompt
Explain why averaging the coordinates gives the halfway point. Then say why doing it separately in each direction is enough.
Hint: Start on a number line.
Answer:
On a line the point halfway between two numbers is their average, since it is the same distance above the smaller as below the larger — that is what an average is. The horizontal position of the midpoint is exactly the halfway point of the two horizontal positions, so the same reasoning applies.
Doing it separately works because horizontal and vertical position are independent: moving halfway across and halfway up gets you to the middle of the segment, whatever its slope. That independence is what makes coordinate geometry work at all, and it is why nearly every formula in these lessons treats the two directions one at a time.
Section
Section 2
Concept
The distance formula subtracts the coordinates and the midpoint formula adds them. They answer different questions about the same two points.
\[ \text{midpoint: add} \qquad \text{distance: subtract} \]
One gives a point and the other a length.
Figure (svg): Two columns comparing the midpoint formula with the distance formula
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-736 — the midpoint formula's structure, contrasted with the distance formula of Lesson 12.7
Picture it
Where, or how far.
Figure (svg): Two columns comparing the midpoint formula with the distance formula
Both are used on the same pairs of points and confusing them is easy for that reason. Asking whether the answer should be a point or a number settles which to use.
Worked example
Comparing the two questions directly.
\[ \text{For } (-2, 3) \text{ and } (4, 2), \text{ find the midpoint and the distance.} \]
Find the midpoint
Why: Add and halve each pair.
\[ \left(1, \tfrac{5}{2}\right) \]
Find the gaps
Why: Subtract each pair.
\[ 6 \text{ and } -1 \]
Find the distance
Why: Square, add, root.
\[ \sqrt{37} \]
Compare the answers
Why: A point and a length.
Figure (svg): Two columns comparing the midpoint formula with the distance formula
\[ \left(1, \tfrac{5}{2}\right), \qquad \sqrt{37} \approx 6.08 \]
Verify: note the two answers' forms
Why: One answer is an ordered pair and the other a single number, so they could not be confused once written down. The confusion happens during the calculation, which is why the arithmetic difference is worth fixing in mind.
Sorting
A point or a length.
Sort into buckets
Sort each question by the formula it requires.
Every distance question asks how far and every midpoint question asks where. Reading which of those two words appears is enough to choose correctly.
Worked example
Reading the wording.
\[ \text{Which formula does each phrase call for: how far apart, and where is the centre?} \]
Take the first phrase
Why: It asks for a separation.
Choose the formula
Why: Subtract, square, add, root.
Take the second
Why: It asks for a location.
Choose the formula
Why: Average both coordinates.
Figure (svg): Two columns comparing the midpoint formula with the distance formula
\[ \text{how far} \to \text{distance}; \quad \text{where} \to \text{midpoint} \]
Verify: check the answer's form each time
Why: A distance is a single number with units and a midpoint is a pair of coordinates. If the answer you produce is the wrong kind of object, the wrong formula was used.
Error analysis
The student found the midpoint of a segment.
Annotate
On: \( \begin{aligned} \text{midpoint} &= \left(\frac{4 - (-2)}{2}, \; \frac{2 - 3}{2}\right) \\ &= \left(3, -\tfrac{1}{2}\right) \end{aligned} \)
This is the distance formula's arithmetic applied to the midpoint's question, which is an easy slip since the two lessons use the same pairs of points. Checking that each coordinate lies between the endpoints' values catches it in a second.
Faded example
Not subtract.
Fill in the blanks
\text+ x = \tfrac1} 4}___ = ___
Why: Averaging requires adding and then halving. Subtracting would give three, which is the half-gap rather than the halfway position.
Elimination
From the midpoint formula.
Eliminate the wrong options
What form does a midpoint take?
Survives elimination: A
Why: A midpoint is a place, so it needs two coordinates to describe it. Noticing that the answer should be a pair rather than a number is a quick way to check the right formula was used.
Socratic
They look quite different.
Discussion prompt
Say why the distance and midpoint formulas are easy to mix up. Then give a way of keeping them apart that does not rely on memory.
Hint: What do they have in common?
Answer:
They take exactly the same input — two points — and are taught one after the other, so the temptation is to reach for whichever formula was used most recently. Both also combine the x values and then the y values, so their shapes are superficially similar.
The reliable separator is the kind of answer wanted: a length is one number and a location is two coordinates. Deciding what the answer should look like before starting also tells you which arithmetic is needed, since only averaging produces a point between the two.
Section
Section 3
Concept
A midpoint is equidistant from the two endpoints, so computing both half-distances gives a check. Equal halves confirm the averaging was done correctly.
This combines both of the chapter's coordinate formulas.
Figure (svg): A midpoint checked by measuring both halves
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-737 — the note that Example 2 uses the distance formula to check a midpoint
Picture it
The definition, made checkable.
Figure (svg): A midpoint checked by measuring both halves
The gaps to each endpoint come out the same in both directions, which is why the two distances match. Equal gaps are actually a quicker check than equal distances.
Worked example
Using the distance formula as a check.
\[ \text{Check that } \left(1, \tfrac{5}{2}\right) \text{ is the midpoint of } (-2, 3) \text{ and } (4, 2). \]
Find the gaps to the first endpoint
Why: Three across, a half down.
\[ 3 \text{ and } \tfrac{1}{2} \]
Find the gaps to the second
Why: Three across, a half up.
\[ 3 \text{ and } \tfrac{1}{2} \]
Compare
Why: The gaps match.
Conclude
Why: It is the midpoint.
\[ \;\checkmark \]
Figure (svg): A midpoint checked by measuring both halves
\[ \sqrt{9 + \tfrac{1}{4}} \text{ each way} \]
Verify: note the quicker version of the check
Why: Since both halves have identical gaps, the distances must be equal and there is no need to compute either root. Comparing the gaps is faster than comparing the distances and just as conclusive.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 737-737
Faded example
The gaps should match.
Fill in the blanks
\text3 (-2,3): \text1/2 3 \text___ \tfrac______; \quad \text___ (4,2): \text___ ___ \text___ ___
Why: Identical gaps in both directions guarantee identical distances, so the roots need never be taken. Matching gaps is the strongest and quickest form of this check.
Worked example
Equidistant is not quite enough on its own.
\[ \text{Why is being equidistant from both endpoints not the whole definition?} \]
Recall the definition
Why: On the segment, equidistant.
Consider other equidistant points
Why: Off the segment entirely.
Note what averaging guarantees
Why: Each coordinate lies between.
Conclude
Why: The formula gives both.
Figure (svg): A midpoint checked by measuring both halves
\[ \text{on the segment and equidistant} \]
Verify: find another equidistant point
Why: Any point on the perpendicular through the midpoint is equally far from both endpoints, and none of those except the midpoint itself lies on the segment. The averaging formula automatically produces a point between the two, so it never has this ambiguity.
Trap
The distance from the midpoint to one endpoint is about 3.04, so the midpoint is correct.
Compute one half-distance and stop
Why: Only one calculation seemed necessary.
One distance says nothing on its own, since any number could be produced by a wrong midpoint. The check is that the two halves agree, which requires both to be computed.
Compute both half-distances and confirm they are equal.
Compare the two, rather than evaluating one
Why: The equality is the check.
Comparing the coordinate gaps is quicker still and equally convincing.
Elimination
Verifying a midpoint.
Eliminate the wrong options
What must be confirmed?
Survives elimination: A
Why: Equidistance is what a midpoint means, so that is what a check must establish. The other options describe properties that midpoints may or may not happen to have.
Prediction
The check fails.
Predict first
What has gone wrong?
Correct: The averaging was done incorrectly.
Subtracting instead of averaging typically puts the point off the segment entirely.
Why: A correctly averaged midpoint is always equidistant, so unequal halves mean the midpoint itself is wrong — most often because the coordinates were subtracted rather than added. A slip in the distance calculation is possible too, but it would usually affect both halves similarly rather than making them differ. The check is designed to catch the first error, which is the one this lesson invites.
Socratic
There is little to get wrong.
Discussion prompt
Say why a midpoint calculation is worth checking despite its simplicity. Then give the quickest form of the check.
Hint: What is the error it invites?
Answer:
The formula is short but sits immediately after one that looks similar and subtracts, so the error it invites is a wrong operation rather than wrong arithmetic. That kind of error produces a clean-looking answer, which is exactly the kind a check is needed for.
The quickest check is to confirm each coordinate of the answer lies between the two corresponding endpoint values. That takes about two seconds, needs no calculation, and catches the subtraction error every time, since subtracting typically produces a coordinate outside the range.
Section
Section 4
Concept
If the midpoint and one endpoint are known, the other endpoint follows: each of its coordinates is twice the midpoint's, less the known endpoint's.
\[ x_2 = 2m_x - x_1 \]
Rearranging the averaging equation gives it.
Figure (svg): Recovering an endpoint from a midpoint
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-739 — the midpoint formula, rearranged
Picture it
Twice the middle, less one end.
Figure (svg): Recovering an endpoint from a midpoint
This is more often the useful direction, since a midpoint and one end are frequently known while the other end is what you want to locate.
Worked example
Rearranging the formula.
\[ \text{A segment has midpoint } (1, 5) \text{ and one endpoint } (-2, 3). \text{ Find the other.} \]
Set up the x equation
Why: The average is one.
\[ \tfrac{-2 + x_2}{2} = 1 \]
Solve it
Why: Multiply by two, then add.
\[ x _{2} = 4 \]
Set up the y equation
Why: The average is five.
\[ \tfrac{3 + y_2}{2} = 5 \]
Solve it
Why: Ten minus three.
\[ y _{2} = 7 \]
Figure (svg): Recovering an endpoint from a midpoint
\[ (4, 7) \]
Verify: average the two endpoints
Why: Negative two and four average to one, and three and seven average to five — which is the given midpoint. Checking by going forwards is the natural test for any backwards calculation.
Faded example
Rearranged from the average.
Fill in the blanks
x_2 = 2(1) - (-2) = 4
Why: Doubling the midpoint gives the sum of the two endpoints, and subtracting the known one leaves the unknown. Both coordinates are handled the same way.
Worked example
Avoiding the equations.
\[ \text{Why is the other endpoint twice the midpoint, less the known endpoint?} \]
Start from the average
Why: The midpoint's definition.
\[ \tfrac{x_1 + x_2}{2} = m \]
Multiply by two
Why: Clear the fraction.
\[ x _{1} + x _{2} = 2 m \]
Subtract the known end
Why: Isolate the unknown.
\[ x _{2} = 2 m - x _{1} \]
Apply it
Why: Twice one, less negative two.
\[ 2 + 2 = 4 \]
Figure (svg): Recovering an endpoint from a midpoint
\[ x_2 = 2m - x_1 \]
Verify: test the shortcut on the y values
Why: Twice five is ten, less three is seven, matching the answer found by solving. The shortcut is the same equation rearranged once and for all, which saves setting it up each time.
Trap
\[ x_2 = 1 + (1 - (-2)) = 4 \;\checkmark \text{ but } y_2 = 5 + (5 - 3) = 7 \;\checkmark \]
Add the gap from the endpoint to the midpoint once more
Why: The midpoint is halfway, so going the same again reaches the end.
This actually works and is a legitimate alternative — but it is easy to apply in the wrong direction, subtracting the gap and landing back near the known endpoint. The doubling form has no direction to get wrong.
\[ x_2 = 2(1) - (-2) = 4 \]
Use twice the midpoint less the known endpoint
Why: One formula, no direction to choose.
Both methods are correct; the second is simply harder to misapply.
Sorting
Three quantities, any one may be missing.
Sort into buckets
Sort each situation by what has to be computed.
The formula relates three quantities, so knowing any two determines the third. Which direction to use is decided by which one is missing.
Hypothesis
Finding a midpoint seems the natural question.
Predict first
When would you know a midpoint but not an endpoint?
Correct: When a centre is known and a far edge must be located.
Reflecting a point through a centre uses exactly this calculation.
Why: A centre is often the fixed, known thing — the middle of a screen, the pivot of a mechanism, the centre of a circle — while an edge or an opposite corner is what must be found. Knowing where something is centred and where one end sits determines the other end completely, which is a common practical situation and is why the rearranged form is worth having.
Socratic
It has a geometric name.
Discussion prompt
Describe geometrically what twice the midpoint minus a point produces. Then say where you have met that idea before.
Hint: Think about reflection.
Answer:
It reflects the known point through the midpoint: the result is the same distance from the midpoint, on the opposite side and in line with it. That is exactly what the far endpoint of a segment is, which is why the same formula answers both questions.
Reflection through a point appears whenever something is described as opposite or diametrically across — the far side of a circle from a given point, or the position of a counterweight balancing a load. The midpoint formula run backwards is the coordinate version of that idea.
Section
Section 5
Concept
Any setting where positions are recorded as coordinates allows midpoints to be computed directly. Placing something between two objects is a single application of the formula.
Games and interfaces do this constantly.
Figure (svg): A midpoint used to place an object on a screen
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 737-739 — Example 3, on locating a midpoint in computer game design
Picture it
Two positions, one average.
Figure (svg): A midpoint used to place an object on a screen
Screen coordinates work exactly like the ones in this chapter, though the vertical axis often points downwards. The averaging is unaffected by that choice.
Worked example
The lesson opener's situation, with positions supplied.
\[ \text{Two objects sit at } (120, 80) \text{ and } (400, 260). \text{ Where is the midpoint?} \]
Average the x values
Why: A hundred and twenty plus four hundred.
\[ 260 \]
Average the y values
Why: Eighty plus two hundred and sixty.
\[ 170 \]
Write the point
Why: The two averages.
\[ (260, 170) \]
Check the ranges
Why: Both lie between.
\[ \;\checkmark \]
Figure (svg): A midpoint used to place an object on a screen
\[ (260, 170) \]
Verify: check both coordinates lie between
Why: Two hundred and sixty is between a hundred and twenty and four hundred, and a hundred and seventy is between eighty and two hundred and sixty. Both fall in range, as a midpoint's coordinates must.
Faded example
x with x, y with y.
Fill in the blanks
\left(\tfrac400}}170, \tfrac______\right) = (260, ___)
Why: Each average uses the two values measuring the same direction. Pairing an x with a y produces a point that is not on the segment at all.
Worked example
Extending the idea beyond halfway.
\[ \text{Where is the point a quarter of the way from } (0, 0) \text{ to } (8, 12)? \]
Find the midpoint
Why: Average the two.
\[ (4, 6) \]
Find the midpoint of the first half
Why: Average again.
\[ (2, 3) \]
Check the position
Why: Half of a half.
Note the pattern
Why: Repeated averaging.
Figure (svg): A midpoint on a number line, as the one-dimensional case
\[ (2, 3) \]
Verify: check against the full segment
Why: Two is a quarter of eight and three is a quarter of twelve, which is what a quarter of the way from the origin should give. Repeated halving reaches any fraction whose denominator is a power of two.
Trap
\[ \left(\tfrac{120 + 80}{2}, \tfrac{400 + 260}{2}\right) = (100, 330) \]
Average the first point's coordinates and the second's
Why: The numbers were paired as they appear.
An x has been averaged with a y, which compares horizontal with vertical position. The result is not on the segment at all — its x of a hundred is outside the range of a hundred and twenty to four hundred.
\[ \left(\tfrac{120 + 400}{2}, \tfrac{80 + 260}{2}\right) = (260, 170) \]
Average x with x and y with y
Why: Like coordinates only.
Writing the two points one above the other with their x values aligned prevents this, exactly as in Lesson 12.7.
Prediction
Beyond the halfway point.
Predict first
What is the simplest approach?
Correct: Take the midpoint, then the midpoint of the first half.
\[ (0,0) \to (4,6) \to (2,3) \]
Why: Halving twice gives a quarter, so applying the formula to the original start point and the midpoint locates the quarter point. Dividing coordinates by four only works when the segment starts at the origin, which is a special case rather than the general method. Repeated halving reaches any fraction whose denominator is a power of two, and other fractions need a slightly different weighted average.
Sorting
Any setting with coordinates.
Sort into buckets
Sort each situation by whether a midpoint calculation applies.
Positioning questions need midpoints and measuring questions need distances. Both arise constantly wherever coordinates are used, which is why the two formulas are usually learnt together.
Socratic
The formulas apply directly there.
Discussion prompt
Say why representing positions as coordinates makes calculations like this possible. Then say what would be needed without them.
Hint: What can be done with numbers that cannot be done with a picture?
Answer:
Coordinates turn positions into numbers, and numbers can be added, averaged and compared by a procedure. A machine has no way of looking at a picture and judging where the middle is, but it can average two pairs of numbers millions of times a second.
Without coordinates the same task would need a geometric construction — bisecting a segment with compasses, say — which is exact and entirely unsuitable for automation. Descartes's idea of putting numbers on positions is what makes every calculation in this chapter mechanical, and it is why coordinate geometry underlies so much of what computers do with space.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Midpoint | Distance | |
|---|---|---|
| Coordinates are | added and halved | subtracted, squared and added |
| The answer is | a point, given as an ordered pair | a single length |
| Answers the question | where is the middle | how far apart are they |
Both take two points as input and produce entirely different kinds of output. Deciding what the answer should look like is the surest way to pick the right one.
Pattern
To find a midpoint, or to use one, these five moves cover it.
Step four is the two-second check that catches the only real error here, which is subtracting when averaging was needed.
OpenStax Intermediate Algebra 2e, §11.1 Distance and Midpoint Formulas; Circles §11.1
Check
Average each coordinate.
Check your understanding
What is the midpoint of (-2, 3) and (4, 2)?
Answer: A
Why: Negative two and four average to one, and three and two average to five halves.
Check
Add, do not subtract.
Check your understanding
How does the midpoint formula differ from the distance formula?
Answer: A
Why: A midpoint is an average, so the coordinates are added and halved, while a distance uses their differences.
Check
Rearrange the formula.
Check your understanding
A segment has midpoint (1, 5) and one endpoint (-2, 3). What is the other endpoint?
Answer: A
Why: Twice the midpoint less the known endpoint gives two minus negative two, which is four, and ten minus three, which is seven.
Real world
This is the game design question from the lesson opener. Objects on a screen have coordinate positions, so placing something midway between two of them is a direct use of the formula.
Discussion prompt
Two objects sit at the points 120, 80 and 400, 260. Find the point midway between them, and then find the point a quarter of the way from the first to the second.
Hint: Average once for the midpoint, then average again.
Answer:
\[ \left(\tfrac{120 + 400}{2}, \tfrac{80 + 260}{2}\right) = (260, 170) \]
Averaging the first point with that midpoint gives the quarter point: a hundred and twenty with two hundred and sixty averages to a hundred and ninety, and eighty with a hundred and seventy averages to a hundred and twenty-five, so it is at a hundred and ninety, a hundred and twenty-five.
Both coordinates of each answer lie between the corresponding values of the two originals, which is the quick check that the averaging was done and not a subtraction. Repeated halving like this reaches any fraction whose denominator is a power of two, which is why interfaces so often position things at halves, quarters and eighths of a span.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is the midpoint of the segment joining (-2, 3) and (4, 2)?
Correct: (1, 5/2).
\[ \left(\tfrac{-2+4}{2}, \tfrac{3+2}{2}\right) = \left(1, \tfrac{5}{2}\right) \]
Why: A midpoint is an average, so the coordinates are added and halved: negative two plus four gives two, halved to one, and three plus two gives five, halved to five halves. The first option applies the distance formula's subtraction, which gives half of each gap rather than the halfway position — and its y coordinate of negative a half is not even between two and three, which the range check catches instantly. The third option adds correctly but forgets to halve, and the fourth stops at the differences. The confusion between adding and subtracting is easy precisely because this lesson follows one that subtracts the same coordinates, which is why the check that each answer coordinate lies between the two endpoint values is worth doing every time.
Explain it
They subtracted the coordinates because that is what the distance formula did.
Discussion prompt
In no more than four sentences, explain what a midpoint actually is. Then give them a check that takes two seconds.
Hint: It is an average.
Answer:
A usable answer: a midpoint is the average of the two endpoints, so you add the coordinates and halve, once for the x values and once for the y values. Subtracting gives you half the gap between them, which is a distance rather than a position.
The check is that each coordinate of your answer must lie between the two you started with. If a coordinate falls outside that range, you subtracted when you should have averaged.
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: Adding is fixed by remembering that a midpoint is an average. Pairing is fixed by writing the points one above the other with their x values aligned. Working backwards is fixed by the rule twice the midpoint less the known endpoint. Choosing between formulas is fixed by asking whether the answer should be a point or a length. 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 draw a number line, mark two values and their average, and write beside it that the midpoint in the plane is this done twice. Underneath, plot two points on axes, compute their midpoint by averaging each coordinate separately, plot the result, and confirm by eye that it lies halfway along the segment. In the middle, write the midpoint and distance formulas side by side, and mark clearly that one adds and the other subtracts, with a note of what kind of answer each produces. Beneath that, check one of your midpoints by computing the coordinate gaps to each endpoint and confirming they match, then explain in a sentence why matching gaps make the distances equal without any roots being taken. In the lower half, take a midpoint and one endpoint and recover the other, both by solving the averaging equation and by the shortcut of twice the midpoint less the known end, and note that this is a reflection through the midpoint. Finally, in the margin, write the two-second range check.
Every midpoint on your page should have both coordinates lying between the corresponding endpoint values. That single condition catches the subtraction error, which is the only mistake this formula really invites.
Recap
Five things, and the first is the whole idea.
| If the question says | Your first move is |
|---|---|
| Find the midpoint | Average each pair of coordinates |
| Find the distance | Subtract, square, add and root |
| Check a midpoint | Compare the gaps to each endpoint |
| Find the other endpoint | Twice the midpoint, less the known one |
| Place something halfway | Average both screen coordinates |
Lesson 12.9 closes the course by looking at how mathematical statements are justified rather than computed. Conditional statements, converses and indirect proof are what turn the results of this book into a connected body of reasoning.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.8 The Midpoint Formula §12.8, pp. 736-739 — everything on these slides traces back here
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