A Line Segment Has an Endpoint at (2, 5): If the Midpoint Is (6, 11), the Other Endpoint Is (10, 17) — Worked With the Addition Back-Solve

The other endpoint is (10, 17). To get it, double the midpoint: twice (6, 11) gives (12, 22), then subtract the known endpoint (2, 5) to get (10, 17). The boundary that matters is the formula itself — the midpoint is the average of the two endpoints, so the missing endpoint is always 2M − A, where M is the midpoint and A is the endpoint you already have.

How the Midpoint Formula Works in Reverse

The midpoint formula is a pair of averages, and that's the whole engine behind every problem shaped like this one. For a segment with endpoints A(x₁, y₁) and B(x₂, y₂), the midpoint M sits at ((x₁ + x₂)/2, (y₁ + y₂)/2). Read it as the mean of the coordinates, not a point halfway along a ruler. Once you see it as an average, the reverse move stops feeling like a trick.

Solving for the unknown endpoint is a one-line rearrangement in each coordinate. If M = (x₁ + x₂)/2, then x₂ = 2M_x − x₁. The same identity holds for the y-coordinate. This is why the back-solve costs almost nothing: you multiply the midpoint by 2 and subtract the endpoint, coordinate by coordinate. Memorize 2M − A and the problem collapses to two subtractions.

The relationship also tells you where the answer must land. Because the midpoint is the average, the known endpoint and the unknown endpoint sit symmetric about M — the same distance from M on opposite sides. That symmetric picture gives a built-in error check long before you finish the arithmetic. If your candidate answer looks closer to the midpoint than the known endpoint does, something went wrong.

One common snag is sign errors, especially with negative coordinates. Subtracting a negative number adds it, and mixing the order — A − 2M instead of 2M − A — flips the answer to the wrong symmetric point across the known endpoint. Run the check described below every time, and these slips become obvious rather than silent.

Running the Example: (2, 5), Midpoint (6, 11), Answer (10, 17)

Take the two coordinates separately, because the midpoint formula never mixes x with y. With M = (6, 11) and A = (2, 5), the x-coordinate gives 2 × 6 − 2 = 10, and the y-coordinate gives 2 × 11 − 5 = 17. Both land together as the point (10, 17), and the x-value should be computed first simply because alphabetical order keeps the work tidy.

Verification is the part most students skip. Average (2, 5) and (10, 17): the x-mean is (2 + 10)/2 = 6 and the y-mean is (5 + 17)/2 = 11, which reproduces the given midpoint exactly. The check costs two additions and two halvings, and it catches reversed order, sign errors, and dropped numbers in a single pass.

Look at the distances while you're there. From A(2, 5) to M(6, 11) the change is +4 in x and +6 in y. From M(6, 11) to B(10, 17) the change is again +4 in x and +6 in y. The two halves of the segment are identical vectors — that's the symmetry at work, not a coincidence of this particular problem.

So the full set of facts for this segment: endpoint A = (2, 5), midpoint M = (6, 11), endpoint B = (10, 17). Keep those three points visible on one line of scratch work before moving on, because every variation below reuses them.

Back-Solving One Coordinate at a Time

Every midpoint reverse problem splits into two independent one-variable equations, and treating them that way removes most of the confusion. Write (x + x_known)/2 = x_mid for the horizontal piece and (y + y_known)/2 = y_mid for the vertical piece. Neither equation can contaminate the other, so you can solve them in any order and in any mood.

The algebra is two steps per coordinate. Multiply both sides by 2 to clear the fraction, then subtract the known coordinate. For our example, x: (x + 2)/2 = 6 → x + 2 = 12 → x = 10. For y: (y + 5)/2 = 11 → y + 5 = 22 → y = 17. State each result as you finish it so the final pairing step is mechanical.

This separation matters when the question gives you only part of the data. Some problems hand you the midpoint's x but describe the y informally, and you solve whatever is given. Others supply a fractional midpoint such as (6.5, 11), where doubling gives 13 and subtracting 2 gives 11 for the missing x. Fractions are just division — keep them as fractions through the doubling step and convert at the end.

If the unknown coordinate is described with a variable expression rather than a single number — say the endpoint is (x, 2x + 1) — you substitute that entire expression into the midpoint equation. The doubling and subtraction still apply, but now one equation links the two coordinates. Solve it as a standard linear equation in one variable, then evaluate the expression to report the actual point. This is the most common extension of the basic problem and the one that trips people who only memorized 2M − A.

Worked Example With Negative Coordinates: (2, 5) to Midpoint (−3, 7)

Negative numbers are where back-solving earns its reputation, so walk a full negative case. Let A = (2, 5) and M = (−3, 7). The x-value is 2 × (−3) − 2 = −6 − 2 = −8. The y-value is 2 × 7 − 5 = 14 − 5 = 9. The other endpoint is (−8, 9).

Verify: the x-mean of 2 and −8 is (2 + (−8))/2 = −6/2 = −3, and the y-mean of 5 and 9 is (5 + 9)/2 = 7. Both match, so the sign choreography held. Notice the x computation required subtracting a positive from a negative, which drifts further left — the answer must be more negative than the midpoint, and it is.

Watch the two classic order mistakes. Computing A − 2M gives 2 − (−6) = 8 in x and 5 − 14 = −9 in y, producing (8, −9) — a real-looking point that fails verification because it's the reflection across A, not the reflection across M. Computing 2A − M gives (4 + 3, 10 − 7) = (7, 3), another plausible-looking decoy. Both errors survive casual inspection, and only the averaging check kills them.

A quick distance sanity check helps before you commit. From A(2, 5) to M(−3, 7) the change is −5 in x and +2 in y. If the unknown endpoint were correct, the second half should repeat that same change: (−3 − 5, 7 + 2) = (−8, 9). Adding the half-vector to the midpoint is a second, independent route to the same answer, and it's often faster than the formula when coordinates are integers.

Two-Point-Guess Check and the Distance Test

The symmetry gives you a free error detector that most textbook solutions never mention. Because M is the average, the vector from A to M must equal the vector from M to B. Compute the first half-step, then add it to M. If that lands on your candidate answer, the candidate is correct; if it doesn't, discard it regardless of how neat the arithmetic looked.

The distance test is the same idea in a different costume. The length AM must equal the length MB, and both must be exactly half of AB. For (2, 5), (6, 11), (10, 17): AM = √(4² + 6²) = √52, MB = √(4² + 6²) = √52, and AB = √(8² + 12²) = √208 = 2√52 — precisely twice. A student who reports an endpoint that makes the halves unequal has made an arithmetic slip even if the midpoint equation "looked right."

Run either check in about ten seconds, which makes it cheap enough to do on every problem. In an exam setting, the vector check is faster because it avoids square roots; in a homework setting where you want a second confirmation, the distance version proves the midpoint sits exactly halfway in length as well as position. Stack the two checks when the answer matters.

This is also the practical way to catch the reversed-symmetry decoys described earlier. The point (8, −9) from the order mistake produced a vector from M of (11, −16), which doesn't match the half-vector (−5, 2). One comparison exposes it.

Which Coordinate You Plug In First: A Quick Decision Table

The order of operations differs depending on what the problem gives you, and the table below is the fastest way to pick the right route. Each row is a data pattern you'll actually meet in coordinate geometry homework and tests, plus the identity that solves it in one shot.

The origin row is a genuine shortcut, not a trick: if one endpoint is (0, 0), then the missing endpoint is simply twice the midpoint. The table never changes the underlying identity — every row is still 2M − A — but naming the shortcut stops you from doing unnecessary work under time pressure. Confirm on the exam's formula sheet whether a midpoint formula is supplied; most standardized tests in the United States provide one, while many classroom tests expect it memorized, so check your own syllabus rather than assume.

If the problem supplies no numeric midpoint at all and only describes it in words, translate the description first. "The midpoint is twice as far from the y-axis as the endpoint" is a constraint, not a point; turn it into coordinates, then return to the table.

Why This Rule Is Linear, Not Quadratic

Part of the confusion in these problems comes from treating the midpoint like a distance formula, which involves squares and square roots. The midpoint relationship is linear — each coordinate is a plain average. There's no squaring anywhere, which is why the back-solve is just doubling and subtracting, and why two different students doing it correctly always land on the same answer with no branching cases.

The distance formula does appear in the check step, and that's where people blend the two ideas. Distances are square-rooted; midpoints are not. If you find yourself taking a square root mid-computation instead of at verification, pause — you've switched formulas. The actual subtraction sequence, in the order it must happen, is:

1: Write down M and A, labeling each coordinate's role. 2: Double each coordinate of M: (2M_x, 2M_y). 3: Subtract A's coordinates one at a time: (2M_x − A_x, 2M_y − A_y). 4: Verify by averaging A and your answer to reproduce M. 5: Optional: confirm AM = MB, or that M + (M − A) equals your answer.

Those five steps fail in predictable places. Failure case one: reversing subtraction order and getting the reflection across A. Failure case two: averaging M and A again instead of doubling, which yields the quarter point rather than the endpoint. If step 4 fails, the culprit is almost always step 3 or step 2, so redo those two before touching anything else.

Because the rule is linear, it extends cleanly to three-dimensional problems. In 3D, the midpoint of endpoints (x₁, y₁, z₁) and (x₂, y₂, z₂) is the coordinate-wise average, and the missing endpoint is 2M − A in each of the three coordinates. No new technique required — just a third subtraction. The same linearity explains the "section formula" generalization students meet later, where a point divides a segment in a given ratio; the midpoint is the special case with ratio 1:1.

Common Mistakes and the Warning Signs That Catch Them

The most frequent mistake is dividing the midpoint by 2 instead of doubling it, a slip that produces a point pulled toward the origin. There is a visible warning sign: if both coordinates of your answer sit between A and M rather than beyond M in the same direction, you've averaged rather than back-solved. The unknown endpoint must lie on the far side of M from A, always.

Confusing the midpoint with the distance between the endpoints causes a different disaster. A student computes the distance from A to M as √52 ≈ 7.21 and tries to add that length to M, producing a scalar answer instead of coordinates. That mixes two incompatible formulas and produces an answer that works in only the rare case where the segment runs along a single axis. The direction of the half-segment is the vector (M − A), not the distance √((M − A)²).

Rounding is a subtle saboteur when midpoints carry fractions or decimals. Doubling 3.5 gives exactly 7, but doubling 3.33 gives 6.66 and then encourages premature rounding that won't re-average cleanly. Keep decimals as exact values or as fractions — three-fourths rather than 0.75 when the problem gives you fractions — and do the final decimal conversion only when reporting the answer.

The final warning sign is a "clean" answer that fails midpoint verification by a small amount. That's not rounding noise; it's a sign error or a dropped coordinate. Because the correct answer always reproduces the given midpoint exactly, any mismatch — even one unit in one coordinate — means an error upstream, and there's never a reason to accept an answer that fails a ten-second check.

FAQ

What if the midpoint is given in decimals like (6.5, 11)?

Double it first: 2 × 6.5 = 13, and 2 × 11 = 22, then subtract the known endpoint coordinate by coordinate. With A = (2, 5), the x becomes 13 − 2 = 11 and the y becomes 22 − 5 = 17, for an answer of (11, 17). The verification is the x-mean of 2 and 11 = 6.5, matching the given decimal exactly. Never round the doubling step; decimals stay exact through subtraction.

Can the answer be the same point as the known endpoint?

Only if the midpoint equals the endpoint, which means the segment has zero length — a single point, not a proper segment. In that degenerate case, 2M − A gives A back, and any sensible problem excludes it. If you hit this on homework, re-read the midpoint, because a copied coordinate with the wrong sign is far likelier than a deliberately degenerate segment.

How do I find the midpoint of the whole segment ending at (10, 17) and (2, 5)?

That's the endpoint question reversed, and it's simpler: average the coordinates directly. The x-mean is (2 + 10)/2 = 6 and the y-mean is (5 + 17)/2 = 11, giving (6, 11) — the midpoint you started with. Averaging always runs from two endpoints; doubling-and-subtracting only enters when one endpoint is unknown.

Is the midpoint formula the same in three dimensions?

Yes, with one added coordinate. For points (x₁, y₁, z₁) and (x₂, y₂, z₂), the midpoint is the coordinate-wise average, and the missing endpoint is 2M − A in all three coordinates. Nothing new is needed — a third subtraction per coordinate. The 3D midpoint is the special case of the section formula at ratio 1:1, so the linearity of the 2D rule carries straight over.

Do I lose points for skipping the verification step?

Skip it and you lose the safety net that catches reversed subtraction, dropped signs, and wrong order. Losing a mark to a silent sign error is common; recapturing it costs only two additions and two divisions. If a standardized test doesn't supply the midpoint formula, memorizing 2M − A as a single identity saves time and reduces the chance of reverting to the wrong rearrangement.

What if the y-coordinate of the known endpoint is missing entirely?

The midpoint formula decouples the coordinates, so solve whichever equation is fully specified, then hop to the other once the missing piece resurfaces. If the known endpoint is (2, y), the midpoint equation for y reads (y + M_y_something)/2 — substitute the midpoint's y, clear the fraction, and subtract. Many textbook fill-in problems disclose the missing coordinate in a hidden clause of the question, so scan the wording for it.