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Lesson 3 of 4

Introduction to Three-dimensional Geometry · Lesson 3 of 4

Midpoints, Centroids and Distance-Based Applications

“Use coordinate averages and distances to construct figures and translate geometric conditions into equations.”

Learning Objectives

• Derive midpoint and centroid coordinates in three dimensions. • Find medians, missing vertices and unknown coordinate parameters. • Verify a spatial parallelogram using diagonal bisection and distinguish it from a rectangle. • Translate equal distances and fixed distance sums into equations for a variable point. • Keep distance sums distinct from squared-distance sums and check algebraic restrictions.

Combining Coordinates with Geometry

A distance formula measures a known segment, but many geometric questions ask for an unknown point or an entire collection of points. We need to translate the condition that determines those positions. Midpoints give coordinate averages; centroids connect three vertices; equal distances and fixed sums give equations.

Choose the unknown coordinates before calculating. If a single point C is missing, write C=(x,y,z). If any point P satisfying a condition is sought, write P=(x,y,z) and allow all triples that make the condition true. These two uses of x,y,z look similar algebraically, but one seeks a particular location and the other seeks a set of locations.

Midpoints Average Each Coordinate

A midpoint lies halfway from one endpoint to the other along the straight segment. Halfway along the segment means halfway through each coordinate change at the same time. This extends the familiar midpoint rule in a plane to three-dimensional space.

From A=(x₁,y₁,z₁) to B=(x₂,y₂,z₂), the full x-change is x₂−x₁. At the midpoint, x=x₁+(x₂−x₁)/2=(x₁+x₂)/2. The same reasoning applies separately to y and z. It works with negative coordinates as well as positive ones, because the calculation uses signed changes rather than adding unsigned distances.

Midpoint of a spatial segmentLaTeX
A and B are the segment’s endpoints. Each coordinate is averaged independently, with equal endpoint weights.
Example — A midpoint across coordinate planes

Problem
Find the midpoint of A=(3,−1,2) and C=(−1,1,2).

  1. 1.Average the x-coordinates: (3−1)/2=1.
  2. 2.Average the y-coordinates: (−1+1)/2=0.
  3. 3.Average the z-coordinates: (2+2)/2=2. Hence M=(1,0,2).
  4. 4.The zero y-coordinate places M in the ZX-plane. The segment has constant z=2; its midpoint retains that coordinate.

The same formula can recover a missing endpoint. If M is the midpoint of AB, then x_A+x_B=2x_M, and similarly for y and z. Subtract A’s coordinates from twice M’s coordinates to get B. This is a coordinate version of taking an equal step beyond the midpoint.

Example — Recover an endpoint

Problem
A=(2,−3,4) and the midpoint of AB is M=(1,2,0). Find B.

  1. 1.Let B=(u,v,w). The midpoint conditions are (2+u)/2=1, (−3+v)/2=2, (4+w)/2=0.
  2. 2.They give u=0, v=7 and w=−4. Thus B=(0,7,−4).
  3. 3.Check the average of A and B: ((2+0)/2,(−3+7)/2,(4−4)/2)=(1,2,0).
  4. 4.The result uses signed coordinates; taking absolute values before averaging would give the wrong endpoint.

Medians and the Centroid

A median of a triangle joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, even when the triangle lies in a tilted plane in space. Its coordinates can be found by combining the opposite-side midpoint with the two-to-one division of a median.

Definition
Centroid

The common point G of a triangle’s three medians. On a median from a vertex A to an opposite-side midpoint M, AG:GM=2:1, so G is two-thirds of the way from A to M.

Let M be the midpoint of BC. Its x-coordinate is (x_B+x_C)/2. Moving two-thirds of the way from A to M gives x_G=x_A+(2/3)(x_M−x_A)=x_A/3+2x_M/3. Substituting the midpoint coordinate yields (x_A+x_B+x_C)/3. The identical calculation applies to y and z, so every vertex enters with the same weight.

Centroid of a spatial triangleLaTeX
Average the three vertex coordinates component by component. The triangle’s centroid is not generally the midpoint of a side or of a median.
A median and the centroid in spacexyzOABCMG
A median and the centroid in space— This illustration uses A=(0,0,4), B=(0,4,0), C=(4,0,0). M is the midpoint of BC, and G divides AM in the ratio 2:1 from A.
Example — Find a missing triangle vertex

Problem
G=(1,1,1) is the centroid of ABC. Given A=(3,−5,7), B=(−1,7,−6), find C.

  1. 1.Write C=(x,y,z). The centroid equations are (3−1+x)/3=1, (−5+7+y)/3=1, (7−6+z)/3=1.
  2. 2.For x, 2+x=3 gives x=1. For y, 2+y=3 gives y=1.
  3. 3.For z, 1+z=3 gives z=2. Therefore C=(1,1,2).
  4. 4.Check the coordinate totals: A+B+C gives (3,3,3); dividing each total by 3 recovers G.
Example — Parameters fixed by a centroid

Problem
The origin is the centroid of P=(2a,2,6), Q=(−4,3b,−10), R=(8,14,2c). Find a,b,c.

  1. 1.For a centroid at the origin, each total coordinate is zero.
  2. 2.The x-total gives 2a−4+8=0, so 2a=−4 and a=−2.
  3. 3.The y-total gives 2+3b+14=0, so 3b=−16 and b=−16/3.
  4. 4.The z-total gives 6−10+2c=0, so 2c=4 and c=2.
  5. 5.With these values the three coordinate averages are zero, as required.
Example — Lengths of all three medians

Problem
For A=(0,0,6), B=(0,4,0), C=(6,0,0), find the median lengths.

  1. 1.The midpoint of BC is M_A=(3,2,0). Thus AM_A²=3²+2²+(0−6)²=49, so AM_A=7.
  2. 2.The midpoint of CA is M_B=(3,0,3). Thus BM_B²=3²+(0−4)²+3²=34, so BM_B=√34.
  3. 3.The midpoint of AB is M_C=(0,2,3). Thus CM_C²=(0−6)²+2²+3²=49, so CM_C=7.
  4. 4.Every median uses a different opposite-side midpoint. None of these lengths is simply half the length of its opposite side.

Parallelograms in Space Need a Geometric Check

A parallelogram is a planar quadrilateral whose opposite sides are parallel. In a plane, familiar side and diagonal properties make it easy to recognise. In space, four points need not lie in a single plane, so we must not assume planarity from a sketch or from two pairs of equal lengths.

A reliable coordinate test is diagonal bisection: the midpoint of AC must equal the midpoint of BD, with four distinct vertices that do not collapse onto one line. If the diagonals meet at their common midpoint, they lie in one plane determined by the two intersecting diagonal lines. Reflecting both endpoints of one side through that midpoint gives the opposite side, establishing the parallelogram relationship. This supplies the spatial justification that side lengths alone would lack.

Shared midpoint condition for a parallelogramLaTeX
The vertices are named in successive order A,B,C,D. Both diagonals then have the same midpoint. Also check that the resulting figure has four distinct vertices and is not collinear.
Example — A parallelogram but not a rectangle

Problem
Verify that A=(1,2,3), B=(−1,−2,−1), C=(2,3,2), D=(4,7,6) form a parallelogram ABCD, but not a rectangle.

  1. 1.The midpoint of AC is ((1+2)/2,(2+3)/2,(3+2)/2)=(3/2,5/2,5/2).
  2. 2.The midpoint of BD is ((−1+4)/2,(−2+7)/2,(−1+6)/2)=(3/2,5/2,5/2). The diagonals bisect each other.
  3. 3.The figure is nondegenerate: AB²=36, BC²=43, AC²=3, so A,B,C are not collinear. For example, the smallest two lengths √3 and 6 do not add to √43.
  4. 4.The distance calculations also give AB=CD=6 and BC=DA=√43, agreeing with the opposite-side property.
  5. 5.The diagonal lengths are AC=√3 and BD=√155, which are unequal. A rectangle must have equal diagonals.
  6. 6.Therefore the figure is a parallelogram and is not a rectangle. The diagonal midpoint check establishes the needed planar structure before the side-length comparisons.
Example — Verify another spatial parallelogram

Problem
Show that A=(−1,2,1), B=(1,−2,5), C=(4,−7,8), D=(2,−3,4) form a parallelogram in the stated order.

  1. 1.The midpoint of AC is (3/2,−5/2,9/2).
  2. 2.The midpoint of BD is also (3/2,−5/2,9/2), so the diagonals bisect each other.
  3. 3.To check the figure is not collinear, compare coordinate changes AB=(2,−4,4) and BC=(3,−5,3). They cannot have one common scaling factor: the first-coordinate ratio is 3/2 but the second-coordinate ratio is 5/4.
  4. 4.Thus the adjacent sides have different directions and the figure is nondegenerate. It is a parallelogram.
  5. 5.Opposite squared side lengths provide a further check: AB²=CD²=36 and BC²=DA²=43.
Example — A missing parallelogram vertex

Problem
For successive vertices A=(3,−1,2), B=(1,2,−4), C=(−1,1,2), find D.

  1. 1.The diagonals must have a common midpoint. Twice the midpoint of AC is A+C=(2,0,4).
  2. 2.Therefore B+D=(2,0,4), so D=(2−1,0−2,4−(−4))=(1,−2,8).
  3. 3.Check the diagonal midpoints: AC has midpoint (1,0,2); BD has midpoint ((1+1)/2,(2−2)/2,(−4+8)/2)=(1,0,2).
  4. 4.The vertex order matters: this formula treats A and C as opposite vertices and B and D as opposite vertices.
Common mistake

Two pairs of equal opposite-side lengths do not by themselves prove a spatial quadrilateral is planar. Use the shared diagonal midpoint and check nondegeneracy. Also, do not rearrange the supplied vertex order when applying A+C=B+D.

Equal Distances Become an Equation

A point equidistant from two fixed points is not usually unique. We seek the entire set of triples that satisfy the equality. Since distances are nonnegative, equality of their squares is equivalent to equality of the distances, and it removes square roots without losing information.

Definition
Locus

The set of all points satisfying a specified geometric condition. An equation for the locus must include exactly those coordinate triples and no extra ones.

Let the variable point be P=(x,y,z). Write PA² and PB² using the distance formula, then impose PA²=PB². On expanding, the x²,y²,z² terms cancel because they appear once on each side. A linear equation in the three coordinates remains. Here the goal is to obtain and interpret that condition, not to begin a general study of plane equations.

Example — Equidistant points in space

Problem
Find the equation for points equidistant from A=(3,4,−5) and B=(−2,1,4).

  1. 1.Write P=(x,y,z). The equality is (x−3)²+(y−4)²+(z+5)²=(x+2)²+(y−1)²+(z−4)².
  2. 2.The left side expands to x²+y²+z²−6x−8y+10z+50.
  3. 3.The right side expands to x²+y²+z²+4x−2y−8z+21.
  4. 4.Subtract the common square terms and collect: −10x−6y+18z+29=0.
  5. 5.Equivalently 10x+6y−18z−29=0. Every point satisfying this equation has PA²=PB² and hence PA=PB.
  6. 6.As a check, the midpoint of AB is (1/2,5/2,−1/2); substituting gives 5+15+9−29=0.
Example — Cancellation in a simpler distance condition

Problem
Find the equation for points equidistant from (1,2,3) and (3,2,−1).

  1. 1.Let P=(x,y,z) and equate (x−1)²+(y−2)²+(z−3)² to (x−3)²+(y−2)²+(z+1)².
  2. 2.The y-expressions already match and cancel. Expanding the remaining terms gives −2x−6z+10=−6x+2z+10.
  3. 3.Thus 4x−8z=0, or x−2z=0.
  4. 4.The equation imposes no restriction on y. This is reasonable because the two fixed points have the same y-coordinate.
  5. 5.Substitution of their midpoint (2,2,1) verifies the condition.

A Fixed Sum of Squared Distances

The condition PA²+PB²=constant is different from PA+PB=constant. The squared version can be expanded directly as a polynomial. The ordinary distance sum still contains square roots and needs more care if we choose to remove them.

Example — A sum of squared distances

Problem
For A=(3,4,5), B=(−1,3,−7), find the equation of points P with PA²+PB²=2k².

  1. 1.For P=(x,y,z), write (x−3)²+(y−4)²+(z−5)²+(x+1)²+(y−3)²+(z+7)²=2k².
  2. 2.The A-expression expands to x²+y²+z²−6x−8y−10z+50.
  3. 3.The B-expression expands to x²+y²+z²+2x−6y+14z+59.
  4. 4.Adding gives 2x²+2y²+2z²−4x−14y+4z+109=2k².
  5. 5.Move the constant to obtain 2x²+2y²+2z²−4x−14y+4z=2k²−109.
  6. 6.If the condition is instead PA²+PB²=k², the same left side equals k²−109. Do not change k² into 2k² merely because the examples use the same endpoints.

A useful check comes from completing squares. For these fixed points the midpoint is M=(1,7/2,−1), and AB²=16+1+144=161. Expanding around M gives PA²+PB²=2PM²+161/2. The smallest sum is therefore 161/2, attained only at M. This explains why not every value of a stated constant produces any points at all, without introducing additional families of surfaces.

Squared-distance sum about the midpointLaTeX
M is the midpoint of A and B. The identity follows by expanding each coordinate about its average: (u−a)²+(u−b)²=2(u−(a+b)/2)²+(a−b)²/2.

A Fixed Sum of Distances

For ordinary distances, the exact equation may legitimately retain square roots. Removing them by squaring is optional, and every squaring step must be accompanied by the relevant sign conditions. The starting condition is a length sum, so a negative branch introduced by algebra is not acceptable.

Example — Translate a distance sum

Problem
Find an equation for points whose distances from A=(4,0,0) and B=(−4,0,0) add to 10.

  1. 1.With P=(x,y,z), the exact condition is √((x−4)²+y²+z²)+√((x+4)²+y²+z²)=10. This alone is a correct equation of the required set.
  2. 2.To simplify, write rA=PA, rB=PB and s=y²+z²≥0. Then rA²−rB²=−16x.
  3. 3.Since rA+rB=10, (rA−rB)10=−16x, so rA−rB=−8x/5.
  4. 4.Solving the two equations gives rA=5−4x/5 and rB=5+4x/5; both must be nonnegative.
  5. 5.Square the first relation and compare with rA²=(x−4)²+s: x²−8x+16+s=25−8x+16x²/25.
  6. 6.Collect terms to obtain 9x²/25+s=9, or x²/25+(y²+z²)/9=1.
  7. 7.For this polynomial equation, s≥0 implies |x|≤5. Therefore 5±4x/5 are both at least 1, so taking square roots recovers the positive distances rA,rB and their sum 10.
  8. 8.The sign check establishes equivalence, rather than merely showing that the original condition implies the polynomial equation. No further classification of the surface is needed here.

The fixed points are eight units apart, so the triangle inequality already gives PA+PB≥8. A prescribed sum below 8 would have no solution. At the value 8, the points satisfying the condition would lie on the segment AB. The supplied value 10 allows off-segment positions, which is consistent with the equation obtained above.

Common mistake

PA²+PB² is not (PA+PB)²: the latter also has the cross term 2PA·PB. When simplifying a radical equation, preserve nonnegative distance branches and verify the result in the original condition. A polynomial consequence is not automatically an equivalent locus equation.

Quiz

Quick check

The midpoint of (−2,4,6) and (4,0,−2) is:

Quick check

The centroid of (0,0,0), (3,0,0), (0,3,3) is:

Quick check

Which is a reliable spatial parallelogram check, with distinct noncollinear vertices?

Quick check

Why may PA=PB be replaced by PA²=PB²?

Quick check

For PA²+PB²=K, the smallest possible K is:

Quick check

In triangle ABC, the median from A joins A to:

Practice Problems

Practice Problems
  1. Find the midpoint of (−4,3,7) and (2,−1,−5), and classify its location relative to coordinate planes.
  2. Given A=(1,−2,3) and midpoint M=(0,4,1), recover B.
  3. A triangle has centroid (2,0,1) and vertices A=(3,1,−2), B=(−1,2,4). Find C and verify the coordinate averages.
  4. For successive parallelogram vertices A=(0,0,0), B=(2,0,1), C=(3,2,2), find D using diagonal bisection.
  5. Find the equation for points equidistant from (2,0,0) and (−2,0,0); explain whether y and z are restricted.
  6. For A=(1,0,0), B=(−1,0,0), write and simplify PA²+PB²=10.
  7. Translate PA+PB=6 for those same fixed points into an exact radical equation and explain why it differs from the squared-distance condition.
  8. Explain why a distance-sum condition smaller than AB has no possible point P.

Key Takeaways

Key Takeaways

• Midpoints average two endpoint coordinates; centroids average three vertex coordinates. • A median joins a vertex to an opposite-side midpoint; its length then follows from the distance formula. • Shared diagonal midpoints verify a nondegenerate spatial parallelogram, including its planar structure. • A parallelogram with unequal diagonals cannot be a rectangle. • Equal distances may be squared equivalently because distances are nonnegative. • Distance sums and squared-distance sums are different conditions and require different algebra. • Check constant ranges and sign restrictions so a locus equation describes exactly the intended points.

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Distance and Relationships Between Points

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Chapter Summary and Practice