Midpoint, centroid, and equidistant problems

Several standard problem types in 3D coordinate geometry reduce to the same toolkit: midpoint formula, centroid formula, and the equation "find a point equidistant from given points". This subtopic drills these patterns.

Definitions and reminders

• Midpoint of $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}, \dfrac{z_1+z_2}{2}\right)$.
• Centroid of triangle with vertices $A, B, C$: average of the three coordinate triples.
• Equidistant locus: the set of points equidistant from two given points $A, B$ is the perpendicular bisector plane of segment $AB$.
• Centroid of a tetrahedron: average of the four vertices.

Geometric facts used

Parallelogram. If $ABCD$ is a parallelogram (in space), then the diagonals $AC$ and $BD$ share a midpoint. So midpoint of $AC$ = midpoint of $BD$.

Median of a triangle. A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, which divides each median in ratio $2 : 1$ (vertex side : midpoint side).

Tetrahedron centroid. The four lines joining each vertex to the centroid of the opposite face are concurrent at the centroid of the tetrahedron.

Worked examples

Example 1. Vertices of a triangle: $A(2, 4, 6), B(0, 6, -2), C(4, 2, 4)$. Find: (a) The midpoint of $BC$. (b) The centroid. (c) The length of the median from $A$.

(a) $M_{BC} = (2, 4, 1)$. (b) $G = ((2+0+4)/3, (4+6+2)/3, (6-2+4)/3) = (2, 4, 8/3)$. (c) Median length $|AM_{BC}| = \sqrt{0 + 0 + 25} = 5$.

Example 2. Given a parallelogram with vertices $A(1, 0, 0), B(2, 1, 2), C(3, 2, 4)$. Find $D$.

Diagonal midpoints equal: midpoint of $AC$ = midpoint of $BD$.

Midpoint $AC = (2, 1, 2)$. So $(B + D)/2 = (2, 1, 2) \Rightarrow D = 2(2, 1, 2) - (2, 1, 2) = (2, 1, 2)$. Wait, that's $B$ itself, meaning $A, B, C$ are collinear , not a parallelogram.

Let me redo with valid points: $A(1, 2, 3), B(4, 5, 6), C(7, 4, 5)$. Midpoint $AC = (4, 3, 4)$. So $D = 2(4, 3, 4) - (4, 5, 6) = (4, 1, 2)$.

Example 3. Find the point on the $z$-axis equidistant from $(1, -2, 3)$ and $(5, 4, 1)$.

Let the point be $(0, 0, t)$. Equate squared distances: $1 + 4 + (t - 3)^2 = 25 + 16 + (t - 1)^2$. $5 + t^2 - 6t + 9 = 41 + t^2 - 2t + 1$. $14 - 6 t = 42 - 2 t \Rightarrow -4 t = 28 \Rightarrow t = -7$.

So $(0, 0, -7)$.

Example 4. The vertices of a triangle are $A(0, 0, 0), B(a, 0, 0), C(0, b, 0)$. Show that the centroid is $(a/3, b/3, 0)$ and that the length of the median from $A$ to midpoint of $BC$ equals $\dfrac{1}{2}\sqrt{a^2 + b^2}$.

Centroid: $(a/3, b/3, 0)$. ✓ Midpoint of $BC$: $(a/2, b/2, 0)$. Length from $A$: $\sqrt{a^2/4 + b^2/4} = \dfrac{1}{2}\sqrt{a^2 + b^2}$.

Example 5. Three vertices of a parallelogram are $(2, -1, 5), (1, 2, 0), (-2, 5, 5)$. Find the fourth.

If these are $A, B, C$ in order, with $D$ opposite $B$: midpoint of diagonals $AC$ and $BD$ coincide. Midpoint $AC = (0, 2, 5)$. So $D = 2(0, 2, 5) - (1, 2, 0) = (-1, 2, 10)$.

Try it yourself

1. Midpoint of $(3, 4, 5)$ and $(-1, 0, 7)$.
2. Centroid of vertices $(0, 0, 0), (2, 0, 0), (0, 3, 0), (0, 0, 4)$. (Tetrahedron centroid , take average of all four.)
3. Find $a, b, c$ if $(a, b, c)$ is the centroid of $(2, 0, 0), (0, 3, 0), (0, 0, 5)$.
4. Find the point on the $x$-axis equidistant from $(0, -2, 5)$ and $(3, 0, -2)$.
5. Find a point equidistant from $(1, 0, 0), (0, 1, 0), (0, 0, 1)$.
6. Three vertices of a parallelogram are $(2, 3, 4), (5, 4, 2), (0, 0, 6)$. Find the fourth (three possible answers , at least one).
7. Show that the centroid of a triangle divides each median in $2:1$.
8. The points $A(1, 2, 3), B(2, 3, 4), C(0, 1, 2)$ are given. Find the length of the median from $A$.
9. Find $\lambda$ such that the points $(1, 2, 3), (4, 5, 6), (7, 8, \lambda)$ are collinear.
10. Find a point equidistant from $(0, 0, 0)$ and lying on the $x$-axis with distance $4$ from origin.
11. Find the equation of the locus of a point equidistant from $(2, 0, 0)$ and $(0, 3, 0)$.
12. The vertices of a tetrahedron are $(0, 0, 0), (4, 0, 0), (0, 4, 0), (0, 0, 4)$. Find its centroid.

Pitfalls / Tricks

• Always square distances before equating , avoids square roots.
• For a parallelogram, check that the diagonal midpoints coincide. If not, the labelling of $A, B, C$ might need reordering.
• The centroid of a tetrahedron uses four vertices in the average.
• Insight. The equidistant locus is always a plane (perpendicular bisector of the segment joining the two points). In Class XII you will write its equation explicitly.