Introduction to Three Dimensional Geometry • Topic 3 of 3

Section Formula in Three Dimensions

Suppose a point $R$ lies on the line through $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ and divides the join in the ratio $m : n$. The section formula gives the coordinates of $R$ — again the plane version with a third coordinate handled identically.

Internal division (the point $R$ lies between $P$ and $Q$):

$$R = \left(\dfrac{mx_2 + nx_1}{m + n},\ \dfrac{my_2 + ny_1}{m + n},\ \dfrac{mz_2 + nz_1}{m + n}\right)$$

External division (the point $R$ lies on the line but outside the segment): replace $n$ by $-n$, so the denominator becomes $m - n$.

$$R = \left(\dfrac{mx_2 - nx_1}{m - n},\ \dfrac{my_2 - ny_1}{m - n},\ \dfrac{mz_2 - nz_1}{m - n}\right)$$

Midpoint (the special internal case $m = n = 1$):

$$M = \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2},\ \dfrac{z_1 + z_2}{2}\right)$$

Centroid of a triangle with vertices $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, $C(x_3,y_3,z_3)$ — the plain average of the three vertices:

$$G = \left(\dfrac{x_1 + x_2 + x_3}{3},\ \dfrac{y_1 + y_2 + y_3}{3},\ \dfrac{z_1 + z_2 + z_3}{3}\right)$$
CaseRatio usedCoordinate (e.g. $x$)
Internal division$m : n$$\dfrac{mx_2 + nx_1}{m+n}$
External division$m : n$ (outside)$\dfrac{mx_2 - nx_1}{m-n}$
Midpoint$1 : 1$$\dfrac{x_1 + x_2}{2}$
Centroid (triangle)medians meet at $2:1$$\dfrac{x_1 + x_2 + x_3}{3}$

Finding an unknown ratio. To find the ratio in which a given point or a coordinate plane divides $PQ$, set the ratio as $k : 1$ and use one coordinate of the section formula. For a coordinate plane the relevant coordinate is zero — the $YZ$-plane forces $x = 0$, the $ZX$-plane forces $y = 0$, the $XY$-plane forces $z = 0$ — giving a single equation in $k$. A positive $k$ means the division is internal; a negative $k$ means it is external.

Verifying special quadrilaterals. The midpoint formula gives a clean coordinate test: a quadrilateral is a parallelogram exactly when its two diagonals share the same midpoint (they bisect each other). If, in addition, the diagonals are equal in length the parallelogram is a rectangle; combining equal diagonals with equal adjacent sides confirms a square. This turns a picture-heavy geometry question into a short calculation.

Deeper Insight — one weighted average, four results: Every formula above is the same operation — a weighted average of coordinates — wearing different clothes. Internal division weights $Q$ by $m$ and $P$ by $n$ (note the cross-pairing: the $m$ measuring the pull toward $Q$ multiplies $Q$’s coordinate $x_2$), and dividing by $m+n$ keeps the result a genuine average. The midpoint is the symmetric case $1:1$; external division is the same algebra with one weight made negative, which is why the sign flips to $m - n$. Even the centroid fits: it sits two-thirds of the way along each median, dividing it $2:1$ from the vertex, and that ratio produces the clean average of all three vertices. So you really only need to memorise the internal-division formula carefully — the rest are quick specialisations. A practical warning: keep the order $(P, Q)$ and the ratio $(m, n)$ consistent, because swapping them silently changes which point you compute.

Point R divides segment PQ internally in the ratio m to n Internal Division in Ratio m : n P(x₁, y₁, z₁) R Q(x₂, y₂, z₂) mn Centroid G of a triangle is the average of its three vertices Centroid G = average of A, B, C G ABC
Solved Examples
1
Worked Example
Find the coordinates of the point that divides the segment joining $P(1, -2, 3)$ and $Q(3, 4, -5)$ internally in the ratio $2 : 3$.
Solution
  1. Here $m = 2$, $n = 3$, $P = (1,-2,3)$, $Q = (3,4,-5)$, and $m + n = 5$.
  2. $x = \dfrac{2(3) + 3(1)}{5} = \dfrac{6 + 3}{5} = \dfrac{9}{5}$.
  3. $y = \dfrac{2(4) + 3(-2)}{5} = \dfrac{8 - 6}{5} = \dfrac{2}{5}$.
  4. $z = \dfrac{2(-5) + 3(3)}{5} = \dfrac{-10 + 9}{5} = -\dfrac{1}{5}$.

Answer: $\left(\dfrac{9}{5},\ \dfrac{2}{5},\ -\dfrac{1}{5}\right)$.

2
Worked Example
Find the midpoint of the segment joining $A(4, -3, 2)$ and $B(-2, 5, 8)$.
Solution
  1. $x = \dfrac{4 + (-2)}{2} = 1$.
  2. $y = \dfrac{-3 + 5}{2} = 1$.
  3. $z = \dfrac{2 + 8}{2} = 5$.

Answer: Midpoint $= (1, 1, 5)$.

3
Worked Example
Find the point which divides the segment joining $P(2, 1, 4)$ and $Q(5, -2, 1)$ externally in the ratio $2 : 1$.
Solution
  1. External division: $m = 2$, $n = 1$, so $m - n = 1$.
  2. $x = \dfrac{2(5) - 1(2)}{1} = 10 - 2 = 8$.
  3. $y = \dfrac{2(-2) - 1(1)}{1} = -4 - 1 = -5$.
  4. $z = \dfrac{2(1) - 1(4)}{1} = 2 - 4 = -2$.

Answer: $(8, -5, -2)$.

4
Worked Example
Find the centroid of the triangle whose vertices are $A(1, 2, 3)$, $B(3, -1, 5)$ and $C(2, 5, 1)$.
Solution
  1. $x = \dfrac{1 + 3 + 2}{3} = \dfrac{6}{3} = 2$.
  2. $y = \dfrac{2 + (-1) + 5}{3} = \dfrac{6}{3} = 2$.
  3. $z = \dfrac{3 + 5 + 1}{3} = \dfrac{9}{3} = 3$.

Answer: Centroid $G = (2, 2, 3)$.

5
Worked Example
Find the ratio in which the $YZ$-plane divides the segment joining $A(-2, 4, 7)$ and $B(3, -5, 8)$.
Solution
  1. Let the plane divide $AB$ in the ratio $k : 1$. On the $YZ$-plane the $x$-coordinate is $0$.
  2. $x = \dfrac{k(3) + 1(-2)}{k + 1} = 0 \Rightarrow 3k - 2 = 0$.
  3. $k = \dfrac{2}{3}$, so the ratio is $\dfrac{2}{3} : 1 = 2 : 3$.

Answer: The $YZ$-plane divides $AB$ in the ratio $2 : 3$ (internally).

6
Worked Example
The midpoint of the segment joining $A(2, -1, 4)$ and $B(x, y, z)$ is $M(3, 1, 2)$. Find $B$.
Solution
  1. $\dfrac{2 + x}{2} = 3 \Rightarrow 2 + x = 6 \Rightarrow x = 4$.
  2. $\dfrac{-1 + y}{2} = 1 \Rightarrow -1 + y = 2 \Rightarrow y = 3$.
  3. $\dfrac{4 + z}{2} = 2 \Rightarrow 4 + z = 4 \Rightarrow z = 0$.

Answer: $B = (4, 3, 0)$.

7
Worked Example
Find the coordinates of the points that trisect the segment joining $P(4, 2, -6)$ and $Q(10, -16, 6)$.
Solution
  1. The trisection points divide $PQ$ in the ratios $1 : 2$ and $2 : 1$.
  2. For $1 : 2$: $x = \dfrac{1(10) + 2(4)}{3} = 6$, $y = \dfrac{1(-16) + 2(2)}{3} = -4$, $z = \dfrac{1(6) + 2(-6)}{3} = -2$.
  3. For $2 : 1$: $x = \dfrac{2(10) + 1(4)}{3} = 8$, $y = \dfrac{2(-16) + 1(2)}{3} = -10$, $z = \dfrac{2(6) + 1(-6)}{3} = 2$.

Answer: The trisection points are $(6, -4, -2)$ and $(8, -10, 2)$.

8
Worked Example
Find the ratio in which the point $R(5, 4, -6)$ divides the segment joining $A(3, 2, -4)$ and $B(9, 8, -10)$.
Solution
  1. Let the ratio be $k : 1$. Using the $x$-coordinate: $5 = \dfrac{9k + 3}{k + 1}$.
  2. $5(k + 1) = 9k + 3 \Rightarrow 5k + 5 = 9k + 3 \Rightarrow 4k = 2 \Rightarrow k = \dfrac{1}{2}$.
  3. Check with $z$: $\dfrac{-10(1/2) - 4}{(1/2) + 1} = \dfrac{-9}{3/2} = -6$. ✓

Answer: $R$ divides $AB$ internally in the ratio $1 : 2$.

9
Worked Example
Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$, $B(1, 2, -4)$ and $C(-1, 1, 2)$. Find the fourth vertex $D$.
Solution
  1. In a parallelogram the diagonals $AC$ and $BD$ bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$.
  2. Midpoint of $AC = \left(\dfrac{3 + (-1)}{2}, \dfrac{-1 + 1}{2}, \dfrac{2 + 2}{2}\right) = (1, 0, 2)$.
  3. Let $D = (x, y, z)$. Midpoint of $BD = \left(\dfrac{1 + x}{2}, \dfrac{2 + y}{2}, \dfrac{-4 + z}{2}\right) = (1, 0, 2)$.
  4. So $1 + x = 2$, $2 + y = 0$, $-4 + z = 4$, giving $x = 1$, $y = -2$, $z = 8$.

Answer: $D = (1, -2, 8)$.

10
Worked Example
Find the ratio in which the $XY$-plane divides the segment joining $A(2, -3, 5)$ and $B(7, 1, -2)$, and the coordinates of the point of division.
Solution
  1. On the $XY$-plane $z = 0$. Let the ratio be $k : 1$.
  2. $z = \dfrac{k(-2) + 1(5)}{k + 1} = 0 \Rightarrow -2k + 5 = 0 \Rightarrow k = \dfrac{5}{2}$, i.e. the ratio is $5 : 2$.
  3. $x = \dfrac{5(7) + 2(2)}{7} = \dfrac{39}{7}$, $y = \dfrac{5(1) + 2(-3)}{7} = -\dfrac{1}{7}$, $z = 0$.

Answer: The $XY$-plane divides $AB$ in the ratio $5 : 2$, at the point $\left(\dfrac{39}{7}, -\dfrac{1}{7}, 0\right)$.

11
Worked Example
Two vertices of a triangle are $B(3, -5, 7)$ and $C(-1, 7, -6)$, and its centroid is $G(1, 1, 1)$. Find the third vertex $A$.
Solution
  1. The centroid is the average of the three vertices, so $x_A + x_B + x_C = 3(1) = 3$, and similarly for $y$ and $z$.
  2. $x_A = 3 - 3 - (-1) = 1$.
  3. $y_A = 3 - (-5) - 7 = 1$.
  4. $z_A = 3 - 7 - (-6) = 2$.

Answer: $A = (1, 1, 2)$.

12
Worked Example
Show that the points $A(0, 4, 1)$, $B(2, 3, -1)$, $C(4, 5, 0)$ and $D(2, 6, 2)$ form a square by using midpoints and side-lengths.
Solution
  1. Midpoint of diagonal $AC = (2, 4.5, 0.5)$; midpoint of diagonal $BD = (2, 4.5, 0.5)$ — equal, so $ABCD$ is a parallelogram.
  2. Sides: $AB = \sqrt{4 + 1 + 4} = 3$ and $BC = \sqrt{4 + 4 + 1} = 3$; adjacent sides are equal, so it is a rhombus.
  3. Diagonals: $AC = \sqrt{16 + 1 + 1} = \sqrt{18}$ and $BD = \sqrt{0 + 9 + 9} = \sqrt{18}$; equal diagonals make the rhombus a square.

Answer: Diagonals bisect each other, all sides equal $3$, and equal diagonals $\sqrt{18}$ — so $ABCD$ is a square.

Key Points

  • Internal division ($m:n$): $\left(\dfrac{mx_2+nx_1}{m+n},\ \dfrac{my_2+ny_1}{m+n},\ \dfrac{mz_2+nz_1}{m+n}\right)$.
  • External division: same formula with $n$ replaced by $-n$ (denominator becomes $m-n$).
  • Midpoint is the case $m=n=1$: the average of the two endpoints' coordinates.
  • Centroid of a triangle is the average of the three vertices; the medians meet there in the ratio $2:1$.
  • Find an unknown ratio by setting it as $k:1$; for a coordinate plane the matching coordinate is $0$. Positive $k$ is internal, negative $k$ is external.
  • A quadrilateral is a parallelogram when its diagonals share a midpoint; equal diagonals make it a rectangle, and equal sides too make it a square.
  • Keep the point-order $(P,Q)$ and ratio $(m,n)$ consistent — swapping them changes the answer.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The midpoint of $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is:
Explanation: Midpoint $=$ average of coordinates.
Q2.The section formula (internal, ratio $m:n$) for the $x$-coordinate is:
Explanation: Standard internal-division formula.
Q3.The midpoint of $(2,4,6)$ and $(4,8,10)$ is:
Explanation: Average each coordinate.
Q4.The centroid of a triangle is the ___ of its vertices.
Explanation: Centroid $=$ average of the three vertices.
Practice more MCQs → 📝 Assignment · 35 marks