Section Formula

Theorem 5 (Section formula, internal division). Let $A(\bar a)$ and $B(\bar b)$ be two points and

$R(\bar r)$ a point on segment $AB$ dividing it internally in the ratio $m:n$ (so $A$-$R$-$B$ in order and

$AR:RB=m:n$). Then

Proof idea: since $AR$ and $RB$ point the same way along the line, $n(\overrightarrow{AR})=m(\overrightarrow{RB})$

in magnitude and direction; writing $\overrightarrow{AR}=\bar r-\bar a$ and

$\overrightarrow{RB}=\bar b-\bar r$ and substituting gives $n(\bar r-\bar a)=m(\bar b-\bar r)$, which rearranges

to $(m+n)\bar r=m\bar b+n\bar a$, i.e. the formula above. Componentwise, if $A(a_1,a_2,a_3),B(b_1,b_2,b_3), R(r_1,r_2,r_3)$, then $r_i=(mb_i+na_i)/(m+n)$ for $i=1,2,3$.

Theorem 6 (Section formula, external division). If $R(\bar r)$ divides $AB$ externally in the ratio

$m:n$ (so $R$ lies on line $AB$ but outside the segment), then

The proof mirrors the internal case, using $n(\overrightarrow{AR})=m(\overrightarrow{BR})$ for the external

configuration.

Special cases. Writing the ratio as $k:1$ is often convenient: $\bar r=\dfrac{k\bar b+\bar a}{k+1}$

(internal) or $\bar r=\dfrac{k\bar b-\bar a}{k-1}$ (external). Taking $m=n$ (i.e. $k=1$) in the internal

formula gives the Midpoint Formula: $\bar r=\dfrac{\bar a+\bar b}{2}$.

Centroid formulas. In $\triangle ABC$, the centroid $G(\bar g)$ divides each median internally in the

ratio $2:1$ and is given by $\bar g=\dfrac{\bar a+\bar b+\bar c}{3}$. In a tetrahedron $ABCD$, the centroid

$G(\bar g)$ divides the segment from each vertex to the centroid of the opposite face in the ratio $3:1$, and

$\bar g=\dfrac{\bar a+\bar b+\bar c+\bar d}{4}$.

Worked examples.

• Dividing $A(2,-6,8),B(-1,3,-4)$ internally in the ratio $1:3$: $\bar r=\dfrac{1\cdot\bar b+3\cdot\bar a}{4}$ gives the point $\left(\tfrac54,\tfrac{15}{4},-5\right)$; dividing externally in the same ratio $1:3$ gives $\bar s=\dfrac{1\cdot\bar b-3\cdot\bar a}{-2}$, the point $\left(\tfrac72,\tfrac{21}{2},-14\right)$.
• If $A(3,2,p),B(q,8,-10),C(-2,-3,1)$ are collinear with $C$ dividing $AB$ in the ratio $t:1$, then $\bar c=\dfrac{t\bar b+\bar a}{t+1}$; matching each coordinate gives three equations, and solving the $y$-equation first for $t$ (since it involves no unknowns besides $t$) lets the other two equations be solved for $p,q$ in turn. A negative value of $t$ found this way means $C$ actually divides $AB$ externally rather than internally.
• If $\bar a,\bar b,\bar c$ are position vectors of a triangle's vertices and $5\bar a-3\bar b-2\bar c=\bar 0$, isolating $\bar c=\dfrac{5\bar a-3\bar b}{2}=\dfrac{5\bar a-3\bar b}{5-3}$ matches the external-division form with ratio $5:3$, showing $C$ divides $BA$ externally in the ratio $5:3$.
• Concurrency of medians (by vectors). With $D,E,F$ the midpoints of $BC,CA,AB$ of $\triangle ABC$, the midpoint formula gives $\bar d=(\bar b+\bar c)/2$ etc.; combining $2\bar d+\bar a,\ 2\bar e+\bar b,\ 2\bar f+\bar c$ (each equal to $\bar a+\bar b+\bar c$) and dividing by 3 shows the point $\bar g=\tfrac13(\bar a+\bar b+\bar c)$ lies on each of $AD,BE,CF$, dividing each in the ratio $2:1$ -- so the three medians are concurrent at the centroid.
• Concurrency of angle bisectors (incenter). Using the section formula on each angle bisector (dividing the opposite side in the ratio of the two adjacent sides) and combining the three resulting expressions shows they all meet at $\bar h=\dfrac{|BC|\,\bar a+|AC|\,\bar b+|AB|\,\bar c}{|BC|+|AC|+|AB|}$, the incenter formula, e.g. giving $H\equiv(0,2,3)$ for the triangle $A(0,3,0),B(0,0,4),C(0,3,4)$.
• Standard extra centre formulas noted for reference: centroid $\bar g=(\bar a+\bar b+\bar c)/3$; incenter $\bar h$ as above; orthocenter $\bar p=\dfrac{\tan A\cdot\bar a+\tan B\cdot\bar b+\tan C\cdot\bar c} {\tan A+\tan B+\tan C}$.
• Given centroid coordinates for a triangle or tetrahedron with some unknown vertex coordinates, the centroid formula is solved coordinate-by-coordinate for the unknowns (e.g. finding $a,b,c$ from a stated centroid of triangle $P(1,3,2),Q(3,b,-4),R(5,1,c)$ equal to $G(a,2,-1)$). …