Exercises — Section formula — internal and external division

Denominator check: $m+n=1+2=3\ne 0$ — the internal formula is safe to use. Internal formula, $m=1,n=2$. Remember $m$ multiplies $B$'s coordinates. $x=\frac{1\cdot 7+2\cdot 1}{1+2}=\frac{9}{3}=3,y=\frac{1\cdot 8+2\cdot 2}{3}=\frac{12}{3}=4.$ Why this mix? The split is $1:2$, so $P$ is a mix of one part $B$ and two parts $A$ out of three parts — that is exactly what the fractions above compute. More weight on $A$ ⇒ answer sits nearer $A$. $P=(3,4)$. Sanity: ratio $1:2$ means $P$ is only $\frac{1}{3}$ of the way from $A$ to $B$, so it sits close to $A(1,2)$. And $(3,4)$ is indeed near $A$. ✓

Midpoint = internal division with $m=n=1$: equal parts of each endpoint, so just average each coordinate: $M=\left(\frac{-4+10}{2},\frac{6+(-2)}{2}\right)=\left(\frac{6}{2},\frac{4}{2}\right)=(3,2).$ $M=(3,2)$. See Midpoint formula for why $m=n=1$ collapses the recipe to a plain average. The $1:0$ case: all weight on $B$, so $P=\frac{1\cdot B+0\cdot A}{1+0}=B=(10,-2)$. A zero in the ratio recovers an endpoint — a useful reality check that the formula behaves.

First check the denominator is alive: external needs $m\ne n$. Here $m=3\ne 1=n$, so $m-n=2\ne 0$ — safe to proceed. External formula: minus in numerator, $m-n$ in denominator. $x=\frac{3\cdot 4-1\cdot 1}{3-1}=\frac{11}{2}=5.5,y=\frac{3\cdot 6-1\cdot 2}{3-1}=\frac{16}{2}=8.$ $P=(5.5,8)$. Sanity: $m>n$ externally means $P$ lands beyond $B$. $B=(4,6)$, and $(5.5,8)$ is further along the same line — past $B$. ✓

Denominator check first: as an internal ratio $m+n=-2+3=1\ne 0$, so no blow-up — proceed. Why a negative ratio = external division (the derivation). Feed the negative ratio straight into the internal formula with $m=-2,n=3$: $x=\frac{(-2)\cdot 5+3\cdot 0}{-2+3}=\frac{-10}{1}=-10,y=\frac{(-2)\cdot 10+3\cdot 0}{1}=\frac{-20}{1}=-20.$ Now do the same point through the external formula reading the ratio as positive $2:3$ ($m=2,n=3$): $x=\frac{2\cdot 5-3\cdot 0}{2-3}=\frac{10}{-1}=-10,y=\frac{2\cdot 10-3\cdot 0}{2-3}=\frac{20}{-1}=-20.$ Identical. That is the proof: putting a minus sign on $n$ (or on $m$) turns every $+n$ into $-n$ and every $m+n$ into $m-n$ — which is literally the external formula. So a negative weight is not some new rule; it is the internal formula pointed backwards, and "backwards along the line" is exactly what external (outside the segment) means. $P=(-10,-20)$. Sanity: here $|m|<|n|$ so $P$ is thrown beyond $A$ (the smaller-weight side). $A=(0,0)$ and $P=(-10,-20)$ sits on the opposite side of $A$ from $B$. ✓

Let the ratio be $k:1$ — one unknown $k$ standing in for $m:n$ scaled so $n=1$. A single symbol is easier to solve. With $m=k,n=1$, the internal formula puts $k$ on $B$ and $1$ on $A$. Solve using the $x$-coordinate. Why cross-multiply? The equation is a fraction equal to a number; multiplying both sides by the denominator $(k+1)$ clears the fraction without changing the truth of the equation: $-1=\frac{k\cdot 6+1\cdot (-3)}{k+1}\Rightarrow -1(k+1)=6k-3\Rightarrow -k-1=6k-3.$ Collect $k$ on one side (both sides stay equal because we only add the same term to each): $-1+3=6k+k\Rightarrow 2=7k\Rightarrow k_x=\frac{2}{7}.$ Now the lie-detector. A valid dividing point must give the same $k$ from the $y$-equation, because one $k$ fixes one point on the line — it cannot have two different values. Solve $y$ independently: $3=\frac{k(-8)+1\cdot 10}{k+1}\Rightarrow 3k+3=-8k+10\Rightarrow 11k=7\Rightarrow k_y=\frac{7}{11}.$ The verdict. $k_x=\frac{2}{7}$ but $k_y=\frac{7}{11}$, and $\frac{2}{7}\ne \frac{7}{11}$. The two coordinates demand different split ratios — impossible for a single point on the line $AB$. That contradiction proves $P(-1,3)$ does not lie on line $AB$, so no dividing ratio exists. See Collinearity of three points: a valid ratio exists iff the $x$- and $y$-equations agree on $k$. Answer: no such ratio; $P$ is off the line $AB$.

Use the known coordinate $y=-2$ to solve for the ratio $k:1$ first. Cross-multiply to clear the fraction: $-2=\frac{k\cdot 4+1\cdot (-4)}{k+1}\Rightarrow -2k-2=4k-4\Rightarrow 2=6k\Rightarrow k=\frac{1}{3}.$ Ratio $=1:3$. Now find $x$ with that ratio ($m=1,n=3$): $x=\frac{1\cdot 6+3\cdot 2}{1+3}=\frac{12}{4}=3.$ $x=3$, ratio $1:3$. Sanity: $1:3$ puts $P$ a quarter of the way from $A(2,-4)$ to $B(6,4)$; a quarter of the $y$-rise $8$ is $2$, landing at $-4+2=-2$ ✓.

Two cut points (see Figure 2): $P$ nearer $A$ divides $1:2$; $Q$ nearer $B$ divides $2:1$. Why those ratios? $P$ ends the first third, so $AP:PB=1:2$; $Q$ ends the second third, so $AQ:QB=2:1$. Both have $m+n=3\ne 0$ — internal formula safe. $P$ (ratio $1:2$): $P=\left(\frac{1\cdot 8+2\cdot 2}{3},\frac{1\cdot 6+2\cdot (-3)}{3}\right)=\left(\frac{12}{3},\frac{0}{3}\right)=(4,0).$ $Q$ (ratio $2:1$): $Q=\left(\frac{2\cdot 8+1\cdot 2}{3},\frac{2\cdot 6+1\cdot (-3)}{3}\right)=\left(\frac{18}{3},\frac{9}{3}\right)=(6,3).$ Trisection points: $P=(4,0)$, $Q=(6,3)$. Sanity: the midpoint of the whole segment is $\left(\frac{2+8}{2},\frac{-3+6}{2}\right)=(5,1.5)$, which should be the midpoint of $P$ and $Q$: $\left(\frac{4+6}{2},\frac{0+3}{2}\right)=(5,1.5)$ ✓.

Average method (the Centroid of a triangle shortcut): $G=\left(\frac{1+5+3}{3},\frac{2+(-2)+6}{3}\right)=\left(\frac{9}{3},\frac{6}{3}\right)=(3,2).$ Why the centroid splits each median $2:1$ (proof sketch). A median runs from a vertex, say $A$, to the midpoint $M$ of the opposite side $BC$. Take the point $G$ that averages all three vertices, $G=\frac{A+B+C}{3}$. Now $M=\frac{B+C}{2}$, so $B+C=2M$, and substituting gives $G=\frac{A+2M}{3}$. Read that last expression as the internal section formula on segment $AM$ with weights $m=2$ (on $M$) and $n=1$ (on $A$), since it is exactly $\frac{2M+1A}{2+1}$. So $G$ divides $AM$ in ratio $AG:GM=2:1$, measured from the vertex. Because the averaging formula is symmetric in $A,B,C$, the identical argument works for the medians from $B$ and from $C$ — every median is split $2:1$ at the same point $G$. (This is why the Centroid of a triangle averaging shortcut and the median-split rule agree.) Median-split verification. Midpoint of $BC$ is $M=\left(\frac{5+3}{2},\frac{-2+6}{2}\right)=(4,2)$. The centroid divides median $AM$ in ratio $2:1$ from vertex $A$, so $m=2$ (toward $M$), $n=1$ (toward $A$), $m+n=3\ne 0$: $G=\left(\frac{2\cdot 4+1\cdot 1}{3},\frac{2\cdot 2+1\cdot 2}{3}\right)=\left(\frac{9}{3},\frac{6}{3}\right)=(3,2).$ Both give $G=(3,2)$ ✓ — the section formula proves the averaging shortcut.

Idea: on the $x$-axis, $y=0$. So the crossing point $P$ has $y$-coordinate $0$. Let ratio $=k:1$ and solve the $y$-equation (cross-multiplying to clear the fraction): $0=\frac{k\cdot 6+1\cdot (-3)}{k+1}\Rightarrow 6k-3=0\Rightarrow k=\frac{1}{2}.$ Ratio $=1:2$. Now the $x$-coordinate with $m=1,n=2$: $x=\frac{1\cdot 5+2\cdot 2}{1+2}=\frac{9}{3}=3.$ The $x$-axis cuts $AB$ in ratio $1:2$ at the point $(3,0)$. Sanity: $A$ is below the axis ($y=-3$), $B$ above ($y=6$); the segment must cross, and $k>0$ confirms internal division. ✓

Assume $B$ divides $AC$ in ratio $k:1$. Solve with $x$ (cross-multiply to clear the fraction): $3=\frac{k\cdot 7+1\cdot 1}{k+1}\Rightarrow 3k+3=7k+1\Rightarrow 2=4k\Rightarrow k=\frac{1}{2}.$ Now the lie-detector: plug $k=\frac{1}{2}$ into the $y$-formula and see if it returns $B$'s $y=4$: $y=\frac{\frac{1}{2}\cdot 10+1\cdot 1}{\frac{1}{2}+1}=\frac{5+1}{3/2}=\frac{6}{3/2}=4✓.$ Both coordinates agree at $k=\frac{1}{2}$, so $B$ does lie on line $AC$, dividing it in ratio $1:2$. Since $B$ is a genuine section point of segment $AC$, the three points $A$, $B$, $C$ lie on one straight line — they are collinear. Slope confirmation (Slope of a line): slope $AB=\frac{4-1}{3-1}=\frac{3}{2}$; slope $BC=\frac{10-4}{7-3}=\frac{6}{4}=\frac{3}{2}$. Equal slopes ⇒ collinear ✓ (see Collinearity of three points). Conclusion: $A$, $B$, $C$ are collinear, with $B$ dividing $AC$ internally $1:2$.

Why we can solve for the endpoint. The section formula is a single equation that ties together five quantities: the ratio, both of $A$'s coordinates, both of $B$'s, and $P$. Nothing in it privileges $P$ as "the output." If instead we know the ratio, $A$, and $P$, then $B$ is simply the one unknown left — so we set the formula equal to $P$ and solve for $B$, exactly as we solved for $P$ in L1. The equation is symmetric in what it lets you treat as known. With $m=3,n=2$ ($m+n=5\ne 0$), set the internal formula equal to $P$'s coordinates and clear the fraction: $4=\frac{3x_2+2\cdot 2}{3+2}=\frac{3x_2+4}{5}\Rightarrow 20=3x_2+4\Rightarrow x_2=\frac{16}{3}.$ $5=\frac{3y_2+2\cdot 3}{5}=\frac{3y_2+6}{5}\Rightarrow 25=3y_2+6\Rightarrow y_2=\frac{19}{3}.$ $B=\left(\frac{16}{3},\frac{19}{3}\right)$. Sanity: $P$ should sit $3/5$ of the way from $A$ to $B$. Check $x$: $2+\frac{3}{5}\left(\frac{16}{3}-2\right)=2+\frac{3}{5}\cdot \frac{10}{3}=2+2=4$ ✓.

L1 — plug into the right box (and note $m:0$ or $0:n$ just gives back an endpoint). L2 — respect the minus signs of external / negative ratios; a negative weight is the external formula in disguise, and $m=n$ externally gives no point. L3 — run the machine backwards, and never trust one coordinate alone. L4 — snap the formula onto centroids and trisections. L5 — the unknown can be anywhere; the equation is symmetric, so set up algebra, don't pattern-match. One recipe, five heights of thinking — and always guard the denominator: internal dies at $m+n=0$, external at $m-n=0$.