Worked examples — Section formula — internal and external division

Before anything, recall the two tools we lean on (both proved in the parent):

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The scenario matrix

Every problem the section formula throws is one of these cells. Each cell is claimed by at least one worked example below.

#	Cell (what makes it distinct)	Covered by
C1	Internal, both points in Quadrant I (all-positive), $m>n$	Ex 1
C2	Internal, points across quadrants (negative coords)	Ex 2
C3	Midpoint — the $m=n=1$ special case	Ex 3
C4	External, $m>n$ → $P$ lands beyond $B$	Ex 4
C5	External, $m<n$ → $P$ lands beyond $A$ (behind the start)	Ex 5
C6	Degenerate external: $m=n$ → no finite point (limit to infinity)	Ex 6
C7	Reverse problem: given $P$, find the ratio (and detect internal vs external by its sign)	Ex 7
C8	Negative-ratio input $-m:n$ silently means external	Ex 8
C9	Real-world word problem (a physical "mix the addresses")	Ex 9
C10	Exam twist: point on an axis → solve for an unknown ratio/coordinate	Ex 10

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The examples

Ex 1 — Cell C1: internal, Quadrant I, $m>n$

1. Read weights: $m=2$ (goes with $B$), $n=1$ (goes with $A$). Why this step? The whole formula hinges on which number multiplies which point; getting it right first prevents the classic swap mistake.
2. $x=\frac{m{x}_2+n{x}_1}{m+n}=\frac{2\cdot 8+1\cdot 2}{2+1}=\frac{18}{3}=6.$ Why this step? This is the internal recipe applied to the $x$-shadows — the horizontal gaps split in the same $2:1$ ratio as the segment.
3. $y=\frac{2\cdot 9+1\cdot 3}{3}=\frac{21}{3}=7.$ Why this step? Same recipe on the $y$-shadows; the line's constant slope forces the same ratio vertically. So $P=(6,7)$.

Verify: Two-thirds of the way: $x=2+\frac{2}{3}(8-2)=2+4=6$ ✓, $y=3+\frac{2}{3}(9-3)=3+4=7$ ✓. $P$ is closer to $B$, matching the forecast.

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Ex 2 — Cell C2: internal across quadrants (negative coordinates)

1. $m=3$ (with $B$), $n=2$ (with $A$), sum $m+n=5$. Why this step? Negative coordinates change nothing about the recipe — the signs are just carried along in the arithmetic.
2. $x=\frac{3\cdot 6+2\cdot (-4)}{5}=\frac{18-8}{5}=\frac{10}{5}=2.$ Why this step? We substitute the actual signed values $x_2=6$, $x_1=-4$; the minus sign is part of $A$'s address.
3. $y=\frac{3\cdot (-3)+2\cdot 5}{5}=\frac{-9+10}{5}=\frac{1}{5}=\mathrm{0.2.}$ Why this step? Same recipe with signed $y$'s. $P=(2,0.2)$.

Verify: Fraction check: $x=-4+\frac{3}{5}(6-(-4))=-4+6=2$ ✓, $y=5+\frac{3}{5}(-3-5)=5-4.8=0.2$ ✓. The near-zero $y$ matches the forecast (we just crossed the $x$-axis).

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Ex 3 — Cell C3: the midpoint special case ($m=n=1$)

1. Set $m=n=1$ in the internal formula. Why this step? Equal weights mean $P$ splits $AB$ into two equal pieces — the definition of a midpoint.
2. $x=\frac{1\cdot 6+1\cdot (-4)}{2}=\frac{2}{2}=1,y=\frac{1\cdot (-3)+1\cdot 5}{2}=\frac{2}{2}=1.$ Why this step? With $m=n$, the recipe collapses to $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ — plain averages.

Verify: $M=(1,1)$ should be equidistant from $A$ and $B$. Using the Distance formula: $MA=\sqrt{(1+4{)}^2+(1-5{)}^2}=\sqrt{25+16}=\sqrt{41}$, $MB=\sqrt{(6-1{)}^2+(-3-1{)}^2}=\sqrt{25+16}=\sqrt{41}$ ✓. Equal distances confirm it is the midpoint.

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Ex 4 — Cell C4: external, $m>n$ → $P$ beyond $B$

1. Use the external recipe: minus in numerator, denominator $m-n$. Here $m=3,n=1$, so $m-n=2$. Why this step? External = internal with $n\to -n$; the flipped sign encodes that $A\to P$ and $P\to B$ now point opposite ways.
2. $x=\frac{3\cdot 4-1\cdot 1}{3-1}=\frac{12-1}{2}=\frac{11}{2}=\mathrm{5.5.}$ Why this step? Direct substitution of the external formula.
3. $y=\frac{3\cdot 6-1\cdot 2}{2}=\frac{18-2}{2}=\frac{16}{2}=8.$ So $P=(5.5,8)$.

Verify: Is $P$ past $B$? Check $\frac{AP}{PB}=3$: $AP=\sqrt{(5.5-1{)}^2+(8-2{)}^2}=\sqrt{20.25+36}=\sqrt{56.25}=7.5$; $PB=\sqrt{(5.5-4{)}^2+(8-6{)}^2}=\sqrt{2.25+4}=\sqrt{6.25}=2.5$. Ratio $7.5:2.5=3:1$ ✓, and $x=5.5>4$ confirms $P$ is beyond $B$.

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Ex 5 — Cell C5: external, $m<n$ → $P$ behind $A$

1. $m=1,n=3$, so $m-n=1-3=-2$ (a negative denominator — legal, just carry it). Why this step? A negative denominator is the algebra's way of pushing $P$ to the opposite side.
2. $x=\frac{1\cdot 4-3\cdot 1}{-2}=\frac{4-3}{-2}=\frac{1}{-2}=-\mathrm{0.5.}$ Why this step? Substitute into the external formula, keeping the sign of $m-n$.
3. $y=\frac{1\cdot 6-3\cdot 2}{-2}=\frac{6-6}{-2}=\frac{0}{-2}=0.$ So $P=(-0.5,0)$.

Verify: $P$ should sit behind $A$. Direction check: going $A(1,2)\to P(-0.5,0)$ moves left-and-down, i.e. away from $B$ — exactly "beyond $A$". Ratio: $AP=\sqrt{1.5^2+2^2}=\sqrt{6.25}=2.5$, $PB=\sqrt{4.5^2+6^2}=\sqrt{56.25}=7.5$, so $AP:PB=2.5:7.5=1:3$ ✓.

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Ex 6 — Cell C6: degenerate external $m=n$ → no finite point

1. Denominator $m-n=1-1=0$. Why this step? Division by zero is the signal that no finite point satisfies the condition.
2. Numerator $x$: $1\cdot 4-1\cdot 1=3\ne 0$, so we get $x=\frac{3}{0}$ — undefined. Why this step? A nonzero-over-zero means the point recedes without bound as the ratio approaches $1:1$.
3. Limit view: take ratio $1:(1-\epsilon )$ and let $\epsilon \to 0^{+}$. Then $x=\frac{4-(1-\epsilon )}{\epsilon }=\frac{3+\epsilon }{\epsilon }\to +\infty$. Why this step? Watching the near-degenerate case shows the point genuinely shoots off to infinity — the parent's "parallel-at-infinity".

Verify: As $\epsilon \to 0^{+}$, $x=\frac{3+\epsilon }{\epsilon }$ grows without bound (e.g. $\epsilon =0.01$ gives $x=301$). No finite $P$ exists ✓.

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Ex 7 — Cell C7: reverse — find the ratio, detect internal vs external

1. Let the ratio be $k:1$ and use the $x$-coordinate: $6=\frac{8k+2}{k+1}$. Why this step? Writing the unknown ratio as $k:1$ turns a two-number problem into one equation in one unknown $k$.
2. Solve: $6(k+1)=8k+2\Rightarrow 6k+6=8k+2\Rightarrow 2k=4\Rightarrow k=2$. Why this step? Clearing the fraction and collecting $k$ isolates the ratio. $k=2>0$ ⇒ internal, ratio $2:1$.
3. Cross-check with $y$: $7=\frac{9k+3}{k+1}\Rightarrow 7k+7=9k+3\Rightarrow 2k=4\Rightarrow k=2$. Why this step? The $y$-equation must give the same $k$ — a built-in consistency test (and confirms $P$ is on line $AB$, see Collinearity of three points).

Verify: Both coordinates gave $k=2$, ratio $2:1$ internal ✓. (If $k$ had come out negative, that would signal external — see Ex 8.)

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Ex 8 — Cell C8: negative-ratio input means external

1. Rewrite $-2:3$ as external $2:3$, i.e. $m=2,n=3$, external recipe. Why this step? A minus sign in the ratio is the algebraic fingerprint of external division; converting it prevents plugging a negative into the internal sum.
2. $x=\frac{m{x}_2-n{x}_1}{m-n}=\frac{2\cdot 5-3\cdot 2}{2-3}=\frac{10-6}{-1}=\frac{4}{-1}=-4.$ Why this step? External formula with the negative denominator $m-n=-1$.
3. $y=\frac{2\cdot 7-3\cdot 1}{-1}=\frac{14-3}{-1}=\frac{11}{-1}=-11.$ So $P=(-4,-11)$.

Verify: Same answer directly from the internal formula with $n=-3$: $x=\frac{2\cdot 5+(-3)\cdot 2}{2+(-3)}=\frac{10-6}{-1}=-4$ ✓, $y=\frac{2\cdot 7+(-3)\cdot 1}{-1}=\frac{11}{-1}=-11$ ✓. Both routes agree, and $P$ is far behind $A$ as forecast.

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Ex 9 — Cell C9: real-world word problem

1. Translate the words: $\frac{AP}{PB}=\frac{2/3}{1/3}=\frac{2}{1}$, so $m=2,n=1$, internal. Why this step? Word problems must first be turned into a ratio $m:n$ before any formula can act.
2. $x=\frac{2\cdot 12+1\cdot 0}{3}=\frac{24}{3}=8,y=\frac{2\cdot 9+1\cdot 0}{3}=\frac{18}{3}=6.$ Why this step? With pump at the origin, $A$'s terms vanish — the formula "mixes the addresses" and the origin contributes nothing.

Verify: Sensor at $(8,6)$. Distance from pump: $\sqrt{8^2+6^2}=\sqrt{100}=10$ km; total pipe $\sqrt{12^2+9^2}=\sqrt{225}=15$ km. Fraction $10/15=2/3$ ✓ — exactly two-thirds along, units in km consistent.

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Ex 10 — Cell C10: exam twist — point on an axis fixes the ratio

1. Let the ratio be $k:1$ and demand $y=0$: $0=\frac{k\cdot 6+1\cdot (-3)}{k+1}$. Why this step? "On the $x$-axis" is the constraint $y=0$; the unknown ratio is the thing that must adjust to meet it.
2. A fraction is zero only when its numerator is zero: $6k-3=0\Rightarrow k=\frac{1}{2}$. Why this step? The denominator $k+1$ can't create the zero (it would blow up), so we set the top to $0$. Ratio $=\frac{1}{2}:1=1:2$, positive ⇒ internal, matching the forecast.
3. Find $x$ at $k=\frac{1}{2}$ (i.e. $m=1,n=2$): $x=\frac{1\cdot 5+2\cdot 2}{1+2}=\frac{5+4}{3}=3.$ Why this step? Once the ratio is known, the standard formula gives the actual crossing point $(3,0)$.

Verify: Point $(3,0)$: is $y=0$? Plug $1:2$: $y=\frac{1\cdot 6+2\cdot (-3)}{3}=\frac{6-6}{3}=0$ ✓. Ratio $1:2$ is positive → internal, and $x=3$ is closer to $A$'s $x=2$ than to $B$'s $x=5$ ✓.

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Recall Did every cell get hit?

C1→Ex1, C2→Ex2, C3→Ex3, C4→Ex4, C5→Ex5, C6→Ex6, C7→Ex7, C8→Ex8, C9→Ex9, C10→Ex10. Ten cells, ten checks. ✓

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Connections

• Parent — full derivation
• Midpoint formula — Ex 3 is its living example.
• Distance formula — used to verify ratios in Ex 3, 4, 5, 9.
• Centroid of a triangle — the $2:1$ median ratio is Ex 1's cousin.
• Collinearity of three points — Ex 7's twin-check ($x$ and $y$ giving the same $k$).
• Similar triangles — why the shadow ratios match, behind every example.
• Slope of a line — constant slope guarantees the axis-projections split equally.