Section formula — internal and external division: Visual walkthrough

Step 1 — Two points, and a spot on the rope between them

WHAT. Put down two points. Call the first $A$ and give it an address $(x_{1}, y_{1})$ . Call the second $B$ with address $(x_{2}, y_{2})$ .

An address here just means: how far right (the first number) and how far up (the second number). So $x_{1}$ is "how far right $A$ is", $y_{1}$ is "how far up $A$ is". Same idea for $B$ .

Now imagine a straight rope from $A$ to $B$ . Somewhere on that rope, between the two ends, we mark a point $P$ with unknown address $(x, y)$ . Our whole job is to find $x$ and $y$ .

WHY. Every point on a straight line can be pinned down by how it splits the line. We need to name that split before we can compute anything.

PICTURE. Look at the blue rope from $A$ to $B$ ; the orange dot $P$ sits partway along it.

Figure — Section formula — internal and external division

Step 2 — Cast shadows onto the $x$ -axis

WHAT. Drop a straight vertical line down from each of $A$ , $P$ , $B$ until it hits the $x$ -axis (the horizontal number line at the bottom). Where each vertical lands is that point's shadow.

Shadow of $A$ lands at $x_{1}$ .
Shadow of $P$ lands at $x$ .
Shadow of $B$ lands at $x_{2}$ .

WHY. Diagonal lengths ( $A P$ , $P B$ ) are awkward to measure. Their horizontal shadows are just plain differences of $x$ -numbers — much easier. The trick of the whole derivation is that the shadow keeps the same split as the rope. We prove that in the next step.

PICTURE. Three grey verticals fall from the rope to the axis. The horizontal gap from $x_{1}$ to $x$ is the shadow of $A P$ ; the gap from $x$ to $x_{2}$ is the shadow of $P B$ .

Step 3 — Why the shadow keeps the same ratio (similar triangles)

WHAT. Build two right-angled triangles.

Triangle 1: run right from $A$ , then up to $P$ . Its horizontal side is $x - x_{1}$ ; its slanted side is $A P$ .
Triangle 2: run right from $P$ , then up to $B$ . Its horizontal side is $x_{2} - x$ ; its slanted side is $P B$ .

WHY. Both triangles sit under the same straight rope, so they lean at the exact same steepness (same slope). Two right triangles with the same slope are similar — same shape, different size. And similar triangles have all matching sides in the same ratio. Therefore: $\frac{horizontal side 1}{horizontal side 2} = \frac{slant side 1}{slant side 2} ⟹ \frac{x - x _{1}}{x _{2} - x} = \frac{A P}{P B} .$

PICTURE. The two shaded triangles are copies of one another — the small one scaled up to the big one. The matching angle at the base is marked equal.

Since $\frac{A P}{P B} = \frac{m}{n}$ from Step 1, we can now write the key line:

Step 4 — Solve for $x$ (pure algebra, no more geometry)

WHAT. Take the boxed equation and unknot it to get $x$ alone.

Cross-multiply (multiply both sides by both bottoms): $n (x - x_{1}) = m (x_{2} - x) .$

Left $n (x - x_{1})$ : the $B$ -side count $n$ times the near shadow.
Right $m (x_{2} - x)$ : the $A$ -side count $m$ times the far shadow.

Open the brackets: $n x - n x_{1} = m x_{2} - m x .$

Gather every $x$ -term on the left (add $m x$ to both sides, add $n x_{1}$ ... actually move constants right): $n x + m x = m x_{2} + n x_{1} .$

Left: both $x$ -terms collected — $x (m + n)$ .
Right: the two known numbers $m x_{2}$ and $n x_{1}$ .

Factor and divide by $(m + n)$ : $x = \frac{m x _{2} + n x _{1}}{m + n} .$

WHY. We wanted $x$ by itself. Cross-multiplying clears fractions; collecting terms isolates $x$ ; dividing frees it.

PICTURE. Read the answer as a weighted mix: slide a weight $m$ onto $B$ 's shadow $x_{2}$ and a weight $n$ onto $A$ 's shadow $x_{1}$ ; the balance point of those weights is $x$ .

Step 5 — The $y$ -coordinate comes free

WHAT. Repeat Steps 2–4, but drop the shadows onto the $y$ -axis (the vertical number line on the left) instead.

WHY. The rope has one steepness. The same similar triangles that split the horizontal shadow $m : n$ split the vertical shadow $m : n$ too — nothing new to prove. So every " $x$ " becomes a " $y$ ": $\frac{y - y _{1}}{y _{2} - y} = \frac{m}{n} ⟹ y = \frac{m y _{2} + n y _{1}}{m + n} .$

PICTURE. Same three points, now casting shadows leftward onto the $y$ -axis; the vertical gaps split in the identical $m : n$ proportion.

Step 6 — When $P$ walks past the end (external division)

WHAT. Now let $P$ leave the segment — keep walking beyond $B$ (or behind $A$ ). It still lies on the same straight line, still satisfies $\frac{A P}{P B} = \frac{m}{n}$ , but now it is outside.

WHY the sign flips. When $P$ is beyond $B$ , the trip $A \to P$ and the trip $P \to B$ point in opposite directions. One of the shadow gaps reverses sign. So $x_{2} - x$ becomes negative while $x - x_{1}$ stays positive. Mechanically, that is exactly what you get by replacing $n$ with $- n$ in every formula.

Substituting $n \to - n$ : $x = \frac{m x _{2} + ( - n ) x _{1}}{m + ( - n )} = \frac{m x _{2} - n x _{1}}{m - n} .$

PICTURE. The orange $P$ now sits to the right of $B$ . Watch the far arrow $P \to B$ flip backwards — that reversed arrow is the negative length.

Step 7 — The degenerate cases (never get caught out)

WHAT & WHY — walk every edge:

$m = n$ , internal → the recipe gives $\frac{x _{1} + x _{2}}{2}$ . This is the midpoint: equal parts on both sides means $P$ sits dead centre.
$m = n$ , external → denominator $m - n = 0$ . Dividing by zero = no finite answer. Geometrically $P$ has run off to infinity: a rope split equally and outside can only be reached "at the far end of forever" (the direction parallel to the line, never meeting it).
Negative ratio given, e.g. $- 2 : 3$ → a minus sign hidden in the ratio is the external signal. Treat $- 2 : 3$ as external $2 : 3$ .
$m = 0$ → $\frac{A P}{P B} = 0$ means $A P = 0$ , so $P$ lands exactly on $A$ . Check: $\frac{0 \cdot x _{2} + n x _{1}}{0 + n} = x_{1}$ . ✓
$n = 0$ → $\frac{A P}{P B} = \infty$ means $P B = 0$ , so $P$ lands on $B$ : $\frac{m x _{2} + 0}{m} = x_{2}$ . ✓

PICTURE. One number line showing all landing spots: $m = 0$ at $A$ , $m = n$ at the middle, growing $m$ pushing toward $B$ , external ratios spilling past $B$ toward the infinity arrow.

The one-picture summary

Everything above, compressed: two points, one rope, similar-triangle shadows on both axes, and the weighted-mix formula that falls out. Internal splits stay between; external splits (dashed) shoot past the ends.

Recall Feynman retelling (say it like you're 12)

I have two houses, $A$ and $B$ , each with an address (how far right, how far up). I want to stand on the straight path between them at a spot that is " $m$ steps for every $n$ steps". I don't walk. Instead I drop a shadow of each house onto the floor (the $x$ -axis) and onto the wall (the $y$ -axis). Because the path is dead straight, the shadows split in the exact same ratio as the path — that's the similar-triangles magic. So to find my spot's right-address I mix $m$ portions of $B$ 's right-number with $n$ portions of $A$ 's right-number and divide by the total $m + n$ ; same trick with the up-numbers. If instead I keep walking past a house, one of my shadow-steps runs backwards — a minus sign — so I subtract and divide by $m - n$ . And if I try to split equally and stay outside, there's nowhere to stand: the answer flies off to infinity.

Connections

2.3.04 Section formula — internal and external division (Hinglish) — the parent recipe this page derives.
Similar triangles — the engine of Step 3.
Slope of a line — constant slope is why both shadows split alike.
Midpoint formula — the $m = n$ special case of Step 7.
Distance formula — verify any ratio independently.
Centroid of a triangle — divides each median $2 : 1$ from the vertex.
Collinearity of three points — three points are collinear exactly when one divides the other two in some ratio.