Section formula — internal and external division: Question bank

Before we start, one shared vocabulary so nothing is undefined:

True or false — justify

The line through

A

and

B

divides the segment

A B

in ratio

2 : 3

internally means

P

is closer to

A

than to

B

True —

A P : P B = 2 : 3

and

2 < 3

, so the piece next to

A

is shorter, hence

P

hugs

A

Internal division in ratio

m : n

always gives a point strictly between

A

and

B

for any positive

m, n

True — with both

m, n > 0

the weights

\frac{m}{m + n}

and

\frac{n}{m + n}

are each between

0

and

1

, so

P

never escapes the segment.

Ratio

3 : 2

and ratio

2 : 3

give the same dividing point

False — they name the same segment split from opposite ends;

3 : 2

sits nearer

B

2 : 3

sits nearer

A

, two different points.

The midpoint is a special case of external division

False — the midpoint is internal with

m = n = 1

; external division with

m = n

gives no finite point at all.

For external division the point

P

can never coincide with

A

B

True — landing on

A

needs ratio

0 : n

and landing on

B

needs

m : 0

; both make

P

an endpoint, which is a boundary of internal, not external, division.

\frac{A P}{P B} = \frac{m}{n}

then automatically

\frac{B P}{P A} = \frac{n}{m}

True — flipping which distance you write on top just inverts the fraction; it is the same split described from

B

's side.

Swapping the labels

A

and

B

leaves the dividing point unchanged if you also swap

m

and

n

True — renaming endpoints and simultaneously renaming the ratio pieces describes the identical geometric split, so

P

stays put.

A ratio of

2 : 2

and a ratio of

1 : 1

describe different points

False — only the reduced ratio matters; both mean equal pieces, i.e. the midpoint.

Spot the error

A student writes internal

P = (\frac{m x _{1} + n x _{2}}{m + n}, \dots)

— what went wrong?

They paired

m

with

A

's coordinate

x_{1}

; but

\frac{x - x _{1}}{x _{2} - x} = \frac{m}{n}

solves to put

m

on the far point

x_{2}

. Ratio hits the opposite point.

For external division a student uses denominator

m + n

— why is that wrong?

External means the two projected lengths point opposite ways, so one is signed negative; replacing

n

- n

turns the sum

m + n

into the difference

m - n

A student claims "ratio

- 2 : 3

is impossible because a length can't be negative" — fix the statement

Distances are positive, but a signed ratio is a bookkeeping device: a negative sign flags that

P

lies outside the segment, i.e. it is external division of magnitude

2 : 3

Someone solves a "find the ratio" problem using only the

x

-equation and never checks

y

— is that safe?

It gives the right

k

if the three points are actually collinear; the

y

-equation is the built-in collinearity check, so skipping it risks accepting a

P

that isn't even on line

A B

A student verifies a ratio using the Distance formula but writes

A P + P B = A B

for an external point — spot the flaw

That additivity holds only internally; for an external point

P

beyond

B

you have

A P - P B = A B

(or

A P = A B + P B

), because

P

overshoots the segment.

"The centroid divides a median in ratio

1 : 2

" — correct it

It divides each median in ratio

2 : 1

measured from the vertex; the longer piece (

2

) is on the vertex side, the shorter (

1

) touches the midpoint of the opposite side. See Centroid of a triangle.

A student computes external

3 : 3

division and reports a finite point — what's impossible here?

With

m = n

the denominator

m - n = 0

, so the coordinates blow up;

P

runs off to infinity along the line, there is no finite answer.

Why questions

Why does the same ratio

m : n

appear in both the

x

-shadow and the

y

-shadow of the segment?

Because the line has one constant slope, so equal fractions of horizontal travel correspond to equal fractions of vertical travel — the shadows are just scaled copies via Similar triangles.

Why can we cross-multiply

\frac{x - x _{1}}{x _{2} - x} = \frac{m}{n}

without worrying about sign for internal division?

Internally

P

is between

A

and

B

, so

x - x_{1}

and

x_{2} - x

have the same sign (both positive going

A \to B

), keeping the ratio genuinely positive.

Why does replacing

n

- n

correctly model external division?

Going

A \to P

then

P \to B

reverses direction when

P

is outside, so one projected length is measured backwards; a minus sign on

n

encodes that reversed step.

Why is the midpoint formula "unweighted" — a plain average — while the general formula has weights?

Equal pieces (

m = n

) give equal weights

\frac{1}{2}, \frac{1}{2}

, and a weighted mix with equal weights is exactly the ordinary average. This links to Midpoint formula.

Why must

P

A

B

be collinear for any section-formula question to make sense?

The ratio

\frac{A P}{P B}

measures splitting along one straight line; if

P

were off the line, "dividing the segment" has no meaning — that's the heart of Collinearity of three points.

Why does a larger

m

(with

n

fixed) push

P

toward

B

rather than

A

Bigger

m

means the piece

A P

is longer, so

P

has travelled further from

A

— which is toward

B

Why can two different ratios still land on the same point when one is internal and one external?

They can't for the same pair

A, B

— internal points live strictly inside, external strictly outside; the sign of the ratio cleanly separates the two families.

Edge cases

What point does internal ratio

0 : n

(with

n > 0

) give?

A P = 0

, so

P = A

itself — the weight collapses entirely onto

A

's coordinates.

What point does internal ratio

m : 0

(with

m > 0

) give?

P B = 0

, so

P = B

— all weight sits on

B

As internal ratio

m : n

runs from

0 : 1

up to

1 : 0

, what path does

P

trace?

It slides continuously from

A

(at

0 : 1

) all the way to

B

(at

1 : 0

), covering every point of the segment exactly once.

What happens to

P

as external ratio

m : n

approaches

m = n

The denominator

m - n \to 0

, so

∣ P ∣ \to \infty

;

P

races off to infinity along the line — the "parallel at infinity" limit.

For external ratio

m : n

with

m > n

, which side of the segment is

P

on?

Beyond

B

— the larger weight on

B

's side of the signed formula drags

P

past

B

For external ratio

m : n

with

m < n

, which side is

P

on?

Beyond

A

— now the denominator

m - n

is negative and the geometry places

P

on the far side of

A

Is there any ratio that places

P

exactly at the midpoint via external division?

No — external division can never reach an interior point, and the midpoint is the most interior point of all.

If someone gives a ratio like

4 : 6

, does reducing it to

2 : 3

change the point?

No — only the proportion matters, so

4 : 6

and

2 : 3

name the identical point; always reduce to keep arithmetic clean.

Connections

Midpoint formula — the $m = n = 1$ boundary case probed above.
Centroid of a triangle — the $2 : 1$ -from-vertex trap.
Similar triangles — why both shadows share the ratio.
Distance formula — the independent check behind the internal/external additivity trap.
Slope of a line — the constant-slope reason the shadows match.
Collinearity of three points — why $P, A, B$ must lie on one line.