Is the point that divides a line in a given ratio ever the midpoint?

Only when the ratio is 1 to 1. The point dividing AB with AP to PB equal to m to n sits a fraction m over m plus n of the way from A, and that fraction equals one half only when m equals n. For 1 to 3 the point is a quarter of the way from A, not a half, so it is not the midpoint.

Does the order of the ratio matter?

Yes. AP to PB equal to 1 to 3 puts the small part next to A, so the point is a quarter of the way from A. AP to PB equal to 3 to 1 puts the large part next to A, so the point is three quarters of the way from A. The two are different points, on opposite sides of the midpoint.

How do you find the point that divides AB in the ratio m to n?

Count the total parts m plus n, take the fraction m over m plus n of the way from the named first point A, and compute P equals A plus that fraction times B minus A. For A at (2, 3), B at (14, 15) and the ratio 1 to 3, B minus A is (12, 12), a quarter of which is (3, 3), so P is (5, 6).

GCSE Maths Higher · AQA · Coordinate geometry

Dividing a line in a given ratio: count the parts, do not take the midpoint

To find the point that divides $AB$ with $AP : PB = m : n$ , counted from A, you move a fraction $\tfrac{m}{m+n}$ of the way along, so

P = A + \tfrac{m}{m+n}\,(B - A).

The ratio

1 : 3

means a quarter of the way from A, not a half.

The instinct is to read "divide the line" as "halve the line" and reach for the midpoint, or to step the right fraction but from the wrong end so the ratio is reversed. Both ignore what the ratio is actually telling you. The fix is a fixed habit: count the total parts, take that fraction of the journey, and start from the named first point.

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How to spot it in your own work

You took the midpoint of the segment, halving it regardless of the ratio.
You stepped the right fraction but from the wrong end, so $1 : 3$ turned into $3 : 1$ .
You miscounted the parts, splitting $1 : 3$ into 3 parts instead of 4.
You ignored the order of the ratio: $1 : 3$ from A differs from $3 : 1$ from A.

An exam question that triggers it

Here is the structure of JUN24 Paper 3 Question 15a:

$A$ is the point $(-5, 9)$ and $C$ is the point $(3, -7)$ .

$B$ lies on the line segment $AC$ so that $AB : BC = 1 : 3$ . Work out the coordinates of $B$ .

$AB : BC = 1 : 3$ makes four equal parts, so $B$ is a quarter of the way from A. With $C - A = (8, -16)$ , a quarter is $(2, -4)$ , so $B = (-3, 5)$ . The misconception takes the midpoint $(-1, 1)$ or measures three quarters from A to give $(1, -3)$ .

Why students fall for this

The phrase "divide the line" sounds like "split it down the middle", and the midpoint is the most-practised point on a segment. Under time pressure that strong prior fires before the ratio is read, so a $1 : 3$ question is answered with the halfway point.

The ratio also carries a direction that is easy to lose. A point that divides $AB$ is counted from A, the named first point, but students often step the fraction from whichever end they reach first, which reverses the ratio.

Counting the parts is a third slip. $AP : PB = 1 : 3$ makes $1 + 3 = 4$ equal parts, not 3, so the fraction is a quarter and not a third. Adding the parts wrongly shifts the whole answer.

The fix: Count the parts, take the fraction, and start from the named point

Step 1: count the total parts. For $AP : PB = m : n$ there are $m + n$ equal parts.

Step 2: the point is $\tfrac{m}{m+n}$ of the way. P is $m$ parts along, so the fraction of the journey is $\tfrac{m}{m+n}$ .

Step 3: count from the named first point. A point dividing $AB$ is measured from $A$ , so use the displacement $B - A$ , not $A - B$ .

Step 4: compute and sense-check. Work out $P = A + \tfrac{m}{m+n}\,(B - A)$ and check P lies on the correct side of the midpoint: a first part smaller than the second keeps P near $A$ .

Worked example

L1 structure: $A(2, 3)$ , $B(14, 15)$ . Find P with $AP : PB = 1 : 3$ .

Count the parts. $1 + 3 = 4$ , so P is a quarter of the way from A.
Take the fraction of the step. $B - A = (12, 12)$ , and a quarter of that is $(3, 3)$ .
$P = (2 + 3,\; 3 + 3) = (5, 6)$
The midpoint $(8, 9)$ halves regardless of the ratio; $(11, 12)$ is three quarters from the wrong end; $(6, 7)$ splits into 3 parts not 4.

JUN24·3·15a structure: $A(-5, 9)$ , $C(3, -7)$ . Find B with $AB : BC = 1 : 3$ .

Quarter of the way. $C - A = (8, -16)$ , a quarter is $(2, -4)$ .
$B = (-5 + 2,\; 9 - 4) = (-3, 5)$
Contrast with the midpoint. The midpoint of $AC$ is $\left(\tfrac{-5+3}{2}, \tfrac{9-7}{2}\right) = (-1, 1)$ , a different point. They agree only when the ratio is $1 : 1$ .

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Common questions

Is the ratio point ever the midpoint?: Only when the ratio is $1 : 1$ . The fraction $\tfrac{m}{m+n}$ equals $\tfrac{1}{2}$ only when $m = n$ , so for $1 : 3$ the point is a quarter of the way, at $(5, 6)$ in the example, not the midpoint $(8, 9)$ .
Does the order of the ratio matter?: Yes. $AP : PB = 1 : 3$ is a quarter of the way from A, while $AP : PB = 3 : 1$ is three quarters of the way from A. They are different points on opposite sides of the midpoint, so always count from the named first point.
Which displacement do I use, $B - A$ or $A - B$ ?: Use $B - A$ , because the point is counted from A towards B. Using $A - B$ steps the wrong way: in the JUN24 example that gives $(-7, 13)$ instead of the correct $(-3, 5)$ .

Related misconceptions

Perpendicular and parallel gradientsRelating the gradients of two lines with the wrong rule: reusing the same gradient for a perpendicular line, or flipping without changing the sign, instead of taking the negative reciprocal.
Reading a reordered linear equationMisreading the gradient and intercept of a line written out of the usual order, such as y = 5 - 3x, by taking the constant as the gradient or dropping the sign on the coefficient of x.

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