Year 12 Coordinate Geometry: Point on a line given a length

In this kind of question, you are not told a clean fraction of a segment. Instead, you are told an actual distance, such as 5 units from one endpoint or 3 centimetres beyond a point in the same direction. That means the vector method still works, but you first need to control the vector's length. The usual tool is the unit vector, which has length 1 and points in the required direction.

Length information is easiest to use after normalising the direction vector. Find the direction, turn it into a unit vector, then scale by the required distance.

Main method

Find the direction vector for the line.
Find its magnitude using the distance formula.
Divide by the magnitude to get a unit vector.
Multiply the unit vector by the required distance.
Add or subtract that scaled vector depending on the wording.

This method is especially useful when the point is not a neat midpoint or ratio point. It is also the safest method when the line is diagonal, because it handles both coordinates together instead of trying to guess how much x and y should each change.

Worked example 1

Problem: A point

P

lies on the line from

A (1, 2)

B (7, 5)

. Find the point

P

such that

AP = 5

Find the direction vector: $vector A B = (6, 3)$ .
Its length is $sqrt (6^{2} + 3^{2}) = sqrt (45) = 3 sqrt (5)$ .
The unit direction vector is $(6, 3) / (3 sqrt (5)) = (2 / sqrt (5), 1 / sqrt (5))$ .
A vector of length 5 in that direction is $5 (2 / sqrt (5), 1 / sqrt (5)) = (2 sqrt (5), sqrt (5))$ .
Add this to $A$ : $P = (1 + 2 sqrt (5), 2 + sqrt (5))$ .
This is exact, so it is better than rounding unless the question asks for decimals.

Answer:

P = (1 + 2 sqrt (5), 2 + sqrt (5))

Worked example 2

Problem: Find the point

Q

on the same line as

A (2, 1)

and

B (10, 7)

such that

Q

is 5 units from

A

in the direction of

B

Direction vector: $vector A B = (8, 6)$ .
Magnitude: $sqrt (8^{2} + 6^{2}) = sqrt (100) = 10$ .
Unit vector: $(8, 6) / 10 = (4 / 5, 3 / 5)$ .
Scale to length 5: $5 x (4 / 5, 3 / 5) = (4, 3)$ .
Add to $A$ : $Q = (2, 1) + (4, 3) = (6, 4)$ .

Answer:

Q = (6, 4)

Two possible points

Some questions ask for points on a line at a given distance from a fixed point without specifying a direction. In that case there are often two answers, one on each side of the starting point. The unit vector gives one direction, and the negative of that unit vector gives the other. Always read carefully to see whether the question says "from A toward B" or simply "on the line through A and B".

Common mistakes

Scaling the original direction vector directly without first adjusting its length.
Rounding too early and losing accuracy.
Moving the right distance but in the wrong direction.
Using a point on the segment when the required point lies beyond the endpoint.

Quick check

After finding the point, you can verify it by checking the distance from the starting point. If the required distance was 5, then the displacement vector from the start to your answer should have length exactly 5. This final check often catches arithmetic slips.