To find a point that divides a line in a ratio, turn the ratio into a fraction of the whole vector, then add that piece to the start point. The key step is adding the ratio parts to find the total.
This lesson builds on adding displacement vectors and links to finding a midpoint and a segment length in coordinate geometry.
Why is the fraction not the ratio number?
A ratio AP:PB = 1:2 compares two parts. The whole line AB is 1 + 2 = 3 parts. AP is 1 of those 3 parts, so AP = ⅓AB.
If AP:PB = 3:1 then the whole is 4 parts and AP = ¾AB. The fraction always has the first part on the upper line and the sum of the parts on the lower line.
How do I find the point step by step?
- Find AB by subtracting: end point minus start point, upper and lower separately.
- Add the ratio parts to get the total.
- Write AP as a fraction of AB and multiply both numbers of AB by it.
- Add AP to A to get the position of P.
- Check that P lies between A and B, and that AP and PB have the ratio you were given.
Worked example
A(−2, 1) and B(10, −8). The point P lies on AB with AP:PB = 1:2. Find the coordinates of P.
Step 1, vector AB: (10 − (−2), −8 − 1) = (12, −9).
Step 2, total parts: 1 + 2 = 3, so AP = ⅓AB.
Step 3, vector AP: ⅓ × (12, −9) = (4, −3).
Step 4, position of P: x = −2 + 4 = 2 and y = 1 + (−3) = −2.
Answer: P(2, −2).
Check: PB = B − P = (10 − 2, −8 − (−2)) = (8, −6). This is twice AP = (4, −3), so AP:PB = 1:2. Correct.
The mistake to watch for
A frequent error is using the ratio numbers as if they were the fraction: taking AP:PB = 1:2 to mean AP = ½AB.
Mistaken answer: AP = ½ × (12, −9) = (6, −4.5), so P = (4, −3.5).
The student forgot that the whole is 3 parts, not 2.
Half of AB would make AP equal to PB, a ratio of 1:1. The correction is always to write the total first: add the parts, then use that number as the denominator. Here the check also fails, since PB = (6, −4.5) equals AP rather than being double it.
Check yourself
Try these, then open each answer.
1. A(0, 3), B(10, −7). P is on AB with AP:PB = 2:3. Find P.
Show answer
AB = (10, −10). Total parts = 5, so AP = ⅖AB = (4, −4).
P = (0 + 4, 3 + (−4)) = (4, −1).
Check: PB = (6, −6), which is 3 parts to AP’s 2 parts of size (2, −2).
2. Find the midpoint of A(−4, 6) and B(8, 2).
Show answer
AB = (12, −4). Half is (6, −2). Midpoint = (−4 + 6, 6 + (−2)) = (2, 4).
3. A(4, −2) and B(−8, 10). P is on AB with AP:PB = 1:3. Find P.
Show answer
AB = (−12, 12). Total = 4, so AP = ¼AB = (−3, 3).
P = (4 + (−3), −2 + 3) = (1, 1).
Where this leads next
The next lesson, distinguishing position from displacement, explains the vector OA that starts from the origin. You can then try the vectors and transformations practice set. The non-calculator working trainer helps with fractions of vectors.
If fractions of vectors slow you down, our teachers can show you the structure in online one-to-one Mathematics tuition.