MathsCBSE Class 10 · Coordinate Geometry

Section Formula (Internal Division) for Class 10

Q: Why does m1 multiply the coordinates of B and not A?

Because the point is a weighted average. If AP is the bigger share, P sits further from A and closer to B, so B must carry the bigger weight. The derivation using similar triangles gives this pairing directly.

Q: How can I check my answer quickly?

Make sure each coordinate lies between those of the two ends. For a fuller check, find AP and PB with the distance formula and confirm they are in the given ratio.

Q: Is external division in the Class 10 syllabus for 2026-27?

No. External division is not in the 2026-27 syllabus, and only internal division is examined.

Q: What if the ratio is not given?

Take the ratio as k to 1 and use the condition in the question, such as the point lying on an axis or on a given line, to find k. A positive k means the division is internal. The worked examples on this page show this for the y-axis, the x-axis and a point with an unknown coordinate.

Q: Which NCERT exercise covers the section formula?

Exercise 7.2 of the NCERT textbook covers the section formula, midpoints and related problems. Practise questions of each kind there along with the board-level ones on this page.

Updated 14 October 2026 · Fact-checked

The section formula gives the coordinates of a point that divides a line segment internally in a given ratio. If P divides AB so that AP to PB is m1 to m2, each coordinate of P is a weighted average of the end coordinates. Board papers ask you to find such points, verify them, and use them in longer problems.

Line segment AB on coordinate axes with point P dividing it into parts m and n, with red and green right triangles showing the coordinate di — Notice P splits AB into m and n; the red and green similar right triangles give the ratio that leads to the section formula.
Picture: Huzaifa abedeen · CC BY-SA 4.0 · Wikimedia Commons

Understand Section Formula (Internal Division)

Suppose you have two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , and a point P on the segment AB. If $A P : P B = m_{1} : m_{2}$ , we say P divides AB internally in the ratio $m_{1} : m_{2}$ . The section formula tells you exactly where P is, without drawing anything.

Why does it work? Drop perpendiculars from A, P and B to the x-axis, and draw horizontal lines through A and P. This makes two right triangles, one with hypotenuse AP and one with hypotenuse PB. Their angles match (corresponding angles on parallel lines), so they are similar by AA similarity. Their sides are in the ratio $m_{1} : m_{2}$ , so $\frac{x - x _{1}}{x _{2} - x} = \frac{m _{1}}{m _{2}}$ . Cross-multiplying gives $m_{2} x - m_{2} x_{1} = m_{1} x_{2} - m_{1} x$ , so $x = \frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}$ . The y-coordinate works the same way.

Notice the cross pairing: $m_{1}$ , the part touching A, multiplies the coordinates of B. This feels odd at first, but think of it as a tug of war. A bigger $m_{1}$ means P is further from A, so B pulls harder. The formula is a weighted average of the two ends.

The formula connects the whole chapter. The midpoint is the case $1 : 1$ . Points of trisection use $1 : 2$ and $2 : 1$ . Finding an unknown ratio uses the $k : 1$ form. You can always verify an answer with the distance formula: $A P$ and $P B$ must be in the given ratio.

Conditions: the ratio parts $m_{1}$ and $m_{2}$ are both positive, and P lies between A and B. So each coordinate of P lies between the corresponding coordinates of A and B. In the 2026-27 syllabus only internal division is used, so a quick between-check catches most slips.

Section Formula (Internal Division): formulas to remember

Section formula (internal division): $P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$; P divides $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ with $A P : P B = m_{1} : m_{2}$ . $m_{1}$ pairs with B.
Section formula, k : 1 form: $P = (\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1})$; Use when the ratio is unknown; positive k means internal division.
Midpoint (ratio 1 : 1): $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$; Special case of the section formula.
Fraction-of-the-way form: $P = A + \frac{m _{1}}{m _{1} + m _{2}} (B - A)$; P is $\frac{m _{1}}{m _{1} + m _{2}}$ of the way from A to B. Good for checking answers quickly.
Ratio from a fraction of length: $A P = \frac{p}{q} A B \Rightarrow A P : P B = p : (q - p)$; Convert a statement like $A P = \frac{2}{5} A B$ into a ratio before using the formula.
Ratio when the point is known: $k = \frac{x - x _{1}}{x _{2} - x} (or the same with y)$; Gives AP : PB = k : 1 when P(x, y) is on AB; k > 0 means P lies between A and B. Check the answer with the other coordinate.

How to solve Section Formula (Internal Division) questions

Use this routine for any question asking for a point that divides a segment in a known ratio, or for using that point further.

1Name the points: write $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and the ratio $m_{1} : m_{2}$ with $m_{1}$ for AP.
2If the question gives a fraction such as $A P = \frac{3}{7} A B$ , convert it to the ratio $3 : 4$ first.
3Write the section formula in symbols before substituting; examiners give a mark for the correct formula.
4Substitute with $m_{1}$ multiplying B's coordinates and $m_{2}$ multiplying A's, keeping brackets around negatives.
5Simplify each coordinate separately and write P as an ordered pair.
6Check that each coordinate lies between those of A and B, or verify with the fraction-of-the-way form or distance formula.
7If the question continues (a distance, a shape, an axis), use P in that next step and state units.

Section Formula (Internal Division) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Section formula for internal division

Imagine your school sets up a straight relay track for sports day on a grid marked in metres. The start flag is at $A (- 4, 2)$ and the finish flag is at $B (11, 22)$ . The organisers want a water point W placed so that the runners have covered two-fifths of the track when they reach it.

Find the coordinates of W, and the distance a runner still has to cover from W to the finish.

Try it first, then show the solutionSolution

Two-fifths covered means $A W = \frac{2}{5} A B$ , so $A W : W B = 2 : 3$ .
$x = \frac{2 ( 11 ) + 3 ( - 4 )}{5} = \frac{22 - 12}{5} = 2$ .
$y = \frac{2 ( 22 ) + 3 ( 2 )}{5} = \frac{44 + 6}{5} = 10$ .
So $W (2, 10)$ .
$W B = (11 - 2)^{2} + (22 - 10)^{2} = 81 + 144 = 225 = 15$ .
Check: $A B = 1 5^{2} + 2 0^{2} = 25$ , and $\frac{3}{5} \times 25 = 15$ . Correct.

Answer: W is at $(2, 10)$ and the runner has 15 m left.

Memory hook: Two-fifths of the race done means the water point splits the track 2 : 3.

Section formula for internal division

Suppose a kitchen shelf is drawn on a grid in centimetres, with a straight bracket from $A (2, 3)$ to $B (14, 19)$ . You want to fix two hooks so the bracket is cut into three pieces in the ratio $1 : 2 : 1$ from A.

Find the positions of both hooks and the length of the middle piece.

Try it first, then show the solutionSolution

Total parts $= 1 + 2 + 1 = 4$ , so hook $H_{1}$ is at ratio $1 : 3$ and hook $H_{2}$ at ratio $3 : 1$ from A.
$H_{1} = (\frac{1 ( 14 ) + 3 ( 2 )}{4}, \frac{1 ( 19 ) + 3 ( 3 )}{4}) = (\frac{20}{4}, \frac{28}{4}) = (5, 7)$ .
$H_{2} = (\frac{3 ( 14 ) + 1 ( 2 )}{4}, \frac{3 ( 19 ) + 1 ( 3 )}{4}) = (\frac{44}{4}, \frac{60}{4}) = (11, 15)$ .
$H_{1} H_{2} = (11 - 5)^{2} + (15 - 7)^{2} = 36 + 64 = 10$ .
Check: $A B = 1 2^{2} + 1 6^{2} = 20$ , and the middle piece is $\frac{2}{4} \times 20 = 10$ .

Answer: Hooks at $(5, 7)$ and $(11, 15)$ ; middle piece 10 cm.

Memory hook: Count all the pieces first; each hook's ratio is parts before : parts after.

Tricks to remember Section Formula (Internal Division)

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cross and pull
The ratio number next to A multiplies B. Picture $m_{1}$ reaching across to grab B: the bigger the share AP, the more B pulls P towards itself.
Use it when: Every time you substitute into the section formula.
Parts before : parts after
For any dividing point, count equal parts from A up to the point, then from the point to B. That is the ratio.
Use it when: Converting fractions like $A P = \frac{3}{8} A B$ or multi-point divisions into ratios.
Between-check
Internal division means P is sandwiched: each coordinate of P lies between those of A and B.
Use it when: A 5-second check before moving on in the exam.

Section Formula (Internal Division): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: MCQ: point dividing a segment in 1 : 2 · 1 mark · MCQBoard level

The point which divides the line segment joining $(2, - 3)$ and $(8, 6)$ internally in the ratio $1 : 2$ is
(a) $(6, 3)$
(b) $(4, 0)$
(c) $(5, \frac{3}{2})$
(d) $(4, 3)$

Try it first, then show the solutionSolution

$x = \frac{1 ( 8 ) + 2 ( 2 )}{3} = \frac{12}{3} = 4$ . (½ mark)
$y = \frac{1 ( 6 ) + 2 ( - 3 )}{3} = \frac{0}{3} = 0$ , so the point is $(4, 0)$ . Option (a) is the $2 : 1$ point and (c) is the midpoint. (½ mark)

Answer: (b) $(4, 0)$

Example 2: Assertion-reason: pairing the ratio correctly · 1 mark · Assertion-reasonBoard level

Assertion (A): The point $(- 2, 3)$ divides the line segment joining $A (- 5, 6)$ and $B (4, - 3)$ internally in the ratio $2 : 1$ .
Reason (R): The point dividing the join of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ internally in the ratio $m_{1} : m_{2}$ is $(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is the section formula, so R is true. Using it, the $2 : 1$ point is $(\frac{2 ( 4 ) + 1 ( - 5 )}{3}, \frac{2 ( - 3 ) + 1 ( 6 )}{3}) = (1, 0)$ , not $(- 2, 3)$ . (½ mark)
In fact $(- 2, 3) = (\frac{1 ( 4 ) + 2 ( - 5 )}{3}, \frac{1 ( - 3 ) + 2 ( 6 )}{3})$ is the $1 : 2$ point, so A is false. (½ mark)

Answer: (d) A is false but R is true.

Example 3: Point dividing a segment in a given ratio · 2 marks · Very short answerBoard level

Find the coordinates of the point P which divides the line segment joining $A (- 5, 7)$ and $B (10, - 3)$ internally in the ratio $3 : 2$ .

Try it first, then show the solutionSolution

$m_{1} = 3$ , $m_{2} = 2$ , $A (- 5, 7)$ , $B (10, - 3)$ ; $P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ . (½ mark)
$x = \frac{3 ( 10 ) + 2 ( - 5 )}{5} = \frac{30 - 10}{5} = 4$ . (½ mark)
$y = \frac{3 ( - 3 ) + 2 ( 7 )}{5} = \frac{- 9 + 14}{5} = 1$ . (½ mark)
So $P (4, 1)$ ; check: $- 5 < 4 < 10$ and $- 3 < 1 < 7$ . (½ mark)

Answer: $P (4, 1)$

Example 4: Unknown ratio, then a line through the point · 3 marks · Short answerHarder board level

The point $P (- 4, 6)$ lies on the segment joining $A (- 6, 10)$ and $B (3, - 8)$ . Find the ratio in which P divides AB. P also lies on the line $3 x + k y = 0$ ; find k.

Try it first, then show the solutionSolution

Let $A P : P B = m : n$ . Using x: $\frac{3 m - 6 n}{m + n} = - 4$ , so $3 m - 6 n = - 4 m - 4 n$ , $7 m = 2 n$ , and $m : n = 2 : 7$ . (1)
Check with y: $\frac{2 ( - 8 ) + 7 ( 10 )}{9} = \frac{54}{9} = 6$ , which matches P. (1)
$3 (- 4) + 6 k = 0$ , so $k = 2$ . (1)

Answer: $A P : P B = 2 : 7$ ; $k = 2$ .

Example 5: From a fraction of the length to the point · 3 marks · Short answerBoard level

P and Q lie on the segment joining $A (- 6, 10)$ and $B (9, - 5)$ such that $A P = \frac{2}{5} A B$ and $A Q = \frac{4}{5} A B$ . Find P and Q, and show that $P Q = \frac{2}{5} A B$ .

Try it first, then show the solutionSolution

$A P : P B = 2 : 3$ and $A Q : QB = 4 : 1$ . (½ mark)
$P = (\frac{2 ( 9 ) + 3 ( - 6 )}{5}, \frac{2 ( - 5 ) + 3 ( 10 )}{5}) = (0, 4)$ . (1 mark)
$Q = (\frac{4 ( 9 ) + 1 ( - 6 )}{5}, \frac{4 ( - 5 ) + 1 ( 10 )}{5}) = (6, - 2)$ . (½ mark)
$A B = 1 5^{2} + 1 5^{2} = 152$ and $P Q = 6^{2} + 6^{2} = 62$ . (½ mark)
$\frac{2}{5} \times 152 = 62 = P Q$ . Shown. (½ mark)

Answer: $P (0, 4)$ , $Q (6, - 2)$ , and $P Q = 62$ units $= \frac{2}{5} A B$ .

Example 6: Missing end point from the dividing point · 3 marks · Short answerBoard level

The point $P (2, - 1)$ divides the segment joining $A (- 4, 5)$ and $B (x, y)$ internally in the ratio $3 : 1$ . Find B. Also find the length AB.

Try it first, then show the solutionSolution

By the section formula, $2 = \frac{3 x + 1 ( - 4 )}{4}$ . (½ mark)
$8 = 3 x - 4$ , so $x = 4$ . (½ mark)
$- 1 = \frac{3 y + 1 ( 5 )}{4}$ , so $- 4 = 3 y + 5$ and $y = - 3$ . (1 mark)
$B (4, - 3)$ ; $A B = (4 + 4)^{2} + (- 3 - 5)^{2} = 64 + 64 = 82$ . (½ mark)
Check: $A P = 36 + 36 = 62 = \frac{3}{4} A B$ . (½ mark)

Answer: $B (4, - 3)$ and $A B = 82$ units.

Example 7: Ratio in which the y-axis divides a segment (k : 1 form) · 3 marks · Short answerBoard level

Find the ratio in which the y-axis divides the line segment joining $A (- 3, 4)$ and $B (6, - 2)$ . Also find the point of division.

Try it first, then show the solutionSolution

Let the y-axis divide AB at P in the ratio $k : 1$ . Then $P = (\frac{6 k - 3}{k + 1}, \frac{- 2 k + 4}{k + 1})$ . (½ mark)
P lies on the y-axis, so its x-coordinate is 0: $6 k - 3 = 0$ , giving $k = \frac{1}{2}$ . (1 mark)
So the ratio is $\frac{1}{2} : 1 = 1 : 2$ (k is positive, so the division is internal). (½ mark)
$y = \frac{1 ( - 2 ) + 2 ( 4 )}{3} = \frac{6}{3} = 2$ . (½ mark)
Point of division $(0, 2)$ ; check: $x = \frac{1 ( 6 ) + 2 ( - 3 )}{3} = 0$ . (½ mark)

Answer: The y-axis divides AB in the ratio $1 : 2$ at the point $(0, 2)$ .

Example 8: Points of trisection and where the x-axis cuts the segment · 3 marks · Short answerHarder board level

The line segment joining $A (2, - 2)$ and $B (- 7, 4)$ is divided into three equal parts at P and Q, with P nearer A. Find P and Q. Also find the ratio in which the x-axis divides AB, and show that the x-axis meets AB at P.

Try it first, then show the solutionSolution

P divides AB in $1 : 2$ : $P = (\frac{1 ( - 7 ) + 2 ( 2 )}{3}, \frac{1 ( 4 ) + 2 ( - 2 )}{3}) = (- 1, 0)$ . (1 mark)
Q divides AB in $2 : 1$ : $Q = (\frac{2 ( - 7 ) + 1 ( 2 )}{3}, \frac{2 ( 4 ) + 1 ( - 2 )}{3}) = (- 4, 2)$ . (1 mark)
Let the x-axis divide AB in $k : 1$ . The y-coordinate is 0: $\frac{4 k - 2}{k + 1} = 0$ , so $k = \frac{1}{2}$ and the ratio is $1 : 2$ . (½ mark)
The $1 : 2$ point is P, and $P (- 1, 0)$ has y-coordinate 0, so the x-axis meets AB at P. (½ mark)

Answer: $P (- 1, 0)$ , $Q (- 4, 2)$ ; the x-axis divides AB in the ratio $1 : 2$ , at P.

Example 9: Centroid-type point using the section formula twice · 5 marks · Long answerHarder board level

The vertices of $△ A B C$ are $A (1, 8)$ , $B (- 3, - 2)$ and $C (11, 0)$ . D is the midpoint of BC. G is the point on the median AD with $A G : G D = 2 : 1$ . Find D and G. Then E is the midpoint of CA; show that G also divides the median BE in the ratio $2 : 1$ . Find the length of AD.

Try it first, then show the solutionSolution

$D = (\frac{- 3 + 11}{2}, \frac{- 2 + 0}{2}) = (4, - 1)$ . (1 mark)
$G = (\frac{2 ( 4 ) + 1 ( 1 )}{3}, \frac{2 ( - 1 ) + 1 ( 8 )}{3}) = (3, 2)$ . (1 mark)
$E = (\frac{11 + 1}{2}, \frac{0 + 8}{2}) = (6, 4)$ . (½ mark)
Point dividing BE in $2 : 1$ : $(\frac{2 ( 6 ) + 1 ( - 3 )}{3}, \frac{2 ( 4 ) + 1 ( - 2 )}{3}) = (3, 2)$ . (1 mark)
This is G, so G divides BE in the ratio $2 : 1$ . (½ mark)
$A D = (4 - 1)^{2} + (- 1 - 8)^{2} = 9 + 81 = 90 = 310$ . (1 mark)

Answer: $D (4, - 1)$ , $G (3, 2)$ , G divides BE in $2 : 1$ , and $A D = 310$ units.

Example 10: Unknown ratio and unknown coordinate together · 5 marks · Long answerHarder board level

The point $P (m, 6)$ lies on the line segment joining $A (- 4, 3)$ and $B (6, 8)$ .
(a) Find the ratio $A P : P B$ and the value of m.
(b) Find the ratio in which the y-axis divides AB, and the point where it meets AB.
(c) Verify your ratio in (a) using the distance formula.

Try it first, then show the solutionSolution

(a) Let $A P : P B = k : 1$ . Using y-coordinates: $6 = \frac{8 k + 3}{k + 1}$ . (½ mark)
$6 k + 6 = 8 k + 3$ , so $k = \frac{3}{2}$ and $A P : P B = 3 : 2$ . (1 mark)
$m = \frac{3 ( 6 ) + 2 ( - 4 )}{5} = \frac{10}{5} = 2$ , so $P (2, 6)$ . (½ mark)
(b) Let the y-axis divide AB in $k : 1$ . Then $\frac{6 k - 4}{k + 1} = 0$ , so $k = \frac{2}{3}$ and the ratio is $2 : 3$ . (1 mark)
$y = \frac{2 ( 8 ) + 3 ( 3 )}{5} = \frac{25}{5} = 5$ , so the y-axis meets AB at $(0, 5)$ . (1 mark)
(c) $A P = (2 + 4)^{2} + (6 - 3)^{2} = 45 = 35$ and $P B = (6 - 2)^{2} + (8 - 6)^{2} = 20 = 25$ . (½ mark)
$A P : P B = 35 : 25 = 3 : 2$ . Verified. (½ mark)

Answer: (a) $A P : P B = 3 : 2$ , $m = 2$ (b) ratio $2 : 3$ , point $(0, 5)$ (c) $A P = 35$ , $P B = 25$ , ratio $3 : 2$ .

Example 11: Case study: a zip line at a school camp · 4 marks · Case-basedHOTS / competency

Imagine a school adventure camp sets up a straight zip line on a site plan in metres, from platform $A (0, 30)$ to landing point $B (40, 0)$ . A photographer stands at point P on AB with $A P : P B = 3 : 7$ . A straight rope fence runs along the line $x = 24$ , and a safety marshal M stands where the fence crosses the zip line.
(i) Find the length of the zip line. (1 mark)
(ii) Find the position of P. (1 mark)
(iii) Find the ratio in which M divides AB and the position of M. Who is nearer A, the photographer or the marshal, and by how many metres? (2 marks)

Try it first, then show the solutionSolution

(i) $A B = 4 0^{2} + 3 0^{2} = 2500 = 50$ m. (1 mark)
(ii) $P = (\frac{3 ( 40 ) + 7 ( 0 )}{10}, \frac{3 ( 0 ) + 7 ( 30 )}{10}) = (12, 21)$ . (1 mark)
(iii) Let $A M : M B = k : 1$ . Then $24 = \frac{40 k + 0}{k + 1}$ , so $16 k = 24$ , $k = \frac{3}{2}$ and the ratio is $3 : 2$ . (½ mark)
$M = (\frac{3 ( 40 ) + 2 ( 0 )}{5}, \frac{3 ( 0 ) + 2 ( 30 )}{5}) = (24, 12)$ . (½ mark)
$A P = \frac{3}{10} \times 50 = 15$ m and $A M = \frac{3}{5} \times 50 = 30$ m (check: $2 4^{2} + 1 8^{2} = 30$ ). (½ mark)
So the photographer is nearer A, by $30 - 15 = 15$ m. (½ mark)

Answer: (i) 50 m (ii) $P (12, 21)$ (iii) $A M : M B = 3 : 2$ , $M (24, 12)$ ; the photographer is nearer A, by 15 m.

Where students lose marks in Section Formula (Internal Division)

The slips examiners see most often, and what to write instead.

Mistake: Pairing $m_{1}$ with A instead of B
It feels natural to match the first ratio number with the first point.
Wrong working
For $A (1, 2)$ , $B (7, 8)$ , ratio $1 : 2$ : $x = \frac{1 ( 1 ) + 2 ( 7 )}{3} = 5$ .
Right working
$x = \frac{1 ( 7 ) + 2 ( 1 )}{3} = 3$ , $y = \frac{1 ( 8 ) + 2 ( 2 )}{3} = 4$ , so $(3, 4)$ , nearer A as expected.
Do this: Say aloud: AP's share multiplies B. Then check P is nearer the end with the smaller share.
Usually costs: Usually all marks after the formula step.
Mistake: Using a fraction of length as the ratio
Reading $A P = \frac{1}{4} A B$ as the ratio $1 : 4$ .
Wrong working
$A P = \frac{1}{4} A B$ , so $A P : P B = 1 : 4$ .
Right working
$A P = \frac{1}{4} A B$ leaves $P B = \frac{3}{4} A B$ , so $A P : P B = 1 : 3$ .
Do this: Always find PB as the remaining part before writing the ratio.
Usually costs: Most of a 3-mark question.
Mistake: Sign errors with negative coordinates
Dropping brackets when multiplying a negative coordinate.
Wrong working
$\frac{3 ( - 2 ) + 2 \cdot 4}{5}$ written as $\frac{3 - 2 + 8}{5}$ .
Right working
$\frac{3 ( - 2 ) + 2 ( 4 )}{5} = \frac{- 6 + 8}{5} = \frac{2}{5}$ .
Do this: Put every substituted coordinate in brackets.
Usually costs: ½ to 1 mark, plus follow-on errors.
Mistake: Dividing by the wrong total
Using $m_{1}$ or 2 as the denominator out of habit from the midpoint formula.
Wrong working
Ratio $2 : 3$ : $x = \frac{2 ( 10 ) + 3 ( 0 )}{2} = 10$ .
Right working
$x = \frac{2 ( 10 ) + 3 ( 0 )}{2 + 3} = 4$ .
Do this: Write the denominator $m_{1} + m_{2}$ first, before the numerator.
Usually costs: The answer marks for that coordinate.

Exam tips

Write the formula in symbols before substituting; it often earns a method mark even if arithmetic slips.
Convert any 'fraction of the way' wording into a ratio and show that line in your answer.
In longer questions, the section formula is usually one step inside a shape or median problem, so keep your coordinates exact as fractions.
Use the between-check or a quick distance check before moving on; it costs seconds and catches pairing errors.
In case studies, read each sub-part's ratio from the description carefully and give units with every length.

Practice questions on Section Formula (Internal Division)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Find the coordinates of the point which divides the line segment joining $A (4, - 8)$ and $B (- 6, 7)$ internally in the ratio $2 : 3$ .
Hint
Pair 2 with B and 3 with A; the denominator is 5.
Answer
$x = \frac{2 ( - 6 ) + 3 ( 4 )}{5} = 0$ , $y = \frac{2 ( 7 ) + 3 ( - 8 )}{5} = - 2$ . The point is $(0, - 2)$ .
Q2 · 1 mark · MCQBoard level
If the midpoint of the line segment joining $(a, 4)$ and $(6, b)$ is $(2, 5)$ , then $(a, b)$ is
(a) $(- 2, 6)$
(b) $(2, 6)$
(c) $(- 2, 1)$
(d) $(10, 6)$
Hint
Set each midpoint coordinate equal to the given value.
Answer
(a) $(- 2, 6)$ , since $\frac{a + 6}{2} = 2$ gives $a = - 2$ and $\frac{4 + b}{2} = 5$ gives $b = 6$ .
Q3 · 3 marks · Short answerBoard level
Point $C (1, 3)$ divides the segment joining $A (- 2, 9)$ and $B (x, y)$ such that $A C = \frac{3}{5} A B$ . Find B and the length BC.
Hint
$A C = \frac{3}{5} A B$ means $A C : C B = 3 : 2$ . Solve each coordinate for the unknown.
Answer
$1 = \frac{3 x - 4}{5}$ gives $x = 3$ ; $3 = \frac{3 y + 18}{5}$ gives $y = - 1$ . $B (3, - 1)$ , $B C = 4 + 16 = 25$ units.
Q4 · 3 marks · Short answerHarder board level
A is $(- 1, 7)$ and B is $(4, - 3)$ . The point P divides AB in the ratio $2 : 3$ , and Q is the midpoint of PB. Find P and Q, and show that Q divides AB in the ratio $7 : 3$ .
Hint
Find P by the section formula, then Q by the midpoint formula, then test Q with the ratio $7 : 3$ .
Answer
$P = (\frac{2 ( 4 ) + 3 ( - 1 )}{5}, \frac{2 ( - 3 ) + 3 ( 7 )}{5}) = (1, 3)$ . $Q = (\frac{1 + 4}{2}, \frac{3 - 3}{2}) = (2.5, 0)$ . The point dividing AB in $7 : 3$ is $(\frac{7 ( 4 ) + 3 ( - 1 )}{10}, \frac{7 ( - 3 ) + 3 ( 7 )}{10}) = (2.5, 0)$ , which is Q. (P is $\frac{2}{5}$ of the way along AB and Q halfway from P to B, that is $\frac{7}{10}$ of the way.)
Q5 · 3 marks · Short answerHarder board level
Find the ratio in which the line $2 x + y - 4 = 0$ divides the segment joining $A (2, - 2)$ and $B (3, 7)$ . Also find the point of division and check it.
Hint
Take the ratio as $k : 1$ and substitute the point into the line.
Answer
The point is $(\frac{3 k + 2}{k + 1}, \frac{7 k - 2}{k + 1})$ . Substituting: $2 (3 k + 2) + (7 k - 2) - 4 (k + 1) = 0$ , so $9 k - 2 = 0$ and $k = \frac{2}{9}$ : the ratio is $2 : 9$ . Point $= (\frac{6 + 18}{11}, \frac{14 - 18}{11}) = (\frac{24}{11}, - \frac{4}{11})$ . Check: $\frac{48}{11} - \frac{4}{11} = 4$ .
Q6 · 3 marks · Short answerHarder board level
A is $(- 3, - 1)$ and B is $(9, 7)$ . Find the points P, Q, R that divide AB into four equal parts. Show that Q is the midpoint of PR.
Hint
Use ratios $1 : 3$ , $1 : 1$ and $3 : 1$ from A.
Answer
$P = (\frac{9 - 9}{4}, \frac{7 - 3}{4}) = (0, 1)$ , $Q = (3, 3)$ , $R = (\frac{27 - 3}{4}, \frac{21 - 1}{4}) = (6, 5)$ . Midpoint of PR $= (3, 3) = Q$ .
Q7 · 3 marks · Short answerHarder board level
The line segment joining $A (- 3, 5)$ and $B (6, - 4)$ is trisected at P and Q, with P nearer A. If Q lies on the line $2 x - y + k = 0$ , find k. Also find the ratio in which the y-axis divides AB and show that it meets AB at P.
Hint
Find P and Q with ratios $1 : 2$ and $2 : 1$ , substitute Q into the line, then use $k : 1$ with $x = 0$ .
Answer
$P = (\frac{6 - 6}{3}, \frac{- 4 + 10}{3}) = (0, 2)$ , $Q = (\frac{12 - 3}{3}, \frac{- 8 + 5}{3}) = (3, - 1)$ . $2 (3) - (- 1) + k = 0$ gives $k = - 7$ . For the y-axis, $\frac{6 t - 3}{t + 1} = 0$ gives $t = \frac{1}{2}$ , ratio $1 : 2$ , which is the point $P (0, 2)$ .
Q8 · 3 marks · Short answerHarder board level
Find the ratio in which the line $x + y = 4$ divides the line segment joining $A (- 1, 1)$ and $B (5, 7)$ . Also find the point of intersection.
Hint
Write the point as $(\frac{5 k - 1}{k + 1}, \frac{7 k + 1}{k + 1})$ and substitute into $x + y = 4$ .
Answer
$\frac{5 k - 1 + 7 k + 1}{k + 1} = 4$ gives $12 k = 4 k + 4$ , so $k = \frac{1}{2}$ and the ratio is $1 : 2$ . Point $= (\frac{5 - 2}{3}, \frac{7 + 2}{3}) = (1, 3)$ ; check $1 + 3 = 4$ .
Q9 · 3 marks · Short answerHOTS / competency
Find the error. A student found the point dividing $A (2, 6)$ and $B (12, 1)$ in the ratio $2 : 3$ and wrote $x = \frac{2 ( 2 ) + 3 ( 12 )}{5} = 8$ , $y = \frac{2 ( 6 ) + 3 ( 1 )}{5} = 3$ , so the point is $(8, 3)$ . Explain the error, give the correct point, and verify it using distances.
Hint
Which number should multiply B's coordinates? Check the ratio of AP to PB.
Answer
The student paired 2 with A instead of B, which gives the ratio $3 : 2$ . Correct: $x = \frac{2 ( 12 ) + 3 ( 2 )}{5} = 6$ , $y = \frac{2 ( 1 ) + 3 ( 6 )}{5} = 4$ , point $(6, 4)$ . $A P = 16 + 4 = 25$ , $P B = 36 + 9 = 35$ , ratio $2 : 3$ .
Q10 · 1 mark · Assertion-reasonBoard level
Assertion (A): The x-axis divides the segment joining $(2, - 3)$ and $(5, 6)$ in the ratio $1 : 2$ .
Reason (R): The x-axis divides the segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $- y_{1} : y_{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Set the y-coordinate of the dividing point equal to 0.
Answer
(a). With ratio $k : 1$ , $\frac{k y _{2} + y _{1}}{k + 1} = 0$ gives $k = - \frac{y _{1}}{y _{2}}$ , so R is true. Here $- y_{1} : y_{2} = 3 : 6 = 1 : 2$ , so A is true and follows from R.
Q11 · 3 marks · Short answerHarder board level
Find the ratio in which the point $P (x, 2)$ divides the segment joining $A (12, 5)$ and $B (4, - 3)$ . Also find x and the length AP.
Hint
Use the y-coordinates to find the ratio first.
Answer
With ratio $k : 1$ : $\frac{- 3 k + 5}{k + 1} = 2$ , so $5 k = 3$ and the ratio is $3 : 5$ . $x = \frac{3 ( 4 ) + 5 ( 12 )}{8} = 9$ . $A P = (12 - 9)^{2} + (5 - 2)^{2} = 32$ units.
Q12 · 4 marks · Case-basedHOTS / competency
Case study: On a site plan in metres, a straight water pipe runs from a tank $T (- 2, 8)$ to a house $H (13, - 2)$ . A tap is fixed at the point that divides TH in the ratio $1 : 4$ , and a valve V is fitted where the pipe crosses the boundary line $x = y$ .
(i) Find the length of the pipe. (1 mark)
(ii) Find the position of the tap. (1 mark)
(iii) Find the ratio in which V divides TH, the position of V, and how much farther from T the valve is than the tap. (2 marks)
Hint
For (iii), take the ratio $k : 1$ and use $x = y$ for the dividing point.
Answer
(i) $T H = 1 5^{2} + 1 0^{2} = 325 = 513 \approx 18.03$ m. (1)
(ii) Tap $= (\frac{13 + 4 ( - 2 )}{5}, \frac{- 2 + 4 ( 8 )}{5}) = (1, 6)$ . (1)
(iii) V $= (\frac{13 k - 2}{k + 1}, \frac{- 2 k + 8}{k + 1})$ with $13 k - 2 = - 2 k + 8$ , so $k = \frac{2}{3}$ : ratio $2 : 3$ , and $V = (4, 4)$ . The tap is $\frac{1}{5}$ and the valve $\frac{2}{5}$ of the way along TH, so the valve is $\frac{1}{5} \times 513 = 13 \approx 3.61$ m farther from T. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the section formula for internal division in the ratio $m_{1} : m_{2}$ .Show answer
$(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$
2.Which ratio gives the midpoint?Show answer
$1 : 1$
3.If $A P = \frac{2}{7} A B$ , what is $A P : P B$ ?Show answer
$2 : 5$
4.Which result is used to prove the section formula?Show answer
Similarity of triangles (AA criterion).
5.Point dividing $(0, 0)$ and $(6, 9)$ in $1 : 2$ ?Show answer
$(2, 3)$
6.For internal division, where must P lie?Show answer
Between A and B, so each coordinate lies between those of A and B.

Section Formula (Internal Division): frequently asked questions

Why does m1 multiply the coordinates of B and not A?

Because the point is a weighted average. If AP is the bigger share, P sits further from A and closer to B, so B must carry the bigger weight. The derivation using similar triangles gives this pairing directly.

How can I check my answer quickly?

Make sure each coordinate lies between those of the two ends. For a fuller check, find AP and PB with the distance formula and confirm they are in the given ratio.

Is external division in the Class 10 syllabus for 2026-27?

No. External division is not in the 2026-27 syllabus, and only internal division is examined.

What if the ratio is not given?

Take the ratio as k to 1 and use the condition in the question, such as the point lying on an axis or on a given line, to find k. A positive k means the division is internal. The worked examples on this page show this for the y-axis, the x-axis and a point with an unknown coordinate.

Which NCERT exercise covers the section formula?

Exercise 7.2 of the NCERT textbook covers the section formula, midpoints and related problems. Practise questions of each kind there along with the board-level ones on this page.