Section and Mid-point Formula ICSE Class 10: Ratio, Centroid and Board Questions

What Section and Mid-point Formula is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Section and Mid-point Formula is in the 2. Algebra (vii) Co-ordinate Geometry (b)(i) Section and Mid-point formula part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

The section formula (internal division)

If $P(x,y)$ divides the join of $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m:n$, then
$x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n}.$

Read it as a cross-multiplication: $m$ (the part next to $A$) goes with $B$'s coordinate, and $n$ (the part next to $B$) goes with $A$'s coordinate. If $m>n$, $P$ is nearer $B$.

A quick check: the answer must lie between the two given x-values and between the two given y-values.

Worked example

Find the point $P$ that divides the join of $A(2,-3)$ and $B(7,2)$ internally in the ratio $2:3$.

Try it first, then show the solutionSolution

1. $m=2$, $n=3$, $m+n=5$.
2. $x=\frac{2(7)+3(2)}{5}=\frac{20}{5}=4$.
3. $y=\frac{2(2)+3(-3)}{5}=\frac{-5}{5}=-1$.
4. Check: 4 lies between 2 and 7, and −1 between −3 and 2.

Answer: P(4, −1)

The mid-point formula

The mid-point is the case $m=n=1$:
$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$

It also works backwards. If $M$ is the mid-point of $AB$ and you know $A$ and $M$, then $B=(2x_M-x_A,2y_M-y_A)$: double the mid-point and subtract the known end.

Worked example

(a) Find the mid-point of $(-3,5)$ and $(7,-1)$. (b) $M(1,4)$ is the mid-point of $AB$, where $A$ is $(-2,3)$. Find $B$.

Try it first, then show the solutionSolution

1. (a) $\left(\frac{-3+7}{2},\frac{5+(-1)}{2}\right)=(2,2)$.
2. (b) $B=(2\times 1-(-2),2\times 4-3)=(4,5)$.

Answer: (a) (2, 2) (b) B(4, 5)

Finding the ratio: the k : 1 method

When the ratio is unknown, let it be $k:1$. Then
$x=\frac{k{x}_2+x_1}{k+1},y=\frac{k{y}_2+y_1}{k+1}.$
Use the coordinate you know to find $k$, then write the ratio in whole numbers (for example $k=\frac{2}{3}$ gives $2:3$). Finally use $k$ in the other coordinate to find any unknown.

If $k$ comes out negative, the point is not between $A$ and $B$; recheck your working, because ICSE questions here use internal division.

Worked example

In what ratio does $P(-1,y)$ divide the join of $A(-3,6)$ and $B(2,-4)$? Find $y$.

Try it first, then show the solutionSolution

1. Let the ratio be $k:1$. Using x: $\frac{2k-3}{k+1}=-1$.
2. $2k-3=-k-1$, so $3k=2$ and $k=\frac{2}{3}$: the ratio is $2:3$.
3. $y=\frac{2(-4)+3(6)}{5}=\frac{10}{5}=2$.

Answer: 2 : 3; y = 2

Where a segment meets an axis

A point on the x-axis has $y=0$; a point on the y-axis has $x=0$. So to find the ratio in which an axis divides $AB$:

• x-axis: put $y=0$ in the k : 1 form.
• y-axis: put $x=0$ in the k : 1 form.

Then find the other coordinate of the point of intersection. Write it as a point, such as $(1,0)$, not as '$x=1$'.

Worked example

Find the ratio in which the x-axis divides the join of $A(3,-4)$ and $B(-2,6)$, and the point of intersection.

Try it first, then show the solutionSolution

1. Let the ratio be $k:1$. On the x-axis, $y=0$: $\frac{6k-4}{k+1}=0$.
2. $6k=4$, $k=\frac{2}{3}$: ratio $2:3$.
3. $x=\frac{2(-2)+3(3)}{5}=\frac{5}{5}=1$.

Answer: 2 : 3; the point (1, 0)

The centroid of a triangle

The medians of a triangle (each joining a vertex to the mid-point of the opposite side) meet at the centroid $G$. With vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$:
$G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).$

The centroid divides each median in the ratio $2:1$ from the vertex. The centroid is the balance point of a uniform triangular sheet.

To find a missing vertex from the centroid, use $x_3=3x_G-x_1-x_2$ (and the same for y).

Worked example

Find the centroid of the triangle with vertices $A(1,4)$, $B(-3,2)$ and $C(5,-3)$.

Try it first, then show the solutionSolution

1. Sum of x: $1-3+5=3$; sum of y: $4+2-3=3$.
2. $G=\left(\frac{3}{3},\frac{3}{3}\right)=(1,1)$.

Answer: G(1, 1)

Trisection points and parallelograms

Trisection points divide a segment into three equal parts. The point nearer $A$ divides $AB$ in $1:2$; the point nearer $B$ divides it in $2:1$.

Parallelograms: the diagonals bisect each other, so the diagonals $AC$ and $BD$ of parallelogram $ABCD$ have the same mid-point. This gives the fourth vertex: $D=(x_A+x_C-x_B,y_A+y_C-y_B)$. Take care with the order: $A$, $B$, $C$, $D$ must be consecutive vertices.

Worked example

Find the points of trisection of the join of $A(-4,1)$ and $B(5,7)$.

Try it first, then show the solutionSolution

1. Nearer $A$ (ratio $1:2$): $\left(\frac{5+2(-4)}{3},\frac{7+2(1)}{3}\right)=(-1,3)$.
2. Nearer $B$ (ratio $2:1$): $\left(\frac{2(5)+(-4)}{3},\frac{2(7)+1}{3}\right)=(2,5)$.

Answer: (−1, 3) and (2, 5)

Maths in real life: Section and Mid-point Formula

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.

Topics in this chapter

• Section Formula (Internal Division)
• Mid-point Formula
• Centroid of a Triangle

Section and Mid-point Formula formula sheet

Section formula (internal, m : n)

$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

m is the part next to A(x1, y1); it multiplies B's coordinates.

Unknown ratio k : 1

$x=\frac{k{x}_2+x_1}{k+1}$

Same form for y. Solve for k, then write the ratio in whole numbers.

Mid-point

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Other end: B = 2M − A.

Centroid

$G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

G divides each median 2 : 1 from the vertex.

Points on the axes

$\text{x-axis: }y=0\text{y-axis: }x=0$

Use these to find the ratio in which an axis divides a segment.

Fourth vertex of a parallelogram ABCD

$D=A+C-B$

Diagonals bisect each other; add and subtract coordinate by coordinate.

How the board asks Section and Mid-point Formula

The question patterns to expect, and the method that works for each.

Tricks to remember Section and Mid-point Formula

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

• Cross over

  m : n — m jumps to the far end (B), n jumps to the far end (A). 'Cross over' the ratio.

  Use it when: Writing the section formula without swapping m and n.

• Between check

  An internal point always lies between the two ends in both x and y.

  Use it when: Catching a swapped ratio at once.

• Double minus

  Other end = double the mid-point minus the known end.

  Use it when: Finding an end-point from the mid-point.

• Add three, divide by three

  Centroid = average of the three vertices; on a median it sits two-thirds of the way from the vertex.

  Use it when: Centroid questions.

Section and Mid-point Formula: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Section and Mid-point Formula

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

• ✗Mistake: Swapping the coordinates of points on the axes

  CISCE examiners' comments, 2024: candidates wrote A(6, 0) and D(0, −3) when the points were A(0, 6) on the y-axis and D(−3, 0) on the x-axis.

  Wrong working

  A on the y-axis, 6 units up: A(6, 0)

  Right working

  A(0, 6): a point on the y-axis has x = 0

  ✓Do this: Say it before writing: 'on the y-axis, x is zero'. Write coordinates inside brackets as (x, y).

  Usually costs: The coordinates mark, and every later part that uses them.

• ✗Mistake: Centroid not written as a point

  CISCE examiners' comments, 2024: some candidates left the centroid as (0/3, 0/3), or wrote x = 0 and y = 0 instead of (0, 0).

  ✓Do this: Simplify fully and give the answer as an ordered pair, e.g. G(0, 0).

  Usually costs: 1 mark.

• ✗Mistake: Misapplying the section formula

  CISCE examiners' comments, 2024: incorrect use of the section formula gave a wrong value of k, and calculation errors gave k = −3 instead of 3.

  Wrong working

  $x=\frac{m{x}_1+n{x}_2}{m+n}$

  Right working

  $x=\frac{m{x}_2+n{x}_1}{m+n}$: m goes with the far end

  ✓Do this: Cross over the ratio, then check that the point lies between the two ends. A negative k for an internal point means an error.

  Usually costs: 2 to 3 marks in a 4-mark part.

• ✗Mistake: Confusing the mid-point and centroid formulas

  The CISCE examiners' report for 2024 lists the mid-point formula and the centroid formula among the concepts in which candidates got confused.

  ✓Do this: Two points: divide by 2. Three vertices: divide by 3.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Leaving the ratio as a fraction

  With the k : 1 method, k = 2/3 is not yet the ratio asked for.

  Wrong working

  Ratio = 2/3

  Right working

  k : 1 = 2/3 : 1 = 2 : 3

  ✓Do this: Always finish with a ratio of whole numbers, then go on to find the point if it is asked.

  Usually costs: 1 mark.

Practice questions on Section and Mid-point Formula

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

1. Q1 · 1 mark · MCQBoard level

   The point dividing the join of $A(0,0)$ and $B(9,6)$ internally in the ratio $1:2$ is:
   (a) $(6,4)$
   (b) $(3,2)$
   (c) $(4.5,3)$
   (d) $(2,3)$

   Hint

   1 part out of 3 from A.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The centroid of the triangle with vertices $(2,3)$, $(4,-1)$ and $(0,4)$ is:
   (a) $(3,3)$
   (b) $(2,2)$
   (c) $(6,6)$
   (d) $(2,3)$

   Hint

   Add and divide by 3.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   The ratio in which the x-axis divides the join of $(2,-3)$ and $(5,6)$ is:
   (a) $2:1$
   (b) $1:2$
   (c) $3:5$
   (d) $1:1$

   Hint

   Put y = 0 in the k : 1 form.

   Answer

   (b): $6k-3=0$ gives $k=\frac{1}{2}$.

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The mid-point of the join of $(a,b)$ and $(-a,-b)$ is the origin.
   Reason (R): The mid-point is found by averaging the x-coordinates and averaging the y-coordinates.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Average a and −a.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   $P$ is a point on the segment joining $A(3,-4)$ and $B(-6,2)$ such that $AP=\frac{2}{3}AB$. Find $P$ and state on which axis it lies.

   Hint

   If AP is two-thirds of AB, then AP : PB = 2 : 1.

   Answer

   $AP:PB=2:1$. $P=\left(\frac{2(-6)+3}{3},\frac{2(2)+(-4)}{3}\right)=(-3,0)$, which lies on the x-axis.

6. Q6 · 4 marks · Short answerHarder board level

   $M(1,-1)$ is the mid-point of $AB$, where $A$ is $(-5,4)$. Find $B$, and then the point that divides $AB$ in the ratio $1:2$.

   Hint

   B = 2M − A.

   Answer

   $B=(2+5,-2-4)=(7,-6)$. Point in $1:2$: $\left(\frac{7+2(-5)}{3},\frac{-6+2(4)}{3}\right)=\left(-1,\frac{2}{3}\right)$.

7. Q7 · 3 marks · Long answerBoard level

   Find the point that divides the join of $A(-3,7)$ and $B(5,-1)$ internally in the ratio $3:1$.

   Hint

   m + n = 4.

   Answer

   $\left(\frac{3(5)+(-3)}{4},\frac{3(-1)+7}{4}\right)=(3,1)$

8. Q8 · 4 marks · Long answerHarder board level

   The point $P(m,6)$ divides the join of $A(-4,3)$ and $B(2,8)$ in the ratio $k:1$. Find the ratio and the value of $m$.

   Hint

   Use the y-coordinate first.

   Answer

   $\frac{8k+3}{k+1}=6$ gives $2k=3$, $k=\frac{3}{2}$: ratio $3:2$. $m=\frac{3(2)+2(-4)}{5}=-\frac{2}{5}$.

9. Q9 · 4 marks · Long answerHOTS / competency

   Triangle $ABC$ has vertices $A(1,-2)$, $B(5,4)$ and $C(-3,1)$. $D$, $E$, $F$ are the mid-points of $BC$, $CA$ and $AB$. Find $D$, $E$, $F$ and show that triangles $ABC$ and $DEF$ have the same centroid.

   Hint

   Work out both centroids.

   Answer

   $D(1,2.5)$, $E(-1,-0.5)$, $F(3,1)$. Centroid of $ABC$: $\left(\frac{3}{3},\frac{3}{3}\right)=(1,1)$. Centroid of $DEF$: $\left(\frac{1-1+3}{3},\frac{2.5-0.5+1}{3}\right)=(1,1)$. They are the same.

10. Q10 · 4 marks · Long answerHarder board level

    $A(1,2)$, $B(4,3)$ and $C(6,6)$ are three consecutive vertices of parallelogram $ABCD$. Find $D$ and the point where the diagonals meet.

    Hint

    Diagonals bisect each other.

    Answer

    $D=A+C-B=(3,5)$. The diagonals meet at the mid-point of $AC$: $(3.5,4)$ (the mid-point of $BD$ is also $(3.5,4)$).

11. Q11 · 4 marks · Short answerHOTS / competency

    The segment joining $A(-2,3)$ and $B(4,-6)$ meets the y-axis at $P$ and the x-axis at $Q$. Find the ratio in which each axis divides $AB$, and show that $P$ and $Q$ are the same point.

    Hint

    Find both points.

    Answer

    y-axis: $4k-2=0$, $k=\frac{1}{2}$, ratio $1:2$, $P=\left(0,\frac{-6+2(3)}{3}\right)=(0,0)$. x-axis: $-6k+3=0$, $k=\frac{1}{2}$, ratio $1:2$, $Q=\left(\frac{4+2(-2)}{3},0\right)=(0,0)$. Both are the origin: $AB$ passes through $O$.

12. Q12 · 5 marks · Long answerHOTS / competency

    In triangle $ABC$, $A$ is $(2,6)$ and the centroid is $G(3,2)$. (a) Find the mid-point $D$ of $BC$, using the fact that $G$ divides $AD$ in the ratio $2:1$. (b) If $B$ is $(1,-3)$, find $C$. (c) Check your answer with the centroid formula.

    Hint

    Let D(p, q) and use the section formula on AD.

    Answer

    (a) $\frac{2p+2}{3}=3$ gives $p=3.5$; $\frac{2q+6}{3}=2$ gives $q=0$: $D(3.5,0)$. (b) $C=2D-B=(6,3)$. (c) $\left(\frac{2+1+6}{3},\frac{6-3+3}{3}\right)=(3,2)$ ✓

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. 1.Section formula for the x-coordinate (ratio m : n)?Show answer

   (m x₂ + n x₁) ÷ (m + n)

2. 2.Mid-point of (x₁, y₁) and (x₂, y₂)?Show answer

   ((x₁ + x₂)/2, (y₁ + y₂)/2)

3. 3.Centroid of a triangle?Show answer

   The average of the three vertices.

4. 4.In what ratio does the centroid divide a median?Show answer

   2 : 1 from the vertex.

5. 5.Condition for a point on the x-axis?Show answer

   y = 0

6. 6.Ratios for the two points of trisection of AB?Show answer

   1 : 2 and 2 : 1

7. 7.M is the mid-point of AB. How do you find B?Show answer

   B = 2M − A, coordinate by coordinate.

8. 8.Is external division in the ICSE 2027 syllabus?Show answer

   No: internal section only.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the section formula without swapping m and n?
• ☐Do you check that an internal point lies between the two ends?
• ☐Can you find an end-point from the mid-point?
• ☐Can you find the ratio in which an axis divides a segment?
• ☐Do you write the ratio in whole numbers?
• ☐Can you find a centroid and a missing vertex?
• ☐Do you write every answer as a point (x, y)?

Section and Mid-point Formula: frequently asked questions