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Section and Mid-Point Formula — ICSE Class 10 Maths Important Questions

14 ICSE Class 10 Maths practice questions on Section and Mid-Point Formula, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Reviewed by Classmate AI Team · 30 September 2026

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Key concepts
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Section and Mid-Point Formula — ICSE Class 10 Maths Important Questions

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Quick answer

Frequently-tested ICSE Section and Mid-point Formula questions use the section formula (mx2+nx1m+n,my2+ny1m+n)\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)(mx_2+nx_1/m+n,my_2+ny_1/m+n), the mid-point formula, and the centroid (x1+x2+x33,y1+y2+y33)\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)(x_1+x_2+x_3/3,y_1+y_2+y_3/3). Finding the ratio in which a line or axis divides a segment, and unknown coordinates, appear every year.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Section and Mid-Point Formula

Within the ICSE Class 10 Maths chapter Section and Mid-point Formula you find the point that divides a line segment in a given ratio, the mid-point of a segment, and the centroid of a triangle. You also work backwards to find unknown coordinates or the ratio of division, including division by the coordinate axes. Given the endpoints (x1,y1)(x_1,y_1)(x_1,y_1) and (x2,y2)(x_2,y_2)(x_2,y_2) of a line segment AB, typical questions ask you to find the coordinates of the point dividing the segment joining the points in a ratio such as AP:PB, calculate that ratio when the dividing point is already known, or find where the line joining the points meets the x-axis or y-axis.

Section formula (internal division)Mid-point formulaCentroid of a triangleRatio of divisionApplications (parallelogram, points on axes)

Key concepts & formulas

Section formula

The point dividing the join of (x1,y1)(x_1,y_1)(x_1,y_1) and (x2,y2)(x_2,y_2)(x_2,y_2) internally in the ratio m:nm:nm:n is (mx2+nx1m+n,my2+ny1m+n)\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)(mx_2+nx_1/m+n,my_2+ny_1/m+n).

Mid-point formula

The mid-point of (x1,y1)(x_1,y_1)(x_1,y_1) and (x2,y2)(x_2,y_2)(x_2,y_2) is (x1+x22,y1+y22)\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)(x_1+x_2/2,y_1+y_2/2) (the case m:n=1:1m:n=1:1m:n=1:1).

Centroid

The centroid of a triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1,y_1),(x_2,y_2),(x_3,y_3)(x_1,y_1),(x_2,y_2),(x_3,y_3) is (x1+x2+x33,y1+y2+y33)\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)(x_1+x_2+x_3/3,y_1+y_2+y_3/3).

Ratio of division

Assume the ratio k:1k:1k:1 and use the section formula; a point on the xxx-axis has y=0y=0y=0, and a point on the yyy-axis has x=0x=0x=0.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The mid-point of the segment joining (2,3)(2,3)(2,3) and (4,7)(4,7)(4,7) is:

  1. (a)

    (3,5)(3,5)(3,5)

  2. (b)

    (6,10)(6,10)(6,10)

  3. (c)

    (1,2)(1,2)(1,2)

  4. (d)

    (3,4)(3,4)(3,4)

Show model answer

Answer: (a) (3,5)(3,5)(3,5).

(2+42,3+72)=(3,5).\left(\dfrac{2+4}{2},\dfrac{3+7}{2}\right)=(3,5).(2+4/2,3+7/2)=(3,5).

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Q2MCQEasy1 mark

The centroid of the triangle with vertices (0,0)(0,0)(0,0), (6,0)(6,0)(6,0) and (0,9)(0,9)(0,9) is:

  1. (a)

    (2,3)(2,3)(2,3)

  2. (b)

    (3,3)(3,3)(3,3)

  3. (c)

    (2,2)(2,2)(2,2)

  4. (d)

    (6,9)(6,9)(6,9)

Show model answer

Answer: (a) (2,3)(2,3)(2,3).

(0+6+03,0+0+93)=(2,3).\left(\dfrac{0+6+0}{3},\dfrac{0+0+9}{3}\right)=(2,3).(0+6+0/3,0+0+9/3)=(2,3).

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Q3MCQEasy1 mark

The point dividing the join of (1,2)(1,2)(1,2) and (7,5)(7,5)(7,5) internally in the ratio 1:21:21:2 is:

  1. (a)

    (3,3)(3,3)(3,3)

  2. (b)

    (5,4)(5,4)(5,4)

  3. (c)

    (4,3.5)(4,3.5)(4,3.5)

  4. (d)

    (3,4)(3,4)(3,4)

Show model answer

Answer: (a) (3,3)(3,3)(3,3).

(1⋅7+2⋅13,1⋅5+2⋅23)=(93,93)=(3,3).\left(\dfrac{1\cdot7+2\cdot1}{3},\dfrac{1\cdot5+2\cdot2}{3}\right)=\left(\dfrac{9}{3},\dfrac{9}{3}\right)=(3,3).(1·7+2·1/3,1·5+2·2/3)=(9/3,9/3)=(3,3).

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Q4MCQHOTS1 mark

The ratio in which the xxx-axis divides the join of (2,−3)(2,-3)(2,-3) and (5,6)(5,6)(5,6) is:

  1. (a)

    1:21:21:2

  2. (b)

    2:12:12:1

  3. (c)

    1:31:31:3

  4. (d)

    3:13:13:1

Show model answer

Answer: (a) 1:21:21:2.

Let the ratio be k:1k:1k:1. On the xxx-axis y=0y=0y=0: 6k+(−3)k+1=0⇒6k=3⇒k=12\dfrac{6k+(-3)}{k+1}=0\Rightarrow 6k=3\Rightarrow k=\tfrac126k+(-3)/k+1=0 6k=3 k=12, i.e. 1:21:21:2.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The mid-point of the segment joining (−1,4)(-1,4)(-1,4) and (3,−2)(3,-2)(3,-2) is (1,1)(1,1)(1,1).

Reason (R): The mid-point of (x1,y1)(x_1,y_1)(x_1,y_1) and (x2,y2)(x_2,y_2)(x_2,y_2) is (x1+x22,y1+y22)\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)(x_1+x_2/2,y_1+y_2/2).

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) (−1+32,4+(−2)2)=(1,1)\left(\dfrac{-1+3}{2},\dfrac{4+(-2)}{2}\right)=(1,1)(-1+3/2,4+(-2)/2)=(1,1), so A is true and R is the formula that produces it.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

M(2,−1)M(2,-1)M(2,-1) is the mid-point of ABABAB, where AAA is (−3,4)(-3,4)(-3,4). Find the coordinates of BBB.

Show model answer

Let B=(x,y)B=(x,y)B=(x,y).

−3+x2=2⇒x=7\dfrac{-3+x}{2}=2\Rightarrow x=7-3+x/2=2 x=7 and 4+y2=−1⇒y=−6.\dfrac{4+y}{2}=-1\Rightarrow y=-6.4+y/2=-1 y=-6.

So B=(7,−6).B=(7,-6).B=(7,-6).

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Q7Very ShortModerate2 marks

Find the ratio in which the yyy-axis divides the join of A(−2,5)A(-2,5)A(-2,5) and B(6,−3)B(6,-3)B(6,-3). Also find the point of intersection.

Show model answer

Let the ratio be k:1k:1k:1. On the yyy-axis x=0x=0x=0:

6k+(−2)k+1=0⇒6k=2⇒k=13\dfrac{6k+(-2)}{k+1}=0\Rightarrow 6k=2\Rightarrow k=\dfrac{1}{3}6k+(-2)/k+1=0 6k=2 k=1/3, so the ratio is 1:31:31:3.

y=1⋅(−3)+3⋅51+3=124=3y=\dfrac{1\cdot(-3)+3\cdot5}{1+3}=\dfrac{12}{4}=3y=1·(-3)+3·5/1+3=12/4=3, so the point is (0,3)(0,3)(0,3).

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

The mid-point of the segment joining (2a,4)(2a,4)(2a,4) and (−2,3b)(-2,3b)(-2,3b) is (1,5)(1,5)(1,5). Find the values of aaa and bbb.

Show model answer

xxx-coordinate: 2a+(−2)2=1⇒2a−2=2⇒a=2.\dfrac{2a+(-2)}{2}=1\Rightarrow 2a-2=2\Rightarrow a=2.2a+(-2)/2=1 2a-2=2 a=2.

yyy-coordinate: 4+3b2=5⇒4+3b=10⇒3b=6⇒b=2.\dfrac{4+3b}{2}=5\Rightarrow 4+3b=10\Rightarrow 3b=6\Rightarrow b=2.4+3b/2=5 4+3b=10 3b=6 b=2.

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Q9Short AnswerModerate3 marks

Find the ratio in which the point (−4,6)(-4,6)(-4,6) divides the join of A(−6,10)A(-6,10)A(-6,10) and B(3,−8)B(3,-8)B(3,-8).

Show model answer

Let the ratio be k:1k:1k:1. Using the xxx-coordinate: 3k+(−6)k+1=−4⇒3k−6=−4k−4⇒7k=2⇒k=27.\dfrac{3k+(-6)}{k+1}=-4\Rightarrow 3k-6=-4k-4\Rightarrow 7k=2\Rightarrow k=\dfrac{2}{7}.3k+(-6)/k+1=-4 3k-6=-4k-4 7k=2 k=2/7.

So the ratio is 2:72:72:7. (Check with yyy: −8(2)+10(7)2+7=−16+709=6\dfrac{-8(2)+10(7)}{2+7}=\dfrac{-16+70}{9}=6-8(2)+10(7)/2+7=-16+70/9=6 correct.)

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Q10Short AnswerHOTS3 marks

Two vertices of a triangle are (1,2)(1,2)(1,2) and (3,5)(3,5)(3,5) and its centroid is (3,3)(3,3)(3,3). Find the third vertex.

Show model answer

Let the third vertex be (x,y)(x,y)(x,y).

1+3+x3=3⇒4+x=9⇒x=5.\dfrac{1+3+x}{3}=3\Rightarrow 4+x=9\Rightarrow x=5.1+3+x/3=3 4+x=9 x=5.

2+5+y3=3⇒7+y=9⇒y=2.\dfrac{2+5+y}{3}=3\Rightarrow 7+y=9\Rightarrow y=2.2+5+y/3=3 7+y=9 y=2.

The third vertex is (5,2).(5,2).(5,2).

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Q11Short AnswerModerate3 marks

The point P(m,6)P(m,6)P(m,6) divides the join of A(−4,3)A(-4,3)A(-4,3) and B(6,8)B(6,8)B(6,8). Find the ratio in which PPP divides ABABAB, and the value of mmm.

Show model answer

Let the ratio be k:1k:1k:1. Using the yyy-coordinate:

8k+3k+1=6⇒8k+3=6k+6⇒k=32\dfrac{8k+3}{k+1}=6\Rightarrow 8k+3=6k+6\Rightarrow k=\dfrac{3}{2}8k+3/k+1=6 8k+3=6k+6 k=3/2, so the ratio is 3:23:23:2.

m=3⋅6+2⋅(−4)3+2=18−85=2.m=\dfrac{3\cdot6+2\cdot(-4)}{3+2}=\dfrac{18-8}{5}=2.m=3·6+2·(-4)/3+2=18-8/5=2.

(Check with yyy: 3⋅8+2⋅35=305=6\dfrac{3\cdot8+2\cdot3}{5}=\dfrac{30}{5}=63·8+2·3/5=30/5=6 correct.)

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

Points A(−3,4)A(-3,4)A(-3,4) and B(9,−2)B(9,-2)B(9,-2) are given. Find (i) the mid-point MMM of ABABAB, (ii) the point PPP that divides ABABAB in the ratio 1:21:21:2, and (iii) the point QQQ that divides ABABAB in the ratio 2:12:12:1.

Show model answer

(i) M=(−3+92,4+(−2)2)=(3,1).M=\left(\dfrac{-3+9}{2},\dfrac{4+(-2)}{2}\right)=(3,1).M=(-3+9/2,4+(-2)/2)=(3,1).

(ii) P=(1⋅9+2⋅(−3)3,1⋅(−2)+2⋅43)=(33,63)=(1,2).P=\left(\dfrac{1\cdot9+2\cdot(-3)}{3},\dfrac{1\cdot(-2)+2\cdot4}{3}\right)=\left(\dfrac{3}{3},\dfrac{6}{3}\right)=(1,2).P=(1·9+2·(-3)/3,1·(-2)+2·4/3)=(3/3,6/3)=(1,2).

(iii) Q=(2⋅9+1⋅(−3)3,2⋅(−2)+1⋅43)=(153,03)=(5,0).Q=\left(\dfrac{2\cdot9+1\cdot(-3)}{3},\dfrac{2\cdot(-2)+1\cdot4}{3}\right)=\left(\dfrac{15}{3},\dfrac{0}{3}\right)=(5,0).Q=(2·9+1·(-3)/3,2·(-2)+1·4/3)=(15/3,0/3)=(5,0).

ICSE Class 10 Maths — Section and Mid-Point Formula: Points A(-3,4) and B(9,-2) are given. Find (i) the mid-point M of AB, (ii) the point P that divides AB in the ratio 1:2, and (i
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Q13Long AnswerHOTS5 marks

Three vertices of a parallelogram ABCDABCDABCD taken in order are A(1,2)A(1,2)A(1,2), B(4,3)B(4,3)B(4,3) and C(6,6)C(6,6)C(6,6). (i) Using the fact that the diagonals of a parallelogram bisect each other, find the coordinates of DDD. (ii) Find the coordinates of the point where the diagonals intersect.

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(i) In parallelogram ABCDABCDABCD the diagonals ACACAC and BDBDBD bisect each other, so their mid-points coincide.

Mid-point of AC=(1+62,2+62)=(72,4).AC=\left(\dfrac{1+6}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{7}{2},4\right).AC=(1+6/2,2+6/2)=(7/2,4).

Let D=(x,y)D=(x,y)D=(x,y). Mid-point of BD=(4+x2,3+y2).BD=\left(\dfrac{4+x}{2},\dfrac{3+y}{2}\right).BD=(4+x/2,3+y/2).

Equating: 4+x2=72⇒x=3\dfrac{4+x}{2}=\dfrac{7}{2}\Rightarrow x=34+x/2=7/2 x=3 and 3+y2=4⇒y=5.\dfrac{3+y}{2}=4\Rightarrow y=5.3+y/2=4 y=5. So D=(3,5).D=(3,5).D=(3,5).

(ii) The diagonals meet at their common mid-point (72,4)\left(\dfrac{7}{2},4\right)(7/2,4), i.e. (3.5, 4).(3.5,\,4).(3.5,\,4).

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Case-based questions (4 marks)

Q14Case-basedModerate4 marks

On a town map drawn with coordinate axes, the library is at A(2,1)A(2,1)A(2,1) and the stadium is at B(10,7)B(10,7)B(10,7).

(i) A bus stop is built at the mid-point of ABABAB. Find its coordinates.

(ii) A shop stands at the point dividing AAA to BBB in the ratio 1:31:31:3. Find its coordinates.

(iii) A cafe stands at the point dividing AAA to BBB in the ratio 3:13:13:1. Find its coordinates.

(iv) Show that the bus stop is also the mid-point of the shop and the cafe.

Show model answer

(i) Bus stop =(2+102,1+72)=(6,4).=\left(\dfrac{2+10}{2},\dfrac{1+7}{2}\right)=(6,4).=(2+10/2,1+7/2)=(6,4).

(ii) Shop =(1⋅10+3⋅24,1⋅7+3⋅14)=(164,104)=(4, 2.5).=\left(\dfrac{1\cdot10+3\cdot2}{4},\dfrac{1\cdot7+3\cdot1}{4}\right)=\left(\dfrac{16}{4},\dfrac{10}{4}\right)=(4,\,2.5).=(1·10+3·2/4,1·7+3·1/4)=(16/4,10/4)=(4,\,2.5).

(iii) Cafe =(3⋅10+1⋅24,3⋅7+1⋅14)=(324,224)=(8, 5.5).=\left(\dfrac{3\cdot10+1\cdot2}{4},\dfrac{3\cdot7+1\cdot1}{4}\right)=\left(\dfrac{32}{4},\dfrac{22}{4}\right)=(8,\,5.5).=(3·10+1·2/4,3·7+1·1/4)=(32/4,22/4)=(8,\,5.5).

(iv) Mid-point of the shop (4, 2.5)(4,\,2.5)(4,\,2.5) and the cafe (8, 5.5)(8,\,5.5)(8,\,5.5) =(4+82,2.5+5.52)=(6,4)=\left(\dfrac{4+8}{2},\dfrac{2.5+5.5}{2}\right)=(6,4)=(4+8/2,2.5+5.5/2)=(6,4), which is the bus stop.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Section and Mid-Point Formula questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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