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Distance Formula — ICSE Class 9 Maths Important Questions

13 ICSE Class 9 Maths practice questions on Distance Formula, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
Questions
5
Topics
3
Key concepts
₹0
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Quick answer

The ICSE Class 9 Distance Formula gives the distance between A(x1,y1)A(x_1,y_1)A(x₁,y₁) and B(x2,y2)B(x_2,y_2)B(x₂,y₂) as AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√((x₂-x₁)²+(y₂-y₁)²). Typical questions find lengths of segments, prove figures are isosceles/right-angled/parallelogram, find points equidistant from two given points, and use distance from the origin x2+y2\sqrt{x^2+y^2}√(x²+y²).

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Distance Formula

Within the ICSE Class 9 Maths chapter Distance Formula you calculate the straight-line distance between two points A(x1,y1)A(x_1,y_1)A(x₁,y₁) and B(x2,y2)B(x_2,y_2)B(x₂,y₂) on the Cartesian plane using AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√((x₂-x₁)²+(y₂-y₁)²), which follows from Pythagoras' theorem. You apply it to find side lengths, classify triangles and quadrilaterals, test collinearity, and locate points that are equidistant from given points, including distance from the origin x2+y2\sqrt{x^2+y^2}√(x²+y²).

The distance formula and its Pythagoras originDistance of a point from the originLengths of sides and classifying trianglesProving quadrilateral types (square, rhombus, parallelogram)Equidistant points and simple loci

Key concepts & formulas

Distance formula

The distance between A(x1,y1)A(x_1,y_1)A(x₁,y₁) and B(x2,y2)B(x_2,y_2)B(x₂,y₂) is AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√((x₂-x₁)²+(y₂-y₁)²). It is derived from Pythagoras' theorem on the right triangle formed with the axes.

Distance from origin

The distance of P(x,y)P(x,y)P(x,y) from the origin O(0,0)O(0,0)O(0,0) is OP=x2+y2OP=\sqrt{x^2+y^2}OP=√(x²+y²), a special case with (x1,y1)=(0,0)(x_1,y_1)=(0,0)(x₁,y₁)=(0,0).

Classifying figures

Compute side lengths: all sides equal and diagonals equal is a square; all sides equal is a rhombus; opposite sides equal is a parallelogram. A triangle is isosceles if two sides are equal and right-angled if a2+b2=c2a^2+b^2=c^2a²+b²=c².

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The distance between the points (0,0)(0,0)(0,0) and (3,4)(3,4)(3,4) is:

  1. (a)

    555

  2. (b)

    777

  3. (c)

    7\sqrt7√7

  4. (d)

    252525

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Answer: (a) 555.

OP=32+42=9+16=25=5OP=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5OP=√(3²+4²)=√(9+16)=√25=5.

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Q2MCQEasy1 mark

The distance between A(2,3)A(2,3)A(2,3) and B(2,8)B(2,8)B(2,8) is:

  1. (a)

    555

  2. (b)

    666

  3. (c)

    11\sqrt{11}√11

  4. (d)

    101010

Show model answer

Answer: (a) 555.

The xxx-coordinates are equal, so AB=(2−2)2+(8−3)2=0+25=5AB=\sqrt{(2-2)^2+(8-3)^2}=\sqrt{0+25}=5AB=√((2-2)²+(8-3)²)=√(0+25)=5.

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Q3MCQModerate1 mark

If the distance between (x,0)(x,0)(x,0) and (0,x)(0,x)(0,x) is 8\sqrt{8}√8 units, then xxx equals:

  1. (a)

    111

  2. (b)

    222

  3. (c)

    444

  4. (d)

    2\sqrt2√2

Show model answer

Answer: (b) 222.

Distance =(x−0)2+(0−x)2=2x2=8=\sqrt{(x-0)^2+(0-x)^2}=\sqrt{2x^2}=\sqrt8=√((x-0)²+(0-x)²)=√2x²=√8. So 2x2=8⇒x2=4⇒x=22x^2=8\Rightarrow x^2=4\Rightarrow x=22x²=8⇒ x²=4⇒ x=2 (taking the positive value).

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Q4MCQHOTS1 mark

The point on the xxx-axis equidistant from A(2,−5)A(2,-5)A(2,-5) and B(−2,9)B(-2,9)B(-2,9) is:

  1. (a)

    (−7,0)(-7,0)(-7,0)

  2. (b)

    (−3,0)(-3,0)(-3,0)

  3. (c)

    (3,0)(3,0)(3,0)

  4. (d)

    (7,0)(7,0)(7,0)

Show model answer

Answer: (a) (−7,0)(-7,0)(-7,0).

Let the point be P(x,0)P(x,0)P(x,0). Then PA2=PB2PA^2=PB^2PA²=PB²:
(x−2)2+25=(x+2)2+81(x-2)^2+25=(x+2)^2+81(x-2)²+25=(x+2)²+81.
x2−4x+4+25=x2+4x+4+81⇒−4x+29=4x+85⇒−8x=56⇒x=−7.x^2-4x+4+25=x^2+4x+4+81\Rightarrow -4x+29=4x+85\Rightarrow -8x=56\Rightarrow x=-7.x²-4x+4+25=x²+4x+4+81⇒ -4x+29=4x+85⇒ -8x=56⇒ x=-7. So P(−7,0)P(-7,0)P(-7,0).

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The triangle with vertices A(0,0)A(0,0)A(0,0), B(4,0)B(4,0)B(4,0) and C(0,3)C(0,3)C(0,3) is right-angled.

Reason (R): In any triangle, if the square of the longest side equals the sum of the squares of the other two sides, the triangle is right-angled.

  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) AB=4AB=4AB=4, AC=3AC=3AC=3, BC=16+9=5BC=\sqrt{16+9}=5BC=√(16+9)=5. Since 32+42=9+16=25=523^2+4^2=9+16=25=5^23²+4²=9+16=25=5², the triangle is right-angled (at AAA), and R (the converse of Pythagoras) correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the distance between the points P(−3,2)P(-3,2)P(-3,2) and Q(1,−1)Q(1,-1)Q(1,-1).

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PQ=(1−(−3))2+(−1−2)2=(4)2+(−3)2PQ=\sqrt{(1-(-3))^2+(-1-2)^2}=\sqrt{(4)^2+(-3)^2}PQ=√((1-(-3))²+(-1-2)²)=√((4)²+(-3)²)

=16+9=25=5=\sqrt{16+9}=\sqrt{25}=5=√(16+9)=√25=5 units.

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Q7Very ShortModerate2 marks

Find the value of aaa if the distance of the point (a,3)(a,3)(a,3) from the origin is 555 units.

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Distance from origin =a2+32=5=\sqrt{a^2+3^2}=5=√(a²+3²)=5.

⇒a2+9=25\Rightarrow a^2+9=25⇒ a²+9=25

⇒a2=16\Rightarrow a^2=16⇒ a²=16

⇒a=±4.\Rightarrow a=\pm4.⇒ a=±4.

So a=4a=4a=4 or a=−4a=-4a=-4.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Show that the points A(1,2)A(1,2)A(1,2), B(5,4)B(5,4)B(5,4) and C(3,8)C(3,8)C(3,8) are the vertices of an isosceles triangle.

Show model answer

Find each side length.

AB=(5−1)2+(4−2)2=16+4=20.AB=\sqrt{(5-1)^2+(4-2)^2}=\sqrt{16+4}=\sqrt{20}.AB=√((5-1)²+(4-2)²)=√(16+4)=√20.

BC=(3−5)2+(8−4)2=4+16=20.BC=\sqrt{(3-5)^2+(8-4)^2}=\sqrt{4+16}=\sqrt{20}.BC=√((3-5)²+(8-4)²)=√(4+16)=√20.

CA=(1−3)2+(2−8)2=4+36=40.CA=\sqrt{(1-3)^2+(2-8)^2}=\sqrt{4+36}=\sqrt{40}.CA=√((1-3)²+(2-8)²)=√(4+36)=√40.

Since AB=BC=20AB=BC=\sqrt{20}AB=BC=√20, two sides are equal, so △ABC\triangle ABC△ ABC is isosceles.

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Q9Short AnswerModerate3 marks

Find the point on the yyy-axis which is equidistant from the points A(6,5)A(6,5)A(6,5) and B(−4,3)B(-4,3)B(-4,3).

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Let the required point be P(0,y)P(0,y)P(0,y) on the yyy-axis.

PA=PB⇒PA2=PB2PA=PB\Rightarrow PA^2=PB^2PA=PB⇒ PA²=PB²:
(0−6)2+(y−5)2=(0+4)2+(y−3)2.(0-6)^2+(y-5)^2=(0+4)^2+(y-3)^2.(0-6)²+(y-5)²=(0+4)²+(y-3)².
36+y2−10y+25=16+y2−6y+9.36+y^2-10y+25=16+y^2-6y+9.36+y²-10y+25=16+y²-6y+9.
61−10y=25−6y.61-10y=25-6y.61-10y=25-6y.
61−25=−6y+10y⇒36=4y⇒y=9.61-25=-6y+10y\Rightarrow 36=4y\Rightarrow y=9.61-25=-6y+10y⇒ 36=4y⇒ y=9.

The point is P(0,9)P(0,9)P(0,9).

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Q10Short AnswerHOTS3 marks

Prove that the points A(−2,−1)A(-2,-1)A(-2,-1), B(1,0)B(1,0)B(1,0), C(4,3)C(4,3)C(4,3) and D(1,2)D(1,2)D(1,2) are the vertices of a parallelogram.

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For a parallelogram, opposite sides must be equal.

AB=(1+2)2+(0+1)2=9+1=10.AB=\sqrt{(1+2)^2+(0+1)^2}=\sqrt{9+1}=\sqrt{10}.AB=√((1+2)²+(0+1)²)=√(9+1)=√10.

DC=(4−1)2+(3−2)2=9+1=10.DC=\sqrt{(4-1)^2+(3-2)^2}=\sqrt{9+1}=\sqrt{10}.DC=√((4-1)²+(3-2)²)=√(9+1)=√10.

BC=(4−1)2+(3−0)2=9+9=18.BC=\sqrt{(4-1)^2+(3-0)^2}=\sqrt{9+9}=\sqrt{18}.BC=√((4-1)²+(3-0)²)=√(9+9)=√18.

AD=(1+2)2+(2+1)2=9+9=18.AD=\sqrt{(1+2)^2+(2+1)^2}=\sqrt{9+9}=\sqrt{18}.AD=√((1+2)²+(2+1)²)=√(9+9)=√18.

Since AB=DC=10AB=DC=\sqrt{10}AB=DC=√10 and BC=AD=18BC=AD=\sqrt{18}BC=AD=√18, both pairs of opposite sides are equal, so ABCDABCDABCD is a parallelogram.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Show that the points A(1,1)A(1,1)A(1,1), B(−1,−1)B(-1,-1)B(-1,-1) and C(−3,3)C(-\sqrt3,\sqrt3)C(-√3,√3) are the vertices of an equilateral triangle. Find the length of each side, leaving the answer in surd form.

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Compute all three sides.

AB=(−1−1)2+(−1−1)2=4+4=8=22.AB=\sqrt{(-1-1)^2+(-1-1)^2}=\sqrt{4+4}=\sqrt8=2\sqrt2.AB=√((-1-1)²+(-1-1)²)=√(4+4)=√8=2√2.

BC=(−3+1)2+(3+1)2.BC=\sqrt{(-\sqrt3+1)^2+(\sqrt3+1)^2}.BC=√((-√3+1)²+(√3+1)²).
Expand: (−3+1)2=3−23+1=4−23(-\sqrt3+1)^2=3-2\sqrt3+1=4-2\sqrt3(-√3+1)²=3-2√3+1=4-2√3 and (3+1)2=3+23+1=4+23(\sqrt3+1)^2=3+2\sqrt3+1=4+2\sqrt3(√3+1)²=3+2√3+1=4+2√3.
Sum =8=8=8, so BC=8=22.BC=\sqrt8=2\sqrt2.BC=√8=2√2.

CA=(1+3)2+(1−3)2=(4+23)+(4−23)=8=22.CA=\sqrt{(1+\sqrt3)^2+(1-\sqrt3)^2}=\sqrt{(4+2\sqrt3)+(4-2\sqrt3)}=\sqrt8=2\sqrt2.CA=√((1+√3)²+(1-√3)²)=√(4+2√3)+(4-2√3)=√8=2√2.

Since AB=BC=CA=22AB=BC=CA=2\sqrt2AB=BC=CA=2√2, all three sides are equal, so △ABC\triangle ABC△ ABC is equilateral with each side 222\sqrt22√2 units.

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Q12Long AnswerHOTS5 marks

The vertices of a quadrilateral are A(0,0)A(0,0)A(0,0), B(4,3)B(4,3)B(4,3), C(1,7)C(1,7)C(1,7) and D(−3,4)D(-3,4)D(-3,4), taken in order.

(i) Find the lengths of all four sides.

(ii) Find the lengths of both diagonals.

(iii) Hence prove that ABCDABCDABCD is a square.

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(i) Sides.
AB=(4−0)2+(3−0)2=16+9=25=5.AB=\sqrt{(4-0)^2+(3-0)^2}=\sqrt{16+9}=\sqrt{25}=5.AB=√((4-0)²+(3-0)²)=√(16+9)=√25=5.
BC=(1−4)2+(7−3)2=9+16=25=5.BC=\sqrt{(1-4)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5.BC=√((1-4)²+(7-3)²)=√(9+16)=√25=5.
CD=(−3−1)2+(4−7)2=16+9=25=5.CD=\sqrt{(-3-1)^2+(4-7)^2}=\sqrt{16+9}=\sqrt{25}=5.CD=√((-3-1)²+(4-7)²)=√(16+9)=√25=5.
DA=(0+3)2+(0−4)2=9+16=25=5.DA=\sqrt{(0+3)^2+(0-4)^2}=\sqrt{9+16}=\sqrt{25}=5.DA=√((0+3)²+(0-4)²)=√(9+16)=√25=5.
All four sides equal 555 units.

(ii) Diagonals.
AC=(1−0)2+(7−0)2=1+49=50.AC=\sqrt{(1-0)^2+(7-0)^2}=\sqrt{1+49}=\sqrt{50}.AC=√((1-0)²+(7-0)²)=√(1+49)=√50.
BD=(−3−4)2+(4−3)2=49+1=50.BD=\sqrt{(-3-4)^2+(4-3)^2}=\sqrt{49+1}=\sqrt{50}.BD=√((-3-4)²+(4-3)²)=√(49+1)=√50.
Both diagonals equal 50=52\sqrt{50}=5\sqrt2√50=5√2 units.

(iii) Conclusion. All four sides are equal (555 units), so ABCDABCDABCD is a rhombus. Since the two diagonals are also equal (525\sqrt25√2), the rhombus is a square. Hence ABCDABCDABCD is a square.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

On a coordinate map (1 unit = 1 km), three mobile towers stand at P(0,0)P(0,0)P(0,0), Q(6,0)Q(6,0)Q(6,0) and R(3,4)R(3,4)R(3,4).

(i) Find the distances PQPQPQ, QRQRQR and RPRPRP.

(ii) What type of triangle is PQRPQRPQR?

(iii) A control room is built at M(3,0)M(3,0)M(3,0), the mid-point of PQPQPQ. Find its distance from RRR.

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(i)
PQ=(6−0)2+(0−0)2=36=6PQ=\sqrt{(6-0)^2+(0-0)^2}=\sqrt{36}=6PQ=√((6-0)²+(0-0)²)=√36=6 km.
QR=(3−6)2+(4−0)2=9+16=25=5QR=\sqrt{(3-6)^2+(4-0)^2}=\sqrt{9+16}=\sqrt{25}=5QR=√((3-6)²+(4-0)²)=√(9+16)=√25=5 km.
RP=(3−0)2+(4−0)2=9+16=25=5RP=\sqrt{(3-0)^2+(4-0)^2}=\sqrt{9+16}=\sqrt{25}=5RP=√((3-0)²+(4-0)²)=√(9+16)=√25=5 km.

(ii) Since QR=RP=5QR=RP=5QR=RP=5 km and PQ=6PQ=6PQ=6 km, two sides are equal, so △PQR\triangle PQR△ PQR is isosceles.

(iii) MR=(3−3)2+(4−0)2=16=4MR=\sqrt{(3-3)^2+(4-0)^2}=\sqrt{16}=4MR=√((3-3)²+(4-0)²)=√16=4 km.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • What is the distance formula in ICSE Class 9?
    The distance between A(x1,y1)A(x_1,y_1)A(x₁,y₁) and B(x2,y2)B(x_2,y_2)B(x₂,y₂) is (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}√((x₂-x₁)²+(y₂-y₁)²), which comes from Pythagoras' theorem. From the origin it is x2+y2\sqrt{x^2+y^2}√(x²+y²), so (3,4)(3,4)(3,4) is 555 units from OOO. This page has 13 questions using it, with answers.
  • How do you show three points form an isosceles triangle?
    Find all three side lengths and show two are equal. For A(1,2)A(1,2)A(1,2), B(5,4)B(5,4)B(5,4) and C(3,8)C(3,8)C(3,8): AB=20AB=\sqrt{20}AB=√20, BC=20BC=\sqrt{20}BC=√20 and AC=40AC=\sqrt{40}AC=√40. Two sides are equal, and since 20+20=4020+20=4020+20=40 the triangle is also right-angled at BBB.
  • How do you find a point on the y-axis equidistant from two points?
    Take it as (0,y)(0,y)(0,y) and equate squared distances. For A(6,5)A(6,5)A(6,5) and B(−4,3)B(-4,3)B(-4,3): 36+(y−5)2=16+(y−3)236+(y-5)^2=16+(y-3)^236+(y-5)²=16+(y-3)², which simplifies to 36=4y36=4y36=4y, so y=9y=9y=9 and the point is (0,9)(0,9)(0,9). Squaring first means you never have to handle square roots.
  • How do you prove points are the vertices of an equilateral triangle?
    Show all three distances are equal. For A(1,1)A(1,1)A(1,1), B(−1,−1)B(-1,-1)B(-1,-1) and C(−3,3)C(-\sqrt3,\sqrt3)C(-√3,√3), each squared distance works out to 888, so every side is 222\sqrt22√2. Keep surds exact rather than using decimals, or the sides may look unequal.
  • Is Distance Formula the last chapter of ICSE Class 9 Maths?
    Yes, Chapter 28, after Graphical Solution (Chapter 27) and Co-ordinate Geometry (Chapter 26). In Class 10, Section and Mid-Point Formula continues the coordinate work and has its own important-questions page.
  • How do I identify a quadrilateral using the distance formula?
    Find all four sides and both diagonals. Opposite sides equal means a parallelogram; all sides equal means a rhombus; add equal diagonals and it is a square; opposite sides and diagonals equal gives a rectangle. State the property you used.
  • Can I download Distance Formula ICSE Class 9 questions?
    Yes. Enter your email in the free PDF box on this page to open all 13 distance-formula questions, the mobile-tower case study included, as a printable worksheet with answers.

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