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Distance Formula — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Distance Formula, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

The ICSE Class 9 Distance Formula gives the distance between A(x1,y1)A(x_1,y_1)A(x_1,y_1) and B(x2,y2)B(x_2,y_2)B(x_2,y_2) as AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√(x_2-x_1)^2+(y_2-y_1)^2. High-yield 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}√x^2+y^2.

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

About Distance Formula

In 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(x_1,y_1) and B(x2,y2)B(x_2,y_2)B(x_2,y_2) on the Cartesian plane using AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√(x_2-x_1)^2+(y_2-y_1)^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}√x^2+y^2.

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(x_1,y_1) and B(x2,y2)B(x_2,y_2)B(x_2,y_2) is AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}AB=√(x_2-x_1)^2+(y_2-y_1)^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=√x^2+y^2, a special case with (x1,y1)=(0,0)(x_1,y_1)=(0,0)(x_1,y_1)=(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^2a^2+b^2=c^2.

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Important questions with answers

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

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

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\sqrt77

  4. (d)

    252525

Show model answer

Answer: (a) 555.

OP=32+42=9+16=25=5OP=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5OP=√3^2+4^2=√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\sqrt22

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=√2x^2=8. So 2x2=8⇒x2=4⇒x=22x^2=8\Rightarrow x^2=4\Rightarrow x=22x^2=8 x^2=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)

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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^2PA^2=PB^2:
(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.x^2-4x+4+25=x^2+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=5^23^2+4^2=9+16=25=5^2, 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=√a^2+3^2=5.

⇒a2+9=25\Rightarrow a^2+9=25a^2+9=25

⇒a2=16\Rightarrow a^2=16a^2=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 ABCABC 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).

Show model answer

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 PA^2=PB^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.(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+y^2-10y+25=16+y^2-6y+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.

Show model answer

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=22.

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=3-23+1=4-23 and (3+1)2=3+23+1=4+23(\sqrt3+1)^2=3+2\sqrt3+1=4+2\sqrt3(3+1)^2=3+23+1=4+23.
Sum =8=8=8, so BC=8=22.BC=\sqrt8=2\sqrt2.BC=8=22.

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+23)+(4-23)=8=22.

Since AB=BC=CA=22AB=BC=CA=2\sqrt2AB=BC=CA=22, all three sides are equal, so △ABC\triangle ABCABC is equilateral with each side 222\sqrt222 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=52 units.

(iii) Conclusion. All four sides are equal (555 units), so ABCDABCDABCD is a rhombus. Since the two diagonals are also equal (525\sqrt252), 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). A control room must be built.

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

(ii) What type of triangle is PQRPQRPQR?

(iii) Find the distance of RRR from the origin.

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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 PQRPQR is isosceles.

(iii) RRR from the origin P(0,0)P(0,0)P(0,0): OR=32+42=25=5OR=\sqrt{3^2+4^2}=\sqrt{25}=5OR=√3^2+4^2=√25=5 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

  • Are these Distance Formula important questions free?
    Yes. All 13 ICSE Class 9 Maths important questions for Distance Formula are free, with full model answers and no login required.
  • Do these Distance Formula questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Distance Formula important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Distance Formula?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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