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Coordinate Geometry — CBSE Class 10 Maths Important Questions

14 CBSE Class 10 Maths practice questions on Coordinate Geometry, 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 · 30 September 2026

Last updated 25 October 2026

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Coordinate Geometry — CBSE Class 10 Maths Important Questions

Distance, Section, Midpoint — Sorted

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Quick answer

This set covers the rationalised Coordinate Geometry syllabus: the distance formula (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}√((x₂ - x₁)² + (y₂ - y₁)²) and the section formula (including midpoint and trisection), applied to distances, ratios of division and geometric-shape problems.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Coordinate Geometry

Coordinate Geometry is a formula-driven chapter. Following the rationalised NCERT syllabus, these questions focus on the distance formula and the section formula (midpoint, ratio of division and points of trisection), together with their use in identifying shapes and solving case-based problems.

Distance formula and distance from the originSection formula for internal divisionMidpoint formula and points of trisectionRatio in which a point, axis or line divides a segmentUsing coordinates to identify geometric shapes

Key concepts & formulas

Distance Formula

The distance between P(x1,y1)P(x_1, y_1)P(x₁, y₁) and Q(x2,y2)Q(x_2, y_2)Q(x₂, y₂) is PQ=(x2−x1)2+(y2−y1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}PQ = √((x₂ - x₁)² + (y₂ - y₁)²). In particular, the distance of a point (x,y)(x, y)(x, y) from the origin is x2+y2\sqrt{x^2 + y^2}√(x² + y²).

Section Formula

The point dividing the join of (x1,y1)(x_1, y_1)(x₁, y₁) and (x2,y2)(x_2, y_2)(x₂, y₂) internally in the ratio m1:m2m_1 : m_2m₁ : m₂ is (m1x2+m2x1m1+m2, m1y2+m2y1m1+m2)\left(\dfrac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)((m₁ x₂ + m₂ x₁)/(m₁ + m₂), (m₁ y₂ + m₂ y₁)/(m₁ + m₂)).

Midpoint and Trisection

The midpoint (ratio 1:11 : 11 : 1) is (x1+x22,y1+y22)\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)((x₁ + x₂)/2, (y₁ + y₂)/2). The points of trisection are found using the ratios 1:21 : 21 : 2 and 2:12 : 12 : 1.

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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 distance of the point P(3,4)P(3, 4)P(3, 4) from the origin is:

  1. (a)

    333 units

  2. (b)

    444 units

  3. (c)

    555 units

  4. (d)

    777 units

Show model answer

Answer: (c) 555 units.

Distance from the origin =x2+y2=32+42=9+16=25=5= \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5= √(x² + y²) = √(3² + 4²) = √(9 + 16) = √25 = 5 units.

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Q2MCQEasy1 mark

The midpoint of the line segment joining A(2,3)A(2, 3)A(2, 3) and B(4,7)B(4, 7)B(4, 7) is:

  1. (a)

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

  2. (b)

    (2,5)(2, 5)(2, 5)

  3. (c)

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

  4. (d)

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

Show model answer

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

Midpoint =(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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Q3MCQModerate1 mark

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

  1. (a)

    (−7,0)(-7, 0)(-7, 0)

  2. (b)

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

  3. (c)

    (0,−7)(0, -7)(0, -7)

  4. (d)

    (2,0)(2, 0)(2, 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. Expanding: −4x+25=4x+81⇒−8x=56⇒x=−7-4x + 25 = 4x + 81 \Rightarrow -8x = 56 \Rightarrow x = -7-4x + 25 = 4x + 81 ⇒ -8x = 56 ⇒ x = -7.

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Q4MCQModerate1 mark

The triangle with vertices A(0,6)A(0, 6)A(0, 6), B(−5,3)B(-5, 3)B(-5, 3) and C(3,1)C(3, 1)C(3, 1) is:

  1. (a)

    equilateral

  2. (b)

    scalene

  3. (c)

    right-angled and isosceles

  4. (d)

    obtuse-angled

Show model answer

Answer: (c) right-angled and isosceles.

AB2=(−5−0)2+(3−6)2=25+9=34AB^2 = (-5 - 0)^2 + (3 - 6)^2 = 25 + 9 = 34AB² = (-5 - 0)² + (3 - 6)² = 25 + 9 = 34, AC2=(3−0)2+(1−6)2=9+25=34AC^2 = (3 - 0)^2 + (1 - 6)^2 = 9 + 25 = 34AC² = (3 - 0)² + (1 - 6)² = 9 + 25 = 34 and BC2=(3+5)2+(1−3)2=64+4=68BC^2 = (3 + 5)^2 + (1 - 3)^2 = 64 + 4 = 68BC² = (3 + 5)² + (1 - 3)² = 64 + 4 = 68.

So AB=ACAB = ACAB = AC (isosceles), and AB2+AC2=34+34=68=BC2AB^2 + AC^2 = 34 + 34 = 68 = BC^2AB² + AC² = 34 + 34 = 68 = BC², so the angle at AAA is a right angle.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The point (0,−3)(0, -3)(0, -3) lies on the yyy-axis.

Reason (R): The xxx-coordinate of every point lying on the yyy-axis is zero.

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

Every point on the yyy-axis has xxx-coordinate 000. Since (0,−3)(0, -3)(0, -3) has xxx-coordinate 000, it lies on the yyy-axis, exactly as R explains.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the value(s) of yyy for which the distance between the points P(2,−3)P(2, -3)P(2, -3) and Q(10,y)Q(10, y)Q(10, y) is 101010 units.

Show model answer

PQ2=(10−2)2+(y+3)2=100⇒64+(y+3)2=100⇒(y+3)2=36PQ^2 = (10 - 2)^2 + (y + 3)^2 = 100 \Rightarrow 64 + (y + 3)^2 = 100 \Rightarrow (y + 3)^2 = 36PQ² = (10 - 2)² + (y + 3)² = 100 ⇒ 64 + (y + 3)² = 100 ⇒ (y + 3)² = 36.

⇒y+3=±6⇒y=3\Rightarrow y + 3 = \pm 6 \Rightarrow y = 3⇒ y + 3 = ± 6 ⇒ y = 3 or y=−9y = -9y = -9.

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Q7Very ShortEasy2 marks

Find the coordinates of the point which divides the line segment joining (−1,7)(-1, 7)(-1, 7) and (4,−3)(4, -3)(4, -3) internally in the ratio 2:32 : 32 : 3.

Show model answer

Using the section formula with m1:m2=2:3m_1 : m_2 = 2 : 3m₁ : m₂ = 2 : 3:

x=2(4)+3(−1)2+3=8−35=1x = \dfrac{2(4) + 3(-1)}{2 + 3} = \dfrac{8 - 3}{5} = 1x = (2(4) + 3(-1))/(2 + 3) = (8 - 3)/5 = 1, y=2(−3)+3(7)5=−6+215=3\quad y = \dfrac{2(-3) + 3(7)}{5} = \dfrac{-6 + 21}{5} = 3y = (2(-3) + 3(7))/5 = (-6 + 21)/5 = 3.

The point is (1,3)(1, 3)(1, 3).

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Find the ratio in which the line segment joining A(1,−3)A(1, -3)A(1, -3) and B(4,5)B(4, 5)B(4, 5) is divided by the xxx-axis. Also find the coordinates of the point of division.

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On the xxx-axis, y=0y = 0y = 0. Let the ratio be k:1k : 1k : 1. Using the yyy-coordinate: 0=5k−3k+1⇒5k=3⇒k=350 = \dfrac{5k - 3}{k + 1} \Rightarrow 5k = 3 \Rightarrow k = \dfrac{3}{5}0 = (5k - 3)/(k + 1) ⇒ 5k = 3 ⇒ k = 3/5.

So the ratio is 3:53 : 53 : 5.

x=3(4)+5(1)3+5=12+58=178x = \dfrac{3(4) + 5(1)}{3 + 5} = \dfrac{12 + 5}{8} = \dfrac{17}{8}x = (3(4) + 5(1))/(3 + 5) = (12 + 5)/8 = 17/8.

The point of division is (178, 0)\left(\dfrac{17}{8},\ 0\right)(17/8, 0).

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Q9Short AnswerModerate3 marks

Find the coordinates of the points of trisection of the line segment joining A(4,−1)A(4, -1)A(4, -1) and B(−2,−3)B(-2, -3)B(-2, -3). (NCERT)

Show model answer

The points of trisection PPP and QQQ divide ABABAB in the ratios 1:21 : 21 : 2 and 2:12 : 12 : 1.

PPP (ratio 1:21 : 21 : 2): x=1(−2)+2(4)3=63=2x = \dfrac{1(-2) + 2(4)}{3} = \dfrac{6}{3} = 2x = (1(-2) + 2(4))/3 = 6/3 = 2,  y=1(−3)+2(−1)3=−53\ y = \dfrac{1(-3) + 2(-1)}{3} = \dfrac{-5}{3}y = (1(-3) + 2(-1))/3 = -5/3. So P(2,−53)P\left(2, -\dfrac{5}{3}\right)P(2, -5/3).

QQQ (ratio 2:12 : 12 : 1): x=2(−2)+1(4)3=0x = \dfrac{2(-2) + 1(4)}{3} = 0x = (2(-2) + 1(4))/3 = 0,  y=2(−3)+1(−1)3=−73\ y = \dfrac{2(-3) + 1(-1)}{3} = \dfrac{-7}{3}y = (2(-3) + 1(-1))/3 = -7/3. So Q(0,−73)Q\left(0, -\dfrac{7}{3}\right)Q(0, -7/3).

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Q10Short AnswerHOTS3 marks

Show that the points A(1,7)A(1, 7)A(1, 7), B(4,2)B(4, 2)B(4, 2), C(−1,−1)C(-1, -1)C(-1, -1) and D(−4,4)D(-4, 4)D(-4, 4) are the vertices of a square.

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AB=(4−1)2+(2−7)2=9+25=34AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{9 + 25} = \sqrt{34}AB = √((4 - 1)² + (2 - 7)²) = √(9 + 25) = √34.

BC=(−1−4)2+(−1−2)2=25+9=34BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{25 + 9} = \sqrt{34}BC = √((-1 - 4)² + (-1 - 2)²) = √(25 + 9) = √34.

CD=(−4+1)2+(4+1)2=9+25=34CD = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{9 + 25} = \sqrt{34}CD = √((-4 + 1)² + (4 + 1)²) = √(9 + 25) = √34.

DA=(1+4)2+(7−4)2=25+9=34DA = \sqrt{(1 + 4)^2 + (7 - 4)^2} = \sqrt{25 + 9} = \sqrt{34}DA = √((1 + 4)² + (7 - 4)²) = √(25 + 9) = √34.

All four sides are equal. The diagonals: AC=(−1−1)2+(−1−7)2=4+64=68AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{4 + 64} = \sqrt{68}AC = √((-1 - 1)² + (-1 - 7)²) = √(4 + 64) = √68 and BD=(−4−4)2+(4−2)2=64+4=68BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{64 + 4} = \sqrt{68}BD = √((-4 - 4)² + (4 - 2)²) = √(64 + 4) = √68.

Since all four sides are equal and the two diagonals are equal, ABCDABCDABCD is a square.

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Q11Short AnswerModerate3 marks

Using the distance formula, show that the points A(1,−1)A(1, -1)A(1, -1), B(2,1)B(2, 1)B(2, 1) and C(4,5)C(4, 5)C(4, 5) are collinear.

Show model answer

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

BC=(4−2)2+(5−1)2=4+16=20=25BC = \sqrt{(4 - 2)^2 + (5 - 1)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}BC = √((4 - 2)² + (5 - 1)²) = √(4 + 16) = √20 = 2√5.

AC=(4−1)2+(5+1)2=9+36=45=35AC = \sqrt{(4 - 1)^2 + (5 + 1)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}AC = √((4 - 1)² + (5 + 1)²) = √(9 + 36) = √45 = 3√5.

Since AB+BC=5+25=35=ACAB + BC = \sqrt{5} + 2\sqrt{5} = 3\sqrt{5} = ACAB + BC = √5 + 2√5 = 3√5 = AC, the point BBB lies on the segment ACACAC, so AAA, BBB and CCC are collinear.

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

If the points A(−2,1)A(-2, 1)A(-2, 1), B(a,0)B(a, 0)B(a, 0), C(4,b)C(4, b)C(4, b) and D(1,2)D(1, 2)D(1, 2) are the vertices of a parallelogram ABCDABCDABCD, find the values of aaa and bbb. Hence find the lengths of the sides ABABAB and BCBCBC.

Show model answer

In a parallelogram the diagonals bisect each other, so the midpoint of ACACAC equals the midpoint of BDBDBD.

Midpoint of AC=(−2+42,1+b2)=(1,1+b2)AC = \left(\dfrac{-2 + 4}{2}, \dfrac{1 + b}{2}\right) = \left(1, \dfrac{1 + b}{2}\right)AC = ((-2 + 4)/2, (1 + b)/2) = (1, (1 + b)/2).

Midpoint of BD=(a+12,0+22)=(a+12,1)BD = \left(\dfrac{a + 1}{2}, \dfrac{0 + 2}{2}\right) = \left(\dfrac{a + 1}{2}, 1\right)BD = ((a + 1)/2, (0 + 2)/2) = ((a + 1)/2, 1).

Equating coordinates: a+12=1⇒a=1\dfrac{a + 1}{2} = 1 \Rightarrow a = 1(a + 1)/2 = 1 ⇒ a = 1; and 1+b2=1⇒b=1\dfrac{1 + b}{2} = 1 \Rightarrow b = 1(1 + b)/2 = 1 ⇒ b = 1.

So the vertices are A(−2,1)A(-2, 1)A(-2, 1), B(1,0)B(1, 0)B(1, 0), C(4,1)C(4, 1)C(4, 1), D(1,2)D(1, 2)D(1, 2).

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

BC=(4−1)2+(1−0)2=9+1=10BC = \sqrt{(4 - 1)^2 + (1 - 0)^2} = \sqrt{9 + 1} = \sqrt{10}BC = √((4 - 1)² + (1 - 0)²) = √(9 + 1) = √10 units.

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Q13Long AnswerModerate5 marks

A(−2,2)A(-2, 2)A(-2, 2) and B(2,8)B(2, 8)B(2, 8) are two points.

(i) Find the coordinates of the points that divide ABABAB into four equal parts.

(ii) Show that the middle one of these points is the midpoint of ABABAB.

(iii) Find the length of ABABAB.

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(i) The three points PPP, QQQ and RRR divide ABABAB in the ratios 1:31 : 31 : 3, 1:11 : 11 : 1 and 3:13 : 13 : 1.

PPP (ratio 1:31 : 31 : 3): (1(2)+3(−2)4,1(8)+3(2)4)=(−1,72)\left(\dfrac{1(2) + 3(-2)}{4}, \dfrac{1(8) + 3(2)}{4}\right) = \left(-1, \dfrac{7}{2}\right)((1(2) + 3(-2))/4, (1(8) + 3(2))/4) = (-1, 7/2).

QQQ (ratio 1:11 : 11 : 1): (−2+22,2+82)=(0,5)\left(\dfrac{-2 + 2}{2}, \dfrac{2 + 8}{2}\right) = (0, 5)((-2 + 2)/2, (2 + 8)/2) = (0, 5).

RRR (ratio 3:13 : 13 : 1): (3(2)+1(−2)4,3(8)+1(2)4)=(1,132)\left(\dfrac{3(2) + 1(-2)}{4}, \dfrac{3(8) + 1(2)}{4}\right) = \left(1, \dfrac{13}{2}\right)((3(2) + 1(-2))/4, (3(8) + 1(2))/4) = (1, 13/2).

(ii) The midpoint of ABABAB is (−2+22,2+82)=(0,5)\left(\dfrac{-2 + 2}{2}, \dfrac{2 + 8}{2}\right) = (0, 5)((-2 + 2)/2, (2 + 8)/2) = (0, 5), which is the point QQQ.

(iii) AB=(2+2)2+(8−2)2=16+36=52=213AB = \sqrt{(2 + 2)^2 + (8 - 2)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}AB = √((2 + 2)² + (8 - 2)²) = √(16 + 36) = √52 = 2√13 units.

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Case-based questions (4 marks)

Q14Case-basedHOTS4 marks

Case Study: In a classroom activity, three friends Ayush, Bhavya and Chetan are seated at points A(3,4)A(3, 4)A(3, 4), B(6,7)B(6, 7)B(6, 7) and C(9,4)C(9, 4)C(9, 4) respectively on a coordinate grid marked on the floor (each unit =1= 1= 1 m). The teacher stands at point TTT, the midpoint of ACACAC.

CBSE Class 10 Maths — Coordinate Geometry: Case Study: In a classroom activity, three friends Ayush, Bhavya and Chetan are seated at points A(3, 4), B(6, 7) and C(9, 4) respectivel

(i) Find the distance between Ayush and Bhavya (ABABAB).

(ii) Find the coordinates of the teacher's position TTT, the midpoint of ACACAC.

(iii) Find the distance BTBTBT, OR find the coordinates of the point that divides ABABAB internally in the ratio 1:21 : 21 : 2.

Show model answer

(i) AB=(6−3)2+(7−4)2=9+9=18=32AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}AB = √((6 - 3)² + (7 - 4)²) = √(9 + 9) = √18 = 3√2 m.

(ii) T=(3+92,4+42)=(6,4)T = \left(\dfrac{3 + 9}{2}, \dfrac{4 + 4}{2}\right) = (6, 4)T = ((3 + 9)/2, (4 + 4)/2) = (6, 4).

(iii) BT=(6−6)2+(7−4)2=0+9=3BT = \sqrt{(6 - 6)^2 + (7 - 4)^2} = \sqrt{0 + 9} = 3BT = √((6 - 6)² + (7 - 4)²) = √(0 + 9) = 3 m.

OR Point dividing ABABAB in the ratio 1:21 : 21 : 2: (1(6)+2(3)3,1(7)+2(4)3)=(123,153)=(4,5)\left(\dfrac{1(6) + 2(3)}{3}, \dfrac{1(7) + 2(4)}{3}\right) = \left(\dfrac{12}{3}, \dfrac{15}{3}\right) = (4, 5)((1(6) + 2(3))/3, (1(7) + 2(4))/3) = (12/3, 15/3) = (4, 5).

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

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Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • What are the important questions on Coordinate Geometry for Class 10?
    The distance formula (finding the distance between two points, or checking if three points are collinear) and the section formula (finding a point that divides a line segment in a given ratio, including the midpoint as a special case) are the two most-tested ideas. A smaller but recurring question type asks for the area of a triangle given its three vertices, using the coordinate geometry area formula.
  • What are the important questions in Coordinate Geometry Class 10?
    Distance-formula questions (an unknown coordinate, equidistant points, proving a square or collinearity) and section-formula questions (a point in a given ratio, the ratio an axis divides a segment in, trisection and four equal parts). This page has 14 with answers, including a classroom seating case study on Ayush, Bhavya and Chetan.
  • Is the area of a triangle formula still in CBSE Class 10 Coordinate Geometry?
    No. The rationalised NCERT chapter covers the distance formula and the section formula (with midpoints and points of trisection) only. Collinearity is therefore shown with distances — AB+BC=ACAB+BC=ACAB+BC=AC — as in one of this page's 3-mark questions, not with the area formula.
  • How do I find the ratio in which the x-axis divides a line segment?
    Any point on the xxx-axis has y=0y=0y=0, so let the ratio be k:1k:1k:1 and set the section-formula yyy-coordinate to zero. For A(1,−3)A(1,-3)A(1,-3) and B(4,5)B(4,5)B(4,5): 5k−3k+1=0\dfrac{5k-3}{k+1}=0(5k-3)/(k+1)=0, so k=35k=\tfrac35k=3/5, a ratio of 3:53:53:5, and the point is (178,0)\left(\tfrac{17}{8},0\right)(17/8,0).
  • How do you prove four points form a square in coordinate geometry?
    Find all four sides and both diagonals with the distance formula. A square needs four equal sides and equal diagonals; equal sides alone give only a rhombus. For A(1,7)A(1,7)A(1,7), B(4,2)B(4,2)B(4,2), C(−1,−1)C(-1,-1)C(-1,-1) and D(−4,4)D(-4,4)D(-4,4) each side is 34\sqrt{34}√34 and each diagonal 68\sqrt{68}√68, so ABCDABCDABCD is a square, one of this page's 3-mark questions.
  • Where can I find Coordinate Geometry Class 10 notes?
    This page's Key concepts & formulas section serves as Coordinate Geometry notes: the distance formula, the section formula, and the midpoint and trisection points as special cases. For the textbook's Exercises 7.1 and 7.2, Classmate AI has a separate NCERT Solutions page for Chapter 7.
  • Where can I get Coordinate Geometry Class 10 extra questions and case studies?
    The page goes past NCERT with a parallelogram whose missing coordinates aaa and bbb must be found, the points dividing A(−2,2)A(-2,2)A(-2,2) to B(2,8)B(2,8)B(2,8) into four equal parts, collinearity of (1,−1)(1,-1)(1,-1), (2,1)(2,1)(2,1) and (4,5)(4,5)(4,5) by distances, and a 4-mark case study on three friends' seats, all with full answers.
  • Where can I practise Coordinate Geometry Class 10 MCQs?
    The opening five questions are objective: the distance of (3,4)(3,4)(3,4) from the origin, a midpoint, the xxx-axis point equidistant from (2,−5)(2,-5)(2,-5) and (−2,9)(-2,9)(-2,9), the type of triangle on three vertices, and an assertion–reason item on points of the yyy-axis. Create Test, free after sign-up, generates timed sets on the chapter.

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