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About Coordinate Geometry — Class 10 CBSE

Apply distance formula, section formula, and area of a triangle in the coordinate plane. This topic is part of the CBSE Class 10 mathematics syllabus (chapter: Chapter 7). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Coordinate Geometry — solved examples for Class 10 CBSE

Example 1easy

Find the distance between the points (0,0)(0, 0) and (3,4)(3, 4).
  1. A)5
  2. B)7
  3. C)6
  4. D)4

Step-by-step solution

  1. d=(3−0)2+(4−0)2=9+16d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16}
    =25=5= \sqrt{25} = 5

Answer: 5

Example 2medium

Find the area of the triangle with vertices (1,2)(1, 2), (4,6)(4, 6), and (5,2)(5, 2).
  1. A)8 sq. units
  2. B)6 sq. units
  3. C)10 sq. units
  4. D)12 sq. units

Step-by-step solution

  1. =12∣1(6−2)+4(2−2)+5(2−6)∣= \frac{1}{2}|1(6-2)+4(2-2)+5(2-6)|
  2. =12∣4+0−20∣=12×16= \frac{1}{2}|4+0-20| = \frac{1}{2} \times 16
    =8 sq. units= 8 \text{ sq. units}

Answer: 8 sq. units

Example 3hard

Find the area of the triangle formed by the points (2,3)(2, 3), (−1,0)(-1, 0), and (2,−4)(2, -4).
  1. A)212\frac{21}{2} sq. units
  2. B)10 sq. units
  3. C)12 sq. units
  4. D)7 sq. units

Step-by-step solution

  1. =12∣2(0−(−4))+(−1)(−4−3)+2(3−0)∣= \frac{1}{2}|2(0-(-4))+(-1)(-4-3)+2(3-0)|
  2. =12∣8+7+6∣=212= \frac{1}{2}|8+7+6| = \frac{21}{2}

Answer: 212\frac{21}{2} sq. units

Practice questions on Coordinate Geometry

  1. Q1.easy

    The midpoint of the line segment joining (2,4)(2, 4) and (6,8)(6, 8) is:
    1. A)(4,6)(4, 6)
    2. B)(3,5)(3, 5)
    3. C)(8,12)(8, 12)
    4. D)(2,2)(2, 2)
    Show answer

    Answer: (4,6)(4, 6)

    Hint: Midpoint = (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right).

  2. Q2.easy

    The distance of the point (5,12)(5, 12) from the origin is:
    1. A)13
    2. B)17
    3. C)7
    4. D)15
    Show answer

    Answer: 13

    Hint: Distance from origin = x2+y2\sqrt{x^2 + y^2}.

  3. Q3.easy

    The coordinates of the point which divides the line segment joining (1,3)(1, 3) and (5,7)(5, 7) in the ratio 1:11:1 are:
    1. A)(3,5)(3, 5)
    2. B)(2,4)(2, 4)
    3. C)(4,6)(4, 6)
    4. D)(6,10)(6, 10)
    Show answer

    Answer: (3,5)(3, 5)

    Hint: Ratio 1:11:1 means the midpoint.

  4. Q4.medium

    Show that the points (1,5)(1, 5), (2,3)(2, 3), and (−2,11)(-2, 11) are collinear by finding the area.
    1. A)Area = 0
    2. B)Area = 1
    3. C)Area = 2
    4. D)Area = 5
    Show answer

    Answer: Area = 0

    Hint: If three points are collinear, the area of the triangle they form is 0.

  5. Q5.medium

    Find the ratio in which the point (3,4)(3, 4) divides the segment joining (1,2)(1, 2) and (7,8)(7, 8).
    1. A)1:21:2
    2. B)2:12:1
    3. C)1:31:3
    4. D)3:13:1
    Show answer

    Answer: 1:21:2

    Hint: Let the ratio be k:1k:1. Use section formula and solve for kk.

  6. Q6.medium

    Find the distance between (a,b)(a, b) and (−a,−b)(-a, -b).
    1. A)2a2+b22\sqrt{a^2+b^2}
    2. B)a2+b2\sqrt{a^2+b^2}
    3. C)2(a+b)2(a+b)
    4. D)a+ba+b
    Show answer

    Answer: 2a2+b22\sqrt{a^2+b^2}

    Hint: Differences: x2−x1=−2ax_2 - x_1 = -2a and y2−y1=−2by_2 - y_1 = -2b.

  7. Q7.hard

    Find the value of kk for which the points (7,−2)(7, -2), (5,1)(5, 1), and (3,k)(3, k) are collinear.
    1. A)4
    2. B)3
    3. C)2
    4. D)5
    Show answer

    Answer: 4

    Hint: Set the area of the triangle = 0.

  8. Q8.hard

    A point PP divides the line segment joining A(1,−5)A(1, -5) and B(−4,5)B(-4, 5) externally in ratio 2:32:3. Find the coordinates of PP.
    1. A)(11,−25)(11, -25)
    2. B)(11,25)(11, 25)
    3. C)(−11,25)(-11, 25)
    4. D)(−2,−1)(-2, -1)
    Show answer

    Answer: (11,−25)(11, -25)

    Hint: For external division, use (mx2−nx1m−n,my2−ny1m−n)\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right).

  9. Q9.hard

    Show that the quadrilateral with vertices A(3,0)A(3, 0), B(6,1)B(6, 1), C(7,4)C(7, 4), and D(4,3)D(4, 3) is a rhombus. What is the side length?
    1. A)10\sqrt{10}
    2. B)13\sqrt{13}
    3. C)5\sqrt{5}
    4. D)8\sqrt{8}
    Show answer

    Answer: 10\sqrt{10}

    Hint: Find the length of each side. If all are equal, it's a rhombus.

These are 9 of the 60 questions available for Coordinate Geometry. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.