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Class 10 Coordinate Geometry: 8 practice questions with answers, step by step

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These are 8 practice questions on NCERT Class 10 Maths Chapter 7, Coordinate Geometry, each with a worked answer and the textbook page it comes from. Four use the distance formula (a distance, a word problem, three points in a line and an equidistant point), and four use the section formula and the mid-point (a point dividing a segment, a mid-point, the ratio cut by an axis and a parallelogram). Try each one on paper before you read the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

The formulas you need

Facts from the NCERT textbook (Mathematics, Textbook for Class X, Reprint 2026-27, Chapter 7, printed pages 99 to 112):

The distance formula

Question 1. Find the distance between the points (3, 2) and (7, 5).

Answer: 5 units. PQ = √((7 − 3)² + (5 − 2)²) = √(16 + 9) = √25 = 5 units. (Distance formula, page 102.)

Question 2. A village B is 24 km east and 7 km north of a village A. How far apart are the two villages?

Answer: 25 km. Put A at the origin O(0, 0); then B is the point (24, 7). The distance is √(24² + 7²) = √(576 + 49) = √625 = 25 km. (The textbook opens the chapter with the same kind of question about two towns, page 100; distance from the origin, page 102.)

Question 3. Do the points A(1, 1), B(3, 5) and C(5, 9) lie on one straight line?

Answer: Yes, they are collinear. AB = √(2² + 4²) = √20 = 2√5, BC = √(2² + 4²) = √20 = 2√5, and AC = √(4² + 8²) = √80 = 4√5. Since AB + BC = 2√5 + 2√5 = 4√5 = AC, the three points lie on a line. (The textbook checks three desks in a classroom the same way, Example 3, pages 103-104.)

Question 4. Find the point on the y-axis that is equidistant from (4, 2) and (−2, 6).

Answer: (0, 5/2). A point on the y-axis is (0, y). Equal distances means equal squared distances: (4 − 0)² + (2 − y)² = (−2 − 0)² + (6 − y)². So 16 + 4 − 4y + y² = 4 + 36 − 12y + y², which gives 20 − 4y = 40 − 12y, so 8y = 20 and y = 5/2. The point is (0, 5/2). (Same method as Example 5, page 104.)

The section formula and the mid-point

Question 5. Find the point that divides the line segment joining (2, −3) and (8, 6) in the ratio 1 : 2.

Answer: (4, 0). With m₁ = 1, m₂ = 2: x = (1 × 8 + 2 × 2)/3 = 12/3 = 4 and y = (1 × 6 + 2 × (−3))/3 = 0/3 = 0. The point is (4, 0), which happens to lie on the x-axis. (Section formula, page 107.)

Question 6. Find the mid-point of the line segment joining (−3, 4) and (7, −2).

Answer: (2, 1). Mid-point = ((−3 + 7)/2, (4 + (−2))/2) = (4/2, 2/2) = (2, 1). (Mid-point, page 107.)

Question 7. In what ratio does the x-axis divide the line segment joining (3, −4) and (−2, 6)? Where does it cut the segment?

Answer: 2 : 3, at (1, 0). Let the ratio be k : 1. The dividing point is ((−2k + 3)/(k + 1), (6k − 4)/(k + 1)). It lies on the x-axis, so its y-coordinate is 0: 6k − 4 = 0, k = 2/3. The ratio is 2/3 : 1, that is 2 : 3. Its x-coordinate is (−2 × 2/3 + 3)/(2/3 + 1) = (5/3)/(5/3) = 1, so the point is (1, 0). (Same method as Example 9, which uses the y-axis, page 110.)

Question 8. A(1, 2), B(5, 3), C(6, 7) and D(p, q) are the vertices of a parallelogram, taken in order. Find p and q.

Answer: p = 2, q = 6. The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD. Mid-point of AC = ((1 + 6)/2, (2 + 7)/2) = (7/2, 9/2). Mid-point of BD = ((5 + p)/2, (3 + q)/2). So 5 + p = 7, giving p = 2, and 3 + q = 9, giving q = 6. D is (2, 6). (Same method as Example 10, page 110.)

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Sources

All opened and read on 21 October 2026. Page numbers are the printed page numbers of the textbook.

The formulas and methods are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.