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Class 10th Mathematics Chapter 7 Coordinate Geometry – Important Questions

Important questions with answers for Class 10th Mathematics Chapter 7 Coordinate Geometry, based on the NCERT CBSE syllabus.

Topics covered in Coordinate Geometry

Frequently asked questions

Where can I find important questions for Class 10th Mathematics Chapter 7 Coordinate Geometry?

This page lists important questions with answers for Coordinate Geometry. Topics covered: Distance formula, Section formula, Midpoint formula, Area of triangle using coordinates.

Are Coordinate Geometry important questions free?

Yes, Notebook gives free NCERT CBSE important questions for Class 10 Mathematics.

How to prepare Coordinate Geometry for Class 10 Mathematics exams?

Read the Coordinate Geometry notes first, revise the key topics (Distance formula, Section formula, Midpoint formula, Area of triangle using coordinates), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 10th Mathematics Chapter 7 Coordinate Geometry important questions for quick revision?

You are on the right page: this important questions page for Coordinate Geometry covers Class 10 Mathematics for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 10th Mathematics Chapter 7 Coordinate Geometry important for the CBSE exam 2026-27?

Coordinate Geometry is part of the Class 10 Mathematics NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 10th Mathematics Chapter 7 Coordinate Geometry Notes · Class 10th Mathematics Chapter 7 Coordinate Geometry PYQ

Coordinate Geometry

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1 Mark
What is the distance of point (3, 4) from the origin?
Ans: Distance = √(3²+4²) = √(9+16) = √25 = 5 units.
2 Marks
Find the midpoint of the segment joining A(3, –2) and B(–1, 4).
Ans: Midpoint = ((3+(–1))/2, (–2+4)/2) = (2/2, 2/2) = (1, 1).
3 Marks
Find the ratio in which the point (–1, k) divides the segment joining (–3, 10) and (6, –8). Also find k.
Ans: Let ratio = m:n. –1 = (6m–3n)/(m+n) → –m–n = 6m–3n → 2n = 7m → m:n = 2:7. k = (–8×2+10×7)/(2+7) = (–16+70)/9 = 54/9 = 6.
5 Marks
Find the area of the quadrilateral whose vertices in order are (–4,–2), (–3,–5), (3,–2), (2,3).
Ans: Split into two triangles: △1: (–4,–2),(–3,–5),(3,–2). Area = ½|–4(–5+2)+–3(–2+2)+3(–2+5)| = ½|12+0+9| = 10.5. △2: (–4,–2),(3,–2),(2,3). Area = ½|–4(–2–3)+3(3+2)+2(–2+2)| = ½|20+15+0| = 17.5. Total = 28 sq units.
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