NCERT Solutions for Class 10 Mathematics Chapter 7 Coordinate Geometry

This worksheet provides NCERT Solutions for Class 10 Mathematics Chapter Coordinate Geometry. This chapter focuses on understanding how to locate points on a plane, calculate distances between points, and use formulas like the distance formula and section formula. It helps students solve geometry problems using coordinates and algebraic methods. This worksheet provides complete and accurate NCERT Solutions that strictly follow the NCERT pattern, helping students build strong analytical and problem-solving skills.

Chapter summary: stories, poems & themes

This chapter is fully concept-based and does not include stories or poems. It focuses on coordinate geometry concepts, numerical problems, and diagram-based understanding. Students learn how to apply formulas to calculate distances, check collinearity, identify shapes, and divide line segments. The main theme is applying mathematical formulas to solve real and graphical problems.

What this NCERT chapter covers?

• Distance formula and its applications 
• Checking collinearity of points 
• Identifying types of triangles and quadrilaterals 
• Section formula for dividing line segments 
• Midpoint formula and its use 
• Solving coordinate-based equations 
• Finding unknown coordinates using conditions 

How to use these NCERT solutions?

Students should first try solving each question independently before referring to the answers. These solutions follow the exact NCERT order and structure, making it easy to understand each step. Parents and teachers can use this worksheet to guide students and explain formulas clearly. Regular practice with these solutions improves accuracy and confidence in solving coordinate geometry problems.

Student tips & learning tricks

• Always apply the correct formula (distance, section, or midpoint) carefully 
• Write each calculation step clearly to avoid mistakes 
• Check signs properly while solving equations 
• Practice solving coordinate problems step by step 
• Recheck calculations to avoid arithmetic errors 

Why NCERT solutions are important?

NCERT Solutions provide a clear and structured approach to learning Mathematics. They strictly follow the CBSE syllabus and help students understand concepts deeply. These solutions improve problem-solving skills, boost confidence, and prepare students effectively for exams and assessments.

Complete answer key – NCERT solutions

Exercise 7.1

1. 
(i) √8 
(ii) √32 
(iii) 2√(a² + b²) 

2. 
Distance = 39 km 

3. 
Points: (1, 5), (2, 3), (–2, –11) 

AB = √[(2–1)² + (3–5)²] = √5 
BC = √[(–2–2)² + (–11–3)²] = √212 
AC = √[(–2–1)² + (–11–5)²] = √265 

Since AB + BC ≠ AC, AB + AC ≠ BC, BC + AC ≠ AB 
Points are NOT collinear 

4. 
Points: (5, –2), (6, 4), (7, –2) 

AB = √[(6–5)² + (4+2)²] = √37 
BC = √[(7–6)² + (–2–4)²] = √37 
AC = √[(7–5)² + (–2+2)²] = 2 

Since AB = BC, triangle is isosceles 

5. 
Using distance formula, calculate all sides 

If AB = BC = CD = DA and diagonals equal → square 

Thus, Champa is correct → ABCD is a square 

6. 
(i) All sides equal → Rhombus 
(ii) Sides not equal, diagonals unequal → No special quadrilateral 
(iii) Opposite sides equal → Parallelogram 

7. 
Let point be (x, 0) 

(x–2)² + (0+5)² = (x+2)² + (0–9)² 

x² – 4x + 4 + 25 = x² + 4x + 4 + 81 

–4x + 29 = 4x + 85 
–8x = 56 
x = –7 

Point = (–7, 0) 

8. 
√[(10–2)² + (y+3)²] = 10 

64 + (y+3)² = 100 
(y+3)² = 36 

y+3 = ±6 
y = 3 or y = –9 

9. 
Q(0,1) equidistant from P(5,–3) and R(x,6) 

QP² = QR² 

25 + 16 = x² + 25 
41 = x² + 25 
x² = 16 
x = ±4 

QR = √41 
PR = √82 

10. 
(x–3)² + (y–6)² = (x+3)² + (y–4)² 

–12x – 4y + 20 = 0 
3x + y = 5 

Exercise 7.2

1. 
x = [2(4) + 3(–1)] / 5 = 1 
y = [2(–3) + 3(7)] / 5 = 3 

Point = (1, 3) 

2. 
First point (1:2) = (2, –5/3) 
Second point (2:1) = (0, –7/3) 

3. 
Use section formula to find both flag positions 
Then apply distance formula 

Midpoint = ((x1+x2)/2, (y1+y2)/2) 

4. 
Using section formula → ratio = 2 : 1 

5. 
0 = [m1(5) + m2(–5)] / (m1 + m2) 

m1 = m2 
Ratio = 1 : 1 

Point = (–1.5, 0) 

6. 
(1+ x)/2 = (4+3)/2 → x = 2 
(2+6)/2 = (y+5)/2 → y = 3 

7. 
(x + 1) / 2 = 2 → x = 3 
(y + 4) / 2 = –3 → y = –10 

Point = (3, –10) 

8. 
P = [3(2) + 4(–2)]/7 , [3(–4) + 4(–2)]/7 

= (–2/7, –20/7) 

9. 
Points: 
(–1, 3.5), (0, 5), (1, 6.5) 

10. 
d1 = √32 
d2 = √72 

Area = ½ × √32 × √72 
= ½ × √2304 
= 24 sq units 

Master coordinate geometry concepts in Class 10 Mathematics with these accurate NCERT Solutions designed for clear understanding and strong exam preparation.
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