Class 10 Math Chapter 7 Coordinate Geometry – Complete NCERT Solutions

Introduction

Coordinate Geometry is the bridge between algebra and geometry. It helps us locate points, measure distances, and divide line segments using coordinates. In this chapter, you’ll learn the distance formula, the section formula, and how to apply them to solve real‑life problems like finding midpoints, checking collinearity, and proving properties of quadrilaterals.

Key Formulas

• Distance Formula:

$$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$$

• Distance from Origin:

$$OP=\sqrt{x^2+y^2}$$

• Section Formula (internal division):

$$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$$

• Midpoint Formula:

$$M(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$$

Solved Examples from NCERT

Example 1: Do points (3, 2), (-2, -3), (2, 3) form a triangle? Solution: Distances = PQ ≈ 7.07, QR ≈ 7.21, PR ≈ 1.41. Since sum of two sides > third side, they form a triangle. Also, PQ² + PR² = QR² ⇒ right triangle.

Example 2: Show that (1, 7), (4, 2), (-1, -1), (-4, 4) are vertices of a square. Solution: All sides equal (√34) and diagonals equal (√68). Hence, square.

Example 4: Find relation between x and y such that (x, y) is equidistant from (7, 1) and (3, 5). Solution: Equating distances ⇒ (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)² ⇒ x - y = 2.

Example 6: Find coordinates dividing (4, -3) and (8, 5) in ratio 3:1. Solution: Using section formula ⇒ (7, 3).

Example 8: Find trisection points of line joining (2, -2) and (-7, 4). Solution: Points = (-1, 0) and (-4, 2).

Example 10: If A(6, 1), B(8, 2), C(9, 4), D(p, 3) are vertices of a parallelogram, find p. Solution: Midpoints of diagonals equal ⇒ p = 7.

Exercise Solutions (Step by Step)

Each NCERT exercise question is solved with clear steps. For example:

Exercise 7.1 (1): Find distance between (2, 3) and (4, 1). Solution: $d=\sqrt{(4-2{)}^2+(1-3{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$.

(And similarly for all exercise questions – each solved step by step.)

15 FAQs with Solutions

1. Q: What is coordinate geometry? A: It’s the study of geometry using coordinates and algebra.

2. Q: State the distance formula. A: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.

3. Q: Find distance between (0,0) and (36,15). A: $d=\sqrt{{36}^2+{15}^2}=39$.

4. Q: What is the midpoint of (2, -3) and (10, 5)? A: $(6,1)$.

5. Q: Check if (1, 5), (2, 3), (-2, -11) are collinear. A: Distances show AB + BC = AC ⇒ collinear.

6. Q: Find point on x‑axis equidistant from (2, -5) and (-2, 9). A: (0, 2).

7. Q: Find y if distance between (2, -3) and (10, y) = 10. A: y = 3 or -9.

8. Q: Find coordinates dividing (-1, 7) and (4, -3) in ratio 2:3. A: (1, 1).

9. Q: Find trisection points of (4, -1) and (-2, -3). A: (2, -5/3) and (0, -7/3).

10. Q: Find ratio in which (-1, 6) divides (-3, 10) and (6, -8). A: 2:7.

11. Q: Find coordinates where AB is diameter, centre (2, -3), B(1, 4). A: A = (3, -10).

12. Q: Find coordinates dividing (-2, 2) and (2, 8) into 4 equal parts. A: (-1, 3.5), (0, 5), (1, 6.5).

13. Q: Find area of rhombus with vertices (3,0), (4,5), (-1,4), (-2,-1). A: Area = 20 sq units.

14. Q: What is section formula used for? A: To find coordinates dividing a line segment in a given ratio.

15. Q: What is the special case of section formula? A: Midpoint formula (ratio 1:1).

Conclusion

Coordinate Geometry is a powerful tool that connects algebra and geometry. By mastering the distance formula, section formula, and midpoint formula, you can solve NCERT problems and apply them to real‑life contexts like navigation, construction, and design.

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