RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry

RD Sharma Solutions for Class 10 Maths Chapter 14 – Coordinate Geometry provide a detailed and concept-based approach to help students master the fundamentals of coordinate-based geometry. 

In this chapter, students explore the use of ordered pairs (coordinates) to study geometry on a plane, known as Coordinate Geometry. The chapter includes both analytical and graphical interpretations of geometric problems.

Created by expert faculty at Infinity Learn, these RD Sharma solutions for class 10 are designed to improve conceptual understanding and problem-solving skills in Class 10 students. They offer step-by-step explanations for all five exercises in the chapter, aligned with the 2025–26 CBSE syllabus.

Key Concepts Covered:

• Calculating the distance between two points using their coordinates.
• Finding the coordinates of a point dividing a line segment in a given ratio.
• Determining the area of a triangle using the coordinates of its vertices.

These RD Sharma solutions aim to build confidence among students and help them solve any type of question from Chapter 14 with clarity and ease. Students can download the free PDF version of the updated solutions from Infinity Learn for offline access.

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Q. Find the distance between the points A(3, 4) and B(7, 1).

Solution: Using distance formula:

D = √[(x₂ − x₁)² + (y₂ − y₁)²]

= √[(7 − 3)² + (1 − 4)²]

= √[16 + 9] = √25

Answer: 5 units

Q. Find the distance between the points (−2, −3) and (4, 1).

Solution: D = √[(4 − (−2))² + (1 − (−3))²]

= √[(6)² + (4)²]

= √[36 + 16] = √52

Answer: √52 units

Q. Find the midpoint of the line joining (2, 3) and (6, 7).

Solution: Midpoint = ((2 + 6)/2, (3 + 7)/2)

= (8/2, 10/2)

= (4, 5)

Answer: (4, 5)

Q. Find the midpoint of the line joining (−4, −2) and (6, 4).

Solution: Midpoint = ((−4 + 6)/2, (−2 + 4)/2)

= (2/2, 2/2) = (1, 1)

Answer: (1, 1)

Q. Find the coordinates of the point dividing the line joining (2, 4) and (6, 8) in the ratio 1:1.

Solution: Using section formula:

= ((1×6 + 1×2)/2, (1×8 + 1×4)/2)

= (8/2, 12/2) = (4, 6)

Answer: (4, 6)

Q. Find the coordinates of the point dividing the line joining (3, −2) and (−5, 4) in the ratio 3:1.

Solution:

= ((3×(−5) + 1×3)/4, (3×4 + 1×(−2))/4)

= (−15 + 3)/4, (12 − 2)/4 = (−12/4, 10/4)

= (−3, 2.5)

Answer: (−3, 2.5)

Q. Find the area of triangle with vertices A(1, 1), B(4, 4), and C(1, 5).

Solution:

Area = 1/2 |1(4−5) + 4(5−1) + 1(1−4)|

= 1/2 |−1 + 16 −3| = 1/2 × 12

Answer: 6 sq units

Q. Find the area of triangle with vertices (0, 0), (4, 0), (4, 3).

Solution:

Area = 1/2 |0(0−3) + 4(3−0) + 4(0−0)|

= 1/2 |0 + 12 + 0| = 1/2 × 12

Answer: 6 sq units

Q. If the distance between points (x, 5) and (1, 1) is 5, find x.

Solution:

√[(x−1)² + 16] = 5 → (x−1)² = 9 → x − 1 = ±3

⇒ x = 4 or x = −2

Answer: x = 4 or −2

Q. Find the point on x-axis equidistant from (2, 3) and (−4, 5).

Solution: Let point be (x, 0).

Distance from (2, 3): √[(x−2)² + 9]

Distance from (−4, 5): √[(x+4)² + 25]

Squaring both: (x−2)² + 9 = (x+4)² + 25

= x² − 4x + 4 + 9 = x² + 8x + 16 + 25

= −12x = 28 → x = −7/3

Answer: (−7/3, 0)

Q. Find the length of diagonal of rectangle with vertices (1, 2), (1, 6), (5, 6), (5, 2).

Solution:

Diagonal = √[(5−1)² + (6−2)²] = √[16 + 16] = √32

Answer: √32 units

Q. Show that points (1, 2), (3, 4), and (5, 6) lie on a straight line.

Solution:

Slope of AB = (4−2)/(3−1) = 1

Slope of BC = (6−4)/(5−3) = 1

Slopes are equal ⇒ points are collinear

Answer: Yes, collinear

Q. Find the centroid of triangle with vertices (1, 2), (3, 4), (5, 0).

Solution:

Centroid = ((1+3+5)/3, (2+4+0)/3) = (9/3, 6/3) = (3, 2)

Answer: (3, 2)

Q. Find the diagonal of a square with side 5 units.

Solution: Diagonal = √2 × side

= √2 × 5 = 5√2

Answer: 5√2 units

Q. Find area of triangle with vertices (2, 3), (4, 7), (6, 3).

Solution: Area = 1/2 |2(7−3) + 4(3−3) + 6(3−7)|

= 1/2 |8 + 0 −24| = 1/2 × 16

Answer: 8 sq units

Q. Prove that triangle with vertices (0, 0), (2, 2), (4, 0) is isosceles.

Solution: AB = BC = √[(2−0)² + (2−0)²] = √8

AC = √[(4−0)²] = 4

AB = BC ⇒ triangle is isosceles

Answer: Yes, it is isosceles

Q. Find the coordinates of a point on the line joining (3, 4) and (6, 10) whose x-coordinate is 3.

Solution: Given x = 3, which is the x-coordinate of the first point.

So, the point is simply: (3, 4)

Answer: (3, 4)