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RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry
By Karan Singh Bisht
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Updated on 11 Jun 2025, 18:18 IST
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.
Download the RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry PDF Here
Access the RD Sharma Solutions for Class 10 Maths Chapter 14 – Co-ordinate Geometry
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)
RD Sharma Solutions for Class 10 Maths Chapter 14 FAQs
Where can I find RD Sharma Solutions for Class 10 Maths Chapter 14?
You can access comprehensive and accurate RD Sharma Solutions for Class 10 Maths Chapter 14 – Statistics on Infinity Learn. These free solutions are created by expert faculty to help students strengthen their understanding of mean, median, mode, and graphical representations of data.
Are Infinity Learn RD Sharma Solutions for Chapter 14 suitable for board exam preparation?
Yes. Infinity Learn provides RD Sharma Chapter 14 solutions that align with the latest CBSE Class 10 syllabus. They include step-by-step answers and practice questions, making them ideal for board exam revision and conceptual clarity.
What topics are covered in Chapter 14 of RD Sharma Class 10 Maths?
Chapter 14 covers key statistics concepts like:
- Mean, median, and mode of grouped data
- Cumulative frequency
- Graphical representations
Infinity Learn solutions simplify each topic for easy understanding and practice.
Why choose Infinity Learn for RD Sharma Class 10 Chapter 14 solutions?
Infinity Learn offers expertly crafted, exam-ready solutions with easy explanations and visual support. Their RD Sharma Class 10 Chapter 14 content boosts accuracy, speed, and problem-solving skills for students aiming for top marks in Maths.