Competitive Exams

IIT JEE, NEET, ESE, GATE, AE/JE, Olympiad

Only IAS

UPSC, State PSC

School Preparation

Foundation (Class 6-12), Commerce, Arts, CuriousJr (1st - 8th), Science, International Boards

Govt Exam

Judiciary, SSC, Defence, Teaching, JAIIB & CAIIB, BIHAR EXAMS WALLAH, UP Exams, Railway, Nursing Exams, Banking, WB Exams

UG & PG Entrance Exams

MBA, IPMAT, IIT JAM, LAW, CUET UG, UGC NET, GMAT, Design & Architecture, Pharma, CSIR NET, CUET PG, NEET PG

FINANCE

CA, CS, Finance Courses, ACCA

Others

Online Degrees

English Proficiency Test

IELTS, TOEFL

Agriculture

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1

Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 before exams for better understanding.

Ananya Gupta25 Jul, 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1: RS Aggarwal Solutions for Class 10 Maths Chapter 16 Co-ordinate Geometry Exercise 16.1 provide detailed explanations and step-by-step solutions to help students understand the fundamentals of coordinate geometry.

This exercise focuses on the basics of plotting points on the Cartesian plane, understanding the concepts of the x-axis and y-axis, and calculating the distance between two points using the distance formula.

These solutions are an excellent resource for reinforcing classroom learning and preparing for exams.

CBSE Compartment Result 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 prepared by subject experts from Physics Wallah provide a comprehensive overview of coordinate geometry.

With clear explanations and step-by-step instructions, these solutions help students grasp the basics of coordinate geometry, making it easier to solve related problems accurately and confidently.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 PDF

The PDF link for RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 is available below.

It includes step-by-step explanations to help students understand and master the concepts, making it an invaluable resource for exam preparation and homework help.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1

Below we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 for the ease of the students –

Q. Find the distance between the points :

(i) A(9, 3) and B (15, 11) (ii) A (7, -4) and B(-5, 1)

(iii) A(-6, -4) and B(9, -12) (iv) A (1, -3 ) and B (4, -6)

(v) P (a+b, a-b) and Q (a-b, a+b)

(vi) P (a sin α, a cos α) and Q (a cos α, - a sin α)

Solution:

Q. Find the distance of each of the following points from the origin:

(i) A(5, -12) (ii) B (-5, 5) (iii) C (-4, -6).

Solution:

(i) The distance of point (5,-12) from the origin is

Origin (0,0) point (5,-12) √[5²+(-12)²] = √25+144 = √169=13

(ii) origin (0,0)

point (-5,5) √(25+25)=√50=5√2

(iii) origin (0,0)

point (-4,-6) √16+36=√52=2√13 Q. Find all possible values of y for which the distance between the points

A (2, -3) and B (10, y) is 10 units.

Solution:

We know that the distance between two points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Given d = 10 units and points are A (2, -3) and B (10, y)

∴ \sqrt{(10 - 2)^{2} + (y - - 3)^{2}} = 10

8^{2} + (y + 3)^{2} = 100

(y + 3)^{2} + 64 = 100

(y + 3)^{2} = 100 - 64

(y + 3)^{2} = 36

(y + 3) = \sqrt{36}

(y + 3) = \pm 6

y = \pm 6 - 3

y = 6 - 3 o r - 6 - 3

y = 3 o r - 9

Q. Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

Solution:

Coordinates of P and Q are (x, 4) and (9, 10) and PQ = 10 units

\to (P Q)^{2} = (10)^{2} u n i t s = 100 u n i t s

⇒

{9 - x}^{2} + {10 - 4}^{2} = 100

⇒ 81–x2–18\timesx+36=100

\to 81 - x^{2} - 18 \times x + 36 = 100

\to x^{2} - (18 \times x) + 17 = 0

→(x–1)(x–17)=0 →x=1orx=17 Hence the value of x is either 1 or 17.

Q. Find the coordinates of the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Solution:

On x axis, the y coordinate is 0 So, let us assume the point to be (x,0) Distance from (-2,5) = Distance from (2,-3)

{(x + 2)}^{2} + 25 = {(x - 2)}^{2} + 9

16= -8x x = -2 So, the point required is (-2,0)

Q. Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, -8).

Solution:

\sqrt{((x - 11)^{2} + (0 + 8)^{2})} = 10

Square both sides and expand

x^{2} - 22 x + 121 + 64 = 100

x^{2} - 22 x + 85 = 0

(x - 17) (x - 5) = 0

x = 17 o r x = 5

Q. Find the point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3).

Solution:

Let the point be (0 , y) So,

\sqrt{6^{2} + (5 - y)^{2}} = \sqrt{(- 4)^{2} + (3 - y)^{2}}

On squaring both sides , we get

36 + 25 + y^{2} - 10 y = 16 + 9 + y^{2} - 6 y

61 - 10 y = 25 - 6 y

10 y - 6 y = 61 - 25

4 y = 36

So, y = 9 So, point = (0, 9)

Q. If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5) prove that 3x = 2y.

Solution:

If P is equidistant from point A and B then,AP:PB=1:1 Distance AP =

\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

= \sqrt{(x - 5)^{2} + (y - 1)^{2}}

A P^{2} = x^{2} + 25 - 10 x + y^{2} + 1 - 2 y

A P^{2} = x^{2} + y^{2} + 26 - 10 x - 2 y

Distance

B P = \sqrt{(- 1 - x)^{2} + (5 - y)^{2}}

B P^{2} = 1 + x^{2} + 2 x + 25 + y^{2} - 10 y

B P^{2} = 26 + x^{2} + y^{2} + 2 x - 10 y

Since P is the midpoint.

A P^{2} = B P^{2}

x^{2} + y^{2} + 26 - 10 x - 2 y = x^{2} + y^{2} + 2 x - 10 y + 26

x^{2} - x^{2} + y^{2} - y^{2} + 26 - 26 - 10 x + 2 x = - 10 y + 2 y

- 12 x = - 8 y

- 3 x = - 2 y

3 x = 2 y

hence proved

Q. Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).

Solution:

Let the required point be P(x,y). Then

A P = B P = C P

That is,

(A P)^{2} = (B P)^{2} = (C P)^{2}

This means,

(A P)^{2} = (B P)^{2}

\Rightarrow (x - 5)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y + 5)^{2}

\Rightarrow x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = x^{2} - 10 x + 25 + y 2 + 10 y + 25

\Rightarrow x^{2} + y^{2} - 10 x - 6 y + 34 = x^{2} + y^{2} - 10 x + 10 y + 50

\Rightarrow x^{2} + y^{2} - 10 x - 6 y - x^{2} - y^{2} + 10 x - 10 y = 50 - 34

\Rightarrow - 16 y = 16

\Rightarrow y = - 16 / 16 = - 1

And

(B P)^{2} = (C P)^{2}

\Rightarrow (x - 5)^{2} + (y + 5)^{2} = (x - 1)^{2} + (y + 5)^{2}

\Rightarrow x^{2} - 10 x + 25 + y^{2} + 10 y + 25 = x^{2} - 2 x + 1 + y^{2} + 10 y + 25

\Rightarrow x^{2} + y^{2} - 10 x + 10 y + 50 = x^{2} + y^{2} - 2 x + 10 y + 26

\Rightarrow x^{2} + y^{2} - 10 x + 10 y - x^{2} - y^{2} + 2 x - 10 y = 26 - 50

\Rightarrow - 8 x = - 24

\Rightarrow y = - 24 / - 8 = 3

Hence the required point is (3,-1)

Q. If the point C(-2, 3) is equidistant from the points A(3, -1) and B(x, 8), find the values of x. Also, find the distance BC.

Solution:

given distance AC=distance BC

Q. (i) If the point P(2, 2) is equidistant from the points (a+b, b-a) and (a-b, a+b), prove that bx = ay. (ii) If the distances of P(x, y) from A(5, 1) and B (-1, 5) are equal then prove that 3x = 2y.

Solution:

(i)

Distance between the points (x, y) and (a+b, b-a) & (a-b, a+b) is equal ⇒

\sqrt{[x - (a + b)]^{2} + [y - (b - a)]^{2}} = \sqrt{[x - (a - b)]^{2} + [y - (a + b)]^{2}}

⇒

x^{2} + (a + b)^{2} - 2 x (a + b) + y^{2} + (b - a)^{2} - 2 y (b - a) = x^{2} + (a - b)^{2} - 2 x (a - b) + y^{2} + (a + b)^{2} - 2 y (a + b)

⇒

- 2 a x - 2 b x - 2 b y + 2 a y = - 2 a x + 2 b x - 2 a y - 2 b y

⇒

a y - b x = b x - a y

⇒

2 a y = 2 b x

⇒

b x = a y

(ii)

It is given that P is equidistant from A and B. So, PA = PB Using distance formula,

then

+ 25 - 10x +

+ 1 - 2y =

+ 1 + 2x +

+ 25 - 10y - 12x = -8y 3x = 2y Hence proved Q. Using the distance formula, show that the given points are collinear:

(i) (1, -1), (5, 2) and 9, 5 (ii) (6, 9), (0, 1) and (-6, -7)

(iii) (-1, -1), (2, 3) and (8, 11) (iv) (-2, 5), (0, 1) and (2, 3).

Solution:

Q. Show that the points A(7 , 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

Solution:

The given points are A(7,10), B(-2,5) and C(3,-4) AB =

\sqrt (- 2 - 7)^{2} + (5 - 10)^{2}

\sqrt (- 9)^{2} + (- 5)^{2}

\sqrt 81 + 25

=√106 units BC =

\sqrt (3 - (- 2))^{2} + (- 4 - 5)^{2}

\sqrt (5)^{2} + (- 9)^{2}

\sqrt 25 + 81

=√106 units AC =

\sqrt (3 - 7)^{2} + (- 4 - 10)^{2}

\sqrt (- 4) 2 + (- 14) 2

\sqrt 16 + 196

\sqrt 212 u n i t s

Since, AB and BC are equal, they form the vertices of an isosceles triangle Also,

(A B)^{2} + (B C)^{2}

\sqrt (106)^{2} + \sqrt (106)^{2}

= √212 and

(A C)^{2}

(\sqrt 212)^{2}

= 212 Thus,

(A B)^{2}

(B C)^{2}

(A C)^{2}

This shows that ΔABC is right angled at B Therefore, the given points A(7,10), B(-2,5) and C(3,-4) are the vertices of an isosceles right-angled triangle.

Q. Show that the points (-3 , 3), (3, 3) and $(- 3 \sqrt{3}, 3 \sqrt{3}$ ) are the vertices of an equilateral triangle.

Solution:

The given points are A (-3,-3), B(3,3) and C(−3√3,3√3). Now, AB =√[(−3−3)^2+(−3−3)^2] = √[(−6)^2+(−6)^2] = √(36+36)=√72 = 6√2 BC = √[(3+3√3)^2+(3−3√3)^2] =√[9+27+18√3+9+27−18√3] = √72=6√2 AC =√[(−3+3√3)^2+(−3−3√3)^2] =√[9+27−18√3+9+27+18√3] = √72= 6√2 AB=BC=AC Hence the given points are the vertices of an equilateral triangle Q. In the given figure P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then y equals

(a) 2 (b) 4 (c) -4 (d) $\frac{- 5}{2}$

Solution:

option a

Q . The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is (a) 20 (b) 12 (c) 6 (d) 16

Solution:

Area of the triangle

= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |

= \frac{1}{2} | 5 (0 - 4) + 8 (4 - 0) + 8 (0 - 0) |

= \frac{1}{2} | - 20 + 32 | = 6 s q . u n i t s

Q. If A(-6, 7) and B(-1, -5) are two given points then the distance 2AB is

(a) 13 (b) 26 (c) 169 (d) 238

Solution:

Q. The distance of P(3, 4) from the x-axis is (a) 3 units (b) 4 units (c) 5 units (d) 1 unit

Solution:

A pair of Cartesian coordinates (a,b) identifies the position of a point. That point is defined as being on the line x=a where it crosses the line y=b. So by definition, that point is |b| from the X axis, and |a| from the y axis. so the answer is 4 unit option b is correct

Q. If P(-1, 1) is the midpoint of the line segment joining A(-3, b) and B(1, b+4) then b=? (a) 1 (b) -1 (c) 2 (d) 0

Solution:

The given ports are A(-3,b) and B(1, b+4)

y = \frac{b + b + 4}{2} 1 = \frac{2 b + 4}{2} 1 = b + 2 b = - 1

Option B

Q. If A(4, 2), B(6, 5) and C(1, 4) be the vertices of $△ A B C$ and AD is a median, then the coordinates of D are (a) $(\frac{5}{2}, 3)$ (b) $(5, \frac{7}{2})$ (c) $(\frac{7}{2}, \frac{9}{2})$ (d) none of these

Solution:

Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC. D(x,y) = {(6+1)/2 , (5+4)/2} = {7/2 ,9/2} (c) (7/2,9/2) is correct answer

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1

Clear Explanations: The solutions provide detailed step-by-step explanations making complex concepts in coordinate geometry easy to understand.
Enhanced Understanding: By working through these solutions students can deepen their understanding of plotting points understanding the Cartesian plane and using the distance formula.
Exam Preparation: These solutions align with the Class 10 syllabus helping students prepare effectively for exams by focusing on the key concepts and types of questions that may appear.
Confidence Building: With comprehensive solutions and explanations, students can build confidence in their ability to tackle coordinate geometry problems, leading to better performance in tests and exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.1 FAQs

What is coordinate geometry?

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. It involves plotting points, lines, and shapes on a Cartesian plane.

What is a Cartesian plane?

A Cartesian plane is a two-dimensional plane defined by a horizontal axis (x-axis) and a vertical axis (y-axis). The point where these axes intersect is called the origin, denoted by (0, 0).

How do you plot a point on the Cartesian plane?

To plot a point, you need an ordered pair of numbers (x, y). The x-coordinate represents the horizontal distance from the origin, and the y-coordinate represents the vertical distance. For example, to plot the point (3, 4), move 3 units to the right along the x-axis and 4 units up along the y-axis.

How do you determine if two lines are perpendicular?

Two lines are perpendicular if the product of their slopes is -1.

🔥 Trending Blogs

CBSE Class 10 Science Half Yearly Important Questions 2025-26 PDF Download

Class 10 Half-Yearly Marathon Endgame 2025 Live Revision Classes

CBSE Subject Codes for Class 10 2026 Subject Wise Class 10 Code List

CBSE Class 10 English Error Correction Questions with Answers

CBSE Sample Papers Class 10 Science 2025-26 PDF Download

Talk to a counsellorHave doubts? Our support team will be happy to assist you!

08448982616

008448982616

Check out these Related Articles

CLASS 10 EXAM

CBSE Sample Papers Class 10 Maths 2025-26 PDF Download Set Wise

Read Full Story

3 Sept 2025, 04:39 pm

CLASS 10 EXAM

CBSE Sample Papers Class 10 English 2025-26 PDF Download

Read Full Story

3 Sept 2025, 04:39 pm

CLASS 10 EXAM

CBSE Class 10th Sample Paper 2024-25 Download Latest PDF

Read Full Story

3 Sept 2025, 04:38 pm

CLASS 10 EXAM

CBSE Sample Papers Class 10 Social Science 2025-26 PDF Download

Read Full Story

3 Sept 2025, 04:28 pm

CLASS 10 EXAM

CBSE Sample Papers Class 10 2025-26 PDF Download with Marking Scheme

Read Full Story

3 Sept 2025, 04:24 pm

CLASS 10 EXAM

CBSE Sample Papers for Class 10 2025-26 PDF with Solutions

Read Full Story

3 Sept 2025, 04:15 pm

CLASS 10 EXAM

CBSE Class 10 Economics Notes Chapter 2 Sectors Of Indian Economy

Read Full Story

21 Aug 2025, 12:32 pm

CLASS 10 EXAM

NCERT Class 10 English Footprints Without Feet Chapter 3 Question Answer

Read Full Story

21 Aug 2025, 12:24 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 English First Flight Chapter 3 Question Answer

Read Full Story

21 Aug 2025, 12:01 pm

Join 15 Million students on the app today!

Live & recorded classes available at ease

Dashboard for progress tracking

Millions of practice questions at your fingertips

Free Learning Resources

PW Books

Notes (Class 10-12)

Class 10 Math's Notes

Class 10 Chemistry Notes

Class 10 Physics Notes

Class 10 Biology Notes

Aptitude & Reasoning

Class 10 Geography

Physics Class 11 Notes

Class 11 Chemistry Notes

Maths Notes Class 11

Zoology Class 11

Class 11 Botany Notes

Physics Class 12 notes

Chemistry Class 12

Maths Notes Class 12

Zoology class 12

Class 12th Botany Notes

PW Study Materials

Arjuna JEE Study Material

JEE 7 PYQs

JEE Mind Maps

Yakeen Study Material

NCERT Punch

Arjuna Neet Study Material

NEET PYQs

CBSE Sample Paper

CBSE Question Bank

Udaan For Class 10th Study Material

Notes (Class 6-9)

Class-6 Theory & Notes

Math's Notes for class 7

Science Notes for class 7

Class 8 Math Notes

Class 8 Chemistry Notes

Class 8 Physics Notes

Class 8 Biology Notes

Class 8 SST Notes

Class 9 Math's Notes

Class 9 Physics Notes

Class 9 Chemistry Notes

Class 9 Biology Notes

Ncert Solutions

NCERT Solutions For Class 6

NCERT Solutions For class 7

NCERT Solutions for class 8

NCERT Solutions for class 9

NCERT Solutions for class 10

NCERT Solutions for class 11

NCERT Solutions for Class 12

Govt Exams

Entrance Exams

Our Other Websites

Class 6th to 12th Online Courses

Class 12 Online Course

Class 11 Online Course

Class 10 Online Course

Class 9 Online Course

Class 8 Online Course

Govt Job Exams Courses

SSC Online Coaching

Bank Exam Online Coaching

TET Online Coaching

JAIIB & CAIIB Online Coaching

Bihar Exam Online Coaching

UPSC Coaching

UPSC Online Coaching

UPSC Offline & Hybrid Coaching

State PSC Online Coaching

UPPSC Online Coaching

BPSC Online Coaching

Defence Exam Coaching

NDA Online Coaching

CDS Online Coaching

AFCAT Online Coaching

Agniveer Online Coaching

Gate Exam Coaching

Civil Engineering Courses

Mechanical Engineering Courses

Formulas

Other Exams

Company Secretary Exam

Know about Physics Wallah

Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.

We Stand Out because

We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.

Our Key Focus Areas

Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.

What Makes Us Different

Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others