Physics Wallah

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 Coordinate Geometry

Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 before exams for better understanding.
authorImageNeha Tanna31 Jul, 2024
Share

Share

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4: The Physics Wallah academic team has produced a comprehensive answer for Chapter 16: Coordinate Geometry in the RS Aggarwal class 10 textbook. One should read the theory of chapter 16 Coordinate Geometry before attempting to solve all of the numerical problems in exercise 16D. This will ensure that you have a firm understanding of Chapter 16 Coordinate Geometry before moving on to the solution.

Complete the NCERT exercise questions and utilize them as a guide. Solutions for Physics Wallah NCERT Class 10 Maths problems in the exercise require assistance to be completed. Class 10 Math NCERT solutions were uploaded by Physics Wallah.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 Coordinate Geometry Overview

Chapter 16 of RS Aggarwal's Class 10 Maths textbook covers Coordinate Geometry, focusing on various concepts related to the coordinate plane. Exercise 16.4 delves into finding the distances between points, the coordinates of the midpoint of a line segment, and the area of a triangle formed by three points. The problems in this exercise require students to apply the distance formula, midpoint formula, and the area of a triangle formula. This exercise helps reinforce the understanding and application of these formulas, aiding students in solving geometric problems on the coordinate plane efficiently.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 PDF

The PDF link for RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 is available below. By downloading this PDF, students can access expert-prepared solutions and improve their problem-solving skills making it a valuable resource for their exam preparation.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 (Ex 16D)

Here we have provided the RS Aggarwal Solutions for Class 10 Maths Chapter 16 Coordinate Geometry Exercise 16.4 to help students prepare better for their exams. These RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 are designed to make it easier for students to understand and practice key concepts in Coordinate Geometry ensuring they are well-prepared and confident for their upcoming tests.
Solution
Using midpoint formula , [ ( x1+x2 ) /2 , ( y1 +y2 ) /2 ] substitute the values , (2,-3y ) = [ (-1 + 5 )/2 ,( y+7 ) /2 ] now on comparing both LHS and RHS -3y = ( y + 7 ) /2 - 3 y × 2 = y + 7 -6 y= y + 7 -7y = 7 ⇒ y = -1 therefore , A = ( -1 , -1 )
Solution
P open parentheses x subscript 1 comma y subscript 1 close parentheses space & Q open parentheses x subscript 2 comma y subscript 2 close parentheses b e space 2 space p o i n t s space t h e n space d i s tan c e space b e t w e e n space t h e m P Q equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root i n space t h i s space q u e s t i o n A B equals A C t h e n square root of open parentheses 3 minus 0 close parentheses squared plus open parentheses p minus 2 close parentheses squared end root equals square root of open parentheses p minus 0 close parentheses squared plus open parentheses 5 minus 2 close parentheses squared end root s q u a r i n g space o n space b o t h space s i d e 3 squared plus open parentheses p minus 2 close parentheses squared equals p squared plus 3 squared p squared minus 4 p plus 4 equals p squared 4 p equals 4 p equals 1
Solution
ABCD is a rectangle whose diagonal is BD. B = (4,0) D = (0,3) Length of BD is given by square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root square root of open parentheses 4 minus 0 close parentheses squared plus open parentheses 0 minus 3 close parentheses squared end root square root of 16 plus 9 end root equals square root of 25 Length of diagonal = BD = 5 units
Solution
W e space h a v e comma P left parenthesis k minus 1 comma 2 right parenthesis semicolon space A left parenthesis 3 comma k right parenthesis semicolon space B left parenthesis k comma 5 right parenthesis space a s space t h e space g i v e n space p o i n t s. N o w comma space P A equals P B rightwards double arrow square root of left parenthesis k minus 1 minus 3 right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared end root equals square root of left parenthesis k minus 1 minus k right parenthesis squared plus left parenthesis 2 minus 5 right parenthesis squared end root rightwards double arrow left parenthesis k minus 4 right parenthesis squared plus left parenthesis 2 minus k right parenthesis squared equals left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared rightwards double arrow k squared plus 16 minus 8 k plus 4 plus k squared minus 4 k equals 1 plus 9 rightwards double arrow 2 k squared minus 12 k plus 20 equals 10 rightwards double arrow 2 k squared minus 12 k plus 10 equals 0 rightwards double arrow k squared minus 6 k plus 5 equals 0 rightwards double arrow k squared minus 5 k minus k plus 5 equals 0 rightwards double arrow k left parenthesis k minus 5 right parenthesis minus 1 left parenthesis k minus 5 right parenthesis equals 0 rightwards double arrow left parenthesis k minus 1 right parenthesis left parenthesis k minus 5 right parenthesis equals 0 rightwards double arrow k minus 1 equals 0 space space o r space space k minus 5 equals 0 rightwards double arrow k equals 1 space space o r space space k equals 5
Solution
l e t space left parenthesis p comma q right parenthesis space d i v i d e space t h e space l i n e space o f space p o i n t s space open parentheses x subscript 1 comma y subscript 1 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 close parentheses space i n space t h e space r a t i o space m colon n t h e n open parentheses p comma q close parentheses equals open parentheses fraction numerator n x subscript 1 plus m y subscript 1 over denominator m plus n end fraction comma fraction numerator n x subscript 2 plus m y subscript 2 over denominator m plus n end fraction close parentheses i n space t h i s space q u e s t i o n p equals x comma q equals 2 comma x subscript 1 equals 12 comma x subscript 2 equals 4 comma y subscript 1 equals 5 comma y subscript 2 equals negative 3 w e space w a n d space t o space f i n d space m colon n t h e n open parentheses x comma 2 close parentheses equals open parentheses fraction numerator 12 n plus 5 m over denominator m plus n end fraction comma fraction numerator 4 n minus 3 m over denominator m plus n end fraction close parentheses s o 2 equals fraction numerator 4 n minus 3 m over denominator m plus n end fraction 2 m plus 2 n equals 4 n minus 3 m 5 m equals 2 n m over n equals 2 over 5 t h e n space t h e space r a t i o space i s space 2 colon 5
Solution
The vertices of the rectangle ABCD are A(2, -1) , B(5, -1), C(5, 6) and D(2, 6). Now, Coordinates of midpoint of AC = (2+5)/2,(−1+6)/2 =7/2,5/2 Coordinates of midpoint of BD= (5+2)/2,(−1+6)/2 =7/2,5/2 Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.
Solution
The given vertices are A(7, -3), B(5, 3) and C(3, -1) Since D and E are the midpoints of BC and AC respectively, therefore Coordinates of D = (5+3)/2,(3–1)/2 =(4,1) Coordinates of E = (7+3)/2,(−3−1)/2 =(5,−2) Now, AD=√(7–4)2+(−3−1)2 =√9+16=√25=5 BE=√(5–5)2+(3+2)2 =√0+25=√25=5 Hence, AD = BE = 5 units
Solution
Ratio given (2:3) = (m:n) A ( 2 , 6 ) = ( x 1 , y 1 ) B ( 5 , 1 ) = ( x 2 , y 2 ) C ( k , 4 ) = ( x , y ) By section formula,we get 2 × 5 + 3 × 2 2 + 3 = k 10 + 6 5 = k 16 = 5 k k = 16 5
Solution
w e space t a k e space t h e space p o i n t space o n space x space a x i s space i s space open parentheses x comma 0 close parentheses A space a n d space B space a l s o space i n space x space a x i s s o space t h e space p o i n t space a l s o space m i d space p o i n t space o f space A B w h e n space P open parentheses x comma y close parentheses space b e space t h e space m i d space p o i n t space o f space Q open parentheses x subscript 1 comma y subscript 1 close parentheses space a n d space R open parentheses x subscript 2 comma y subscript 2 close parentheses t h e n open parentheses x comma y close parentheses equals open parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction comma fraction numerator y subscript 1 plus y subscript 2 over denominator 2 end fraction close parentheses t h e n open parentheses x comma 0 close parentheses equals open parentheses fraction numerator negative 1 plus 5 over denominator 2 end fraction comma fraction numerator 0 plus 0 over denominator 2 end fraction close parentheses open parentheses x comma 0 close parentheses equals open parentheses 2 comma 0 close parentheses open parentheses 2 comma 0 close parentheses space i s space e q u i d i s tan c e space f r o m e space A space a n d space B
Q. Find the distance between the points ( − 8 5 , 2 ) a n d ( 2 5 2 )
Solution
Given points are ( − 8 5 , 2 ) ( 2 5 , 2 ) so x 1 = − 8 5 , y 1 = 2 x 2 = 2 5 , y 2 = 2 We know that Distance formula = √ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = √ ( 2 5 + 8 5 ) 2 + ( 2 − 2 ) 2 = √ ( 10 5 ) 2 + 0 = 10 5 = 2 units Q. Find the value of a, so that the point (3, a) lies on the line represented by 2x-3y = 5
Solution
It is given that , ( 3 , a ) lies on the line 2x - 3y - 5 = 0 Substitute x = 3 , y = a in the equation We get , 2 × 3 - 3a - 5 = 0 6 - 3a - 5 = 0 1 - 3a = 0 1 = 3a 1 3 = a a = 1 3 Q. If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.
Solution
Given, A (4, 3) and B (x, 5) are two points on the circle. Centre of the circle is O (2, 3). ∴ OB = OA (Radius of the circle) ⇒ O B 2 = O A 2 ⇒ ( x – 2 ) 2 + ( 5 – 3 ) 2 = ( 4 – 2 ) 2 + ( 3 – 3 ) 2 [Distance formula] ⇒ ( x – 2 ) 2 + 4 = 4 ⇒ ( x – 2 ) 2 = 0 ⇒ x – 2 = 0 ⇒ x = 2 Thus, the value of x is 2.

Q.

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.
Solution
P open parentheses x subscript 1 comma y subscript 1 close parentheses space & Q open parentheses x subscript 2 comma y subscript 2 close parentheses b e space 2 space p o i n t s space t h e n space d i s tan c e space b e t w e e n space t h e m P Q equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root i n space t h i s space q u e s t i o n A P equals B P square root of open parentheses x minus 7 close parentheses squared plus open parentheses y minus 1 close parentheses squared end root equals square root of open parentheses x minus 3 close parentheses squared plus open parentheses y minus 5 close parentheses squared end root s q u a r i n g space o n space b o t h space s i d e open parentheses x minus 7 close parentheses squared plus open parentheses y minus 1 close parentheses squared equals open parentheses x minus 3 close parentheses squared plus open parentheses y minus 5 close parentheses squared x squared minus 14 x plus 49 plus y squared minus 2 y plus 1 equals x squared minus 6 x plus 9 plus y squared minus 10 y plus 25 minus 14 x minus 2 y plus 50 equals negative 6 x minus 10 y plus 34 8 x minus 8 y equals 16 x minus y equals 2
Q. If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.
Solution
P open parentheses x subscript 1 comma y subscript 1 close parentheses space & Q open parentheses x subscript 2 comma y subscript 2 close parentheses b e space 2 space p o i n t s space t h e n space d i s tan c e space b e t w e e n space t h e m P Q equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root i n space t h i s space q u e s t i o n A P equals B P square root of open parentheses x minus 7 close parentheses squared plus open parentheses y minus 1 close parentheses squared end root equals square root of open parentheses x minus 3 close parentheses squared plus open parentheses y minus 5 close parentheses squared end root s q u a r i n g space o n space b o t h space s i d e open parentheses x minus 7 close parentheses squared plus open parentheses y minus 1 close parentheses squared equals open parentheses x minus 3 close parentheses squared plus open parentheses y minus 5 close parentheses squared x squared minus 14 x plus 49 plus y squared minus 2 y plus 1 equals x squared minus 6 x plus 9 plus y squared minus 10 y plus 25 minus 14 x minus 2 y plus 50 equals negative 6 x minus 10 y plus 34 8 x minus 8 y equals 16 x minus y equals 2
Q. If the centroid of △ ABC having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a+b+c).
Solution
w h e n A open parentheses x subscript 1 comma end subscript y subscript 1 close parentheses comma B open parentheses x subscript 2 comma y subscript 2 close parentheses comma C open parentheses x subscript 3 comma y subscript 3 close parentheses space a r e space t h e space v e r t i c e s space o f space increment A B C t h e n space c e n t r o i d space o f space t h e space increment A B C space i s open parentheses x comma y close parentheses equals open parentheses fraction numerator x subscript 1 plus x subscript 2 plus x subscript 3 over denominator 3 end fraction comma fraction numerator y subscript 1 plus y subscript 2 plus y subscript 3 over denominator 3 end fraction close parentheses a c c o r d i n g space t o space q u e s t i o n open parentheses 0 comma 0 close parentheses equals open parentheses fraction numerator a plus b plus c over denominator 3 end fraction comma fraction numerator b plus c plus a over denominator 3 end fraction close parentheses t h e n fraction numerator a plus b plus c over denominator 3 end fraction equals 0 a plus b plus c equals 0
Q. Find the centroid of △ ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).
Solution
w h e n A open parentheses x subscript 1 comma end subscript y subscript 1 close parentheses comma B open parentheses x subscript 2 comma y subscript 2 close parentheses comma C open parentheses x subscript 3 comma y subscript 3 close parentheses space a r e space t h e space v e r t i c e s space o f space increment A B C t h e n space c e n t r o i d space o f space t h e space increment A B C space i s open parentheses x comma y close parentheses equals open parentheses fraction numerator x subscript 1 plus x subscript 2 plus x subscript 3 over denominator 3 end fraction comma fraction numerator y subscript 1 plus y subscript 2 plus y subscript 3 over denominator 3 end fraction close parentheses a c c o r d i n g space t o space q u e s t i o n open parentheses x comma y close parentheses equals open parentheses fraction numerator 2 minus 4 plus 5 over denominator 3 end fraction comma fraction numerator 2 minus 4 minus 8 over denominator 3 end fraction close parentheses c e n t r o i d space o f space increment A B C equals open parentheses x comma y close parentheses equals open parentheses 1 comma fraction numerator negative 10 over denominator 3 end fraction close parentheses
Q. Find the centroid of △ ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).
Solution
w h e n A open parentheses x subscript 1 comma end subscript y subscript 1 close parentheses comma B open parentheses x subscript 2 comma y subscript 2 close parentheses comma C open parentheses x subscript 3 comma y subscript 3 close parentheses space a r e space t h e space v e r t i c e s space o f space increment A B C t h e n space c e n t r o i d space o f space t h e space increment A B C space i s open parentheses x comma y close parentheses equals open parentheses fraction numerator x subscript 1 plus x subscript 2 plus x subscript 3 over denominator 3 end fraction comma fraction numerator y subscript 1 plus y subscript 2 plus y subscript 3 over denominator 3 end fraction close parentheses a c c o r d i n g space t o space q u e s t i o n open parentheses x comma y close parentheses equals open parentheses fraction numerator 2 minus 4 plus 5 over denominator 3 end fraction comma fraction numerator 2 minus 4 minus 8 over denominator 3 end fraction close parentheses c e n t r o i d space o f space increment A B C equals open parentheses x comma y close parentheses equals open parentheses 1 comma fraction numerator negative 10 over denominator 3 end fraction close parentheses
Q. In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?
Solution
l e t space left parenthesis p comma q right parenthesis space d i v i d e space t h e space l i n e space o f space p o i n t s space open parentheses x subscript 1 comma y subscript 1 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 close parentheses space i n space t h e space r a t i o space m colon n t h e n open parentheses p comma q close parentheses equals open parentheses fraction numerator n x subscript 1 plus m y subscript 1 over denominator m plus n end fraction comma fraction numerator n x subscript 2 plus m y subscript 2 over denominator m plus n end fraction close parentheses i n space t h i s space q u e s t i o n p equals 4 comma q equals 5 comma x subscript 1 equals 2 comma x subscript 2 equals 7 comma y subscript 1 equals 3 comma y subscript 2 equals 8 w e space w a n d space t o space f i n d space m colon n t h e n open parentheses 4 comma 5 close parentheses equals open parentheses fraction numerator 2 n plus 7 m over denominator m plus n end fraction comma fraction numerator 3 n plus 8 m over denominator m plus n end fraction close parentheses s o 4 equals fraction numerator 2 n plus 7 m over denominator m plus n end fraction 4 m plus 4 n equals 2 n plus 7 m 2 n equals 3 m m over n equals 2 over 3 t h e n space t h e space r a t i o space i s space 2 colon 3
Q.
In what ratio does the point C(4,5) divide the join of A(2, 3) and B(7,8)?
Solution
l e t space left parenthesis p comma q right parenthesis space d i v i d e space t h e space l i n e space o f space p o i n t s space open parentheses x subscript 1 comma y subscript 1 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 close parentheses space i n space t h e space r a t i o space m colon n t h e n open parentheses p comma q close parentheses equals open parentheses fraction numerator n x subscript 1 plus m y subscript 1 over denominator m plus n end fraction comma fraction numerator n x subscript 2 plus m y subscript 2 over denominator m plus n end fraction close parentheses i n space t h i s space q u e s t i o n p equals 4 comma q equals 5 comma x subscript 1 equals 2 comma x subscript 2 equals 7 comma y subscript 1 equals 3 comma y subscript 2 equals 8 w e space w a n d space t o space f i n d space m colon n t h e n open parentheses 4 comma 5 close parentheses equals open parentheses fraction numerator 2 n plus 7 m over denominator m plus n end fraction comma fraction numerator 3 n plus 8 m over denominator m plus n end fraction close parentheses s o 4 equals fraction numerator 2 n plus 7 m over denominator m plus n end fraction 4 m plus 4 n equals 2 n plus 7 m 2 n equals 3 m m over n equals 2 over 3 t h e n space t h e space r a t i o space i s space 2 colon 3

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 on Coordinate Geometry offer several benefits to students:

Conceptual Clarity : These solutions provide step-by-step explanations that help students understand the application of the distance formula, midpoint formula, and the area of a triangle formula. This clarity aids in grasping the fundamental concepts of coordinate geometry.

Problem-Solving Skills : By working through these solutions, students enhance their problem-solving skills. The detailed solutions guide them on how to approach and solve different types of coordinate geometry problems systematically.

Exam Preparation : The solutions are aligned with the curriculum and exam pattern, making them an excellent resource for exam preparation. Practicing these problems helps students familiarize themselves with the types of questions that may appear in exams.

Confidence Building : Regular practice with these solutions builds confidence in students. As they solve more problems accurately, their confidence in tackling coordinate geometry questions in exams increases.

Error Reduction : Detailed solutions help identify common mistakes and misunderstandings. By studying the correct methods and explanations, students can avoid these errors in their work.

Time Management : Learning efficient problem-solving techniques from the solutions can help students manage their time better during exams. Understanding the quickest and most accurate methods to solve problems can save valuable time.

RS Aggarwal Solutions for Class 10 Maths Chapter 16 Exercise 16.4 FAQs

What is the use of coordinate geometry in real life?

Coordinate geometry has many practical uses in daily life. All of the maps we use to find places—including physical maps and Google Maps—are based on the coordinate system. Additionally, drawing the land maps to scale is beneficial for large-scale land projects.

What are the rules of coordinate geometry?

Coordinate geometry considers points as ordered pairs that are represented as (x,y), lines can be represented by equations like ax+ by + c = 0, and circles as (x−a)2+(y−b)2=r2, where (a,b) are the coordinates of the center of the circle and r is the radius.

What are the basic things in co ordinate geometry?

Coordinate geometry allows us to find measurements of geometric figures that can be graphed with coordinates.

Who is the father of coordinate geometry?

René Descartes
Popup Close ImagePopup Open Image
Talk to a counsellorHave doubts? Our support team will be happy to assist you!
Popup Image
CTA Image
07019243492
Check out these Related Articles
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions of Class 10 English First Flight Chapter 4 Question Answer
Read Full Story
24 Jul 2025, 09:05 am
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction
Read Full Story
22 Jul 2025, 12:12 pm
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions for Class 10 English First Flight Chapter 2 PDF
Read Full Story
21 Jul 2025, 01:44 pm
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions for Class 10 English First Flight Chapter 3 Question Answer
Read Full Story
19 Jul 2025, 11:13 am
BlogImage
CLASS 10 EXAM
Share
CBSE Class 10 Compartment Admit Card 2025, Check Expected Date
Read Full Story
8 Jul 2025, 04:43 pm
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination
Read Full Story
5 Jul 2025, 02:36 pm
BlogImage
CLASS 10 EXAM
Share
CBSE Class 10 Compartment Exam Date 2025 OUT Download PDF
Read Full Story
28 Jun 2025, 11:20 am
BlogImage
CLASS 10 EXAM
Share
NCERT Solutions for Class 10 Science Chapter 6 Life Processes
Read Full Story
27 Jun 2025, 08:08 am
BlogImage
CLASS 10 EXAM
Share
CBSE Class 10 Board Exams Conducted Twice a Year from 2026 February Mandatory, May Optional
Read Full Story
26 Jun 2025, 12:38 pm
Join 15 Million students on the app today!
Point IconLive & recorded classes available at ease
Point IconDashboard for progress tracking
Point IconMillions of practice questions at your fingertips
Download ButtonDownload Button
Banner Image
Banner Image
Free Learning Resources
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
Privacy Policy
Terms of use
Copyright © 2025 Physicswallah Limited All rights reserved.