# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

Mathematics is one of the subjects that has its own uses. All students need to have a foundational understanding of Mathematics, no matter if they are from the Arts, Sciences, or Commerce stream. Understanding certain fundamental concepts in Mathematics will benefit students in the long run, in addition to helping them do well on their board examinations. However, many students are not interested in studying Mathematics these days because of its difficulty.

Mathematics is a subject that has applications in both academic work and daily life. It might be difficult to get students interested in this subject as it is a challenging one for them to grasp since it requires a lot of mental effort. Students frequently struggle with the concept of Mathematics.

Counting is the foundation or starting point of Mathematics, where humans learned to count the things that are visible to the eye. Pure Mathematics (numerical systems, geometry, matrices, algebra, combinatorics, topology, and calculus) and Applied Mathematics are the two main categories of Mathematics (Engineering, Chemistry, Physics, numerical analysis, etc). Numerous formulas in Mathematics are based on various ideas. By repeatedly answering questions based on these formulas, one can learn them. It is possible to solve some problems quickly using mathematical techniques.

One of the most important topics in Mathematics is called Coordinate Geometry. By using graphs with curves and lines, Coordinate Geometry (also known as Analytic Geometry) explains how Geometry and Algebra are related. Students can answer geometrical problems because it gives them geometrical aspects of Algebra. In this area of Geometry, a pair of ordered numbers is used to describe the location of points on a plane. The study of Geometry using coordinate points is known as Coordinate Geometry. It is possible to determine the distance between two points by dividing a line into m:n segments, locating a line’s midpoint, determining the area of a triangle in the Cartesian plane, and performing other calculations using Coordinate Geometry.

Mathematics has different concepts and formulas that students are not familiar with. It is also considered a complex subject, so students often face problems regarding the same. The National Council of Educational Research and Technology was introduced by the Central Board of Secondary Education. It was created to promote a united educational system for the nation with a national character as well as enable and encourage the various cultural traditions that are practised across the board. Conducting, promoting, and coordinating research in fields connected to school education is one of NCERT’s primary objectives, as well as that of its units. NCERT provides information about the world and addresses all important subjects. Its content is clear and simple for students to understand.

The selection of study materials available to students will make it difficult for them to choose which to employ as they get ready for their examinations. To get excellent marks on their board examinations, several of them frequently spend time looking for the greatest study resources. To assist students in their examination preparation, NCERT has taken on the responsibility of developing and spreading its own study resources. Because they cover the entire CBSE curriculum and are primarily concerned with assisting the students in laying a strong foundation for their higher levels of education, NCERT books are extremely advantageous for CBSE students. In the board examination, CBSE takes precedence over NCERT.

NCERT books are tremendously beneficial for CBSE students because they cover the entire CBSE curriculum and are primarily concerned with assisting the students in laying a strong foundation for their succeeding levels of education. The NCERT series of books, which are complete, informed, and well-written, was written by a team of specialists to help students in their studies. When compared to other study aids, these NCERT books provide students with an ample number of benefits for preparing for their board examinations. The top scorers’ recommendation for the best study materials is the NCERT books.

Subject-matter experts wrote these NCERT books following in-depth study on a specific topic, which helps establish a strong foundation for each subject. Additionally, it offers precise and easy language together with comprehensive and in-depth information about the concept. The NCERT Mathematics solutions offer all key ideas and comprehensive explanations to aid in students’ comprehension. These NCERT study tools offer the proper framework for board examination as well as other tests. These textbooks are highly recommended by nearly all institutions that are affiliated with the CBSE, since NCERT closely follows the CBSE curriculum and provides students with in-depth knowledge.

**NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4**

In Coordinate Geometry, any point, figure, or object’s position is specified using two or more numbers. The location of the object can be specified in a variety of ways, including a line, a plane, three dimensions, etc. The Cartesian coordinate system is broadly used. It is one type of many that use a pair of numbers (coordinates), which are the signed distances between each point and two fixed perpendicular lines known as the axes, to represent each point. There are many different concepts in the chapter Coordinate Geometry which includes graphs, lines, etc. Therefore, students face difficulty in understanding the chapter clearly, as NCERT textbooks do not contain detailed solutions. Students can refer to the Extramarks website for reliable and concise study material.

Extramarks is an online learning platform that concentrates on the K–12, higher education, and Test Prep segments to facilitate studying anywhere and at any time. It makes sure that concepts are learned through interactive video modules on the Learning App. These modules are developed by a team of highly competent in-house subject matter specialists, giving 360-degree coverage of every concept and enabling immersive online learning to improve \ comprehension and retention throughout examination preparation. It offers a variety of school-based technologies, including the Assessment Centre, Smart Class Solutions, and Live Class Platform, to assist students in realising their full potential through engaging education and individualised curriculum-based learning.

Sometimes it happens that students cannot finish their complete course in a timely manner. To ensure that they do not lose any marks in any in-class or board exams, the Extramarks website gives students complete coverage of the course. To help students grasp every subject thoroughly, Extramarks offers them worksheets broken down by chapter, interactive exercises, an infinite number of practise questions, and more. So that they can map their performance, students can assess their level of preparation. Using a self-learning tool, students can study at their own pace using the Extramarks website. They are able to track their level of preparation while practising tests that are customisable. Through the use of engaging graphics and animations, Extramarks helps students study while having interesting modules, preventing them from finding their studies irrelevant. Curriculum-mapped instruction is provided to students by the Extramarks’ website so that they do not have to refer elsewhere for reliable solutions.

The Extramarks website provides students with the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. These NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are developed by Extramarks’ experts to aid Class 10 board examination students in effective study methods. In order for students to solve the NCERT Solutions quickly and easily, these experts create the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. In addition to paying attention to how quickly students can grasp the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, they ensure that they can learn from it. It includes additional practise questions that are connected to the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. Each question in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 is accompanied by an in-depth expert solution. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are developed in accordance with NCERT standards with the goal of covering the complete curriculum of the chapter. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are extremely helpful for achieving successful examination outcomes.

On the Extramarks website, students will find the entire set of notes for Class 10 Maths Exercise 7.4 Solutions. Students can learn the position of a point on a plane using Coordinate Geometry. When the coordinates of the points are given, students will learn how to use the distance formula to calculate the distance between two points and the area of a triangle.

Students who read the in-depth NCERT Solutions offered by our subject-matter experts will be able to score higher marks. It follows the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, which help students prepare in the right way. From the viewpoint of the examination, it comprises all the important questions. It helps students earn good marks in board exams for Mathematics.Students can complete and go over all of the questions in Class 10 Maths Chapter 7 Exercise 7.4 by using the NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.4.

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are available on Extramarks. This online learning website is used as an example to demonstrate how technology may improve the efficiency and clarity of the learning process. In order to promote academic excellence, the Extramarks website is dedicated to safeguarding the growth and success of students. The first and most important step in preparing for Mathematics is to practise the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 since it helps students organise and improve their key concepts. Students can easily handle any complex questions that come up in their school, competitive, or board examinations if they practise the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 from the NCERT textbook. Students may find dependable study materials like the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 on the Extramarks website to aid them in performing better on examinations.

**Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry**

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are provided by Extramarks to students who are registered with its website to aid them in their academics. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are created by subject matter experts at Extramarks to aid students in their preparation for the Class 10 board examination. These experts create the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 to help students rapidly tackle the NCERT questions. Along with ensuring that students can acquire this material rapidly, they also consider the students’ comprehension of the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4.

Extramarks is a website that provides the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 so that students may comprehend all the concepts. The top experts in India, who are specialists in their professions, are available to educate the students. The website has extensive media-based learning modules that go over the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 in great depth. Students who consistently place an emphasis on concepts, conduct in-depth research, and learn quickly see gradual improvements in their grades and learning outcomes. The Extramarks website offers chapter-based worksheets, interactive activities, unlimited practise questions, and more in order to help students fully understand the concept. Students can check their knowledge with adaptive quizzes that have progressively harder difficulty levels, MCQs, and mock exams to form an upward learning graph.

Extramarks provides the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 to assist students to prepare for and do well in the CBSE Class 10 Maths examination. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 is based on the CBSE Board most recent syllabus. Students can prepare effectively for the board examination by practising the key questions for each chapter of the 10th standard Mathematics course. Students will receive complete solutions. Students can consult the NCERT Solutions from the website if they face trouble when attempting to answer the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4.

The solved sample papers, past years’ papers, and other study materials are all available on the Extramarks website related to the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. Students who want to perform well on their examination can use the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. Students can learn independently and at their own speed with the Extramarks online learning platform. They can assess their progress using data supported by AI. To fully master all subjects and themes, Extramarks’ offers chapter-based worksheets, interactive activities, unlimited practise questions, and more.

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are available on the Extramarks’ website, where students may learn all the concepts. Furthermore, Extramarks offers live question-answer sessions where students can ask teachers any question they have about the NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.4.Teachers create worksheets for their students to assist them to identify their areas of weakness, so they may improve on them and perform well on the examination. Extramarks teachers tackle students’ queries about the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 that they are reluctant to ask their school teachers, thereby clearing their concepts.

The formulas and techniques utilised in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 might occasionally be challenging to comprehend. With the help of the easily accessible NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 on Extramarks, students may revise at any time. The steps and concepts required to finish the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 must be known by the students. Many websites provide the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 to students, but it is necessary for students to choose the right study material in order to avoid mistakes in examinations and score high marks. The Extramarks website is an authenticated website that provides reliable NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 to the students who are registered to its website. Therefore, the Extramarks website is a reliable source for students to find the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4.

**NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4 **

Mathematics is a fascinating subject. But it is still a daunting one for some students. They can build such a solid foundation in Mathematics at the Extramarks website that they will find solving any mathematical problems to be simple. Understanding and practising the fundamentals of Mathematics will help students develop their logical thinking and make it easier for them to quickly and effectively understand the concepts in all other subjects.

Based on the curriculum, Extramarks is offering NCERT Solutions such as NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11, NCERT Solutions Class 12. These courses introduce students to the universe of all the key mathematical concepts, such as Coordinate Geometry. It is a good idea for parents to introduce their children to Mathematics in interesting ways because research shows that students are more receptive to learning when it is associated more with a story than with study. Extramarks offers instruction to students up to the 12th grade, during which time they study the most challenging subjects including 3-D Geometry, Vector Algebra, Differential Equations, Matrices, etc.

Extramarks make the textbook concepts interesting by adding graphics and animations. The 2D and 3D animation teams are skilled at making concepts engaging, comprehension and retention during online learning. With the help of animations and creative teaching techniques used by professionals, students will understand the concept thoroughly, and thoughts and concepts will be permanently ingrained in their minds.Making Mathematics enjoyable is intended to help students realise how simple it can be. Students will then be enthusiastic about the subject rather than seeing it as an obstacle. Furthermore, they can challenge themselves by participating in interactive video sessions, regularly taking exams that the specialists have developed, and continuously analysing their performance.

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are designed to aid students in their exam preparation. Along with NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, all of the problems from the NCERT textbooks are present. In order to aid students in acquiring a complete grasp of the concepts taught in Exercise 7.4 Class 10 Maths, the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are described in a practical and detailed manner. Students might feel more confident that they will perform well in their examination because most of the questions in the NCERT frequently appear in the examination. The purpose of the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 is to help students understand the fundamentals and aid students grasp the principles of each topic, thereby boosting their core knowledge.

For students in Class 10, Mathematics is a crucial subject. The structure of the curriculum enables students to acquire new knowledge while developing a conceptual foundation for further study of more complicated subjects. The teachers at Extramarks provide the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 that help students understand the concepts clearly. Students may view the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 PDF file offline by downloading it. The various types of equations and their necessary elements are discussed in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. Using the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 and finishing this chapter will make it straightforward for students to learn the concepts and deal with the exercise questions. To swiftly resolve any questions while working on the issues, students should refer to the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. For all classes and courses, NCERT solutions are available on the Extramarks website.

To avoid problems and questions regarding the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, it is advisable that students study from the Extramarks NCERT textbook solutions. Extramarks is a simple option for students in all classes and subjects.Students have never found Mathematics to be simple, but with Extramarks’ assistance, they will find conceptualising Mathematics to be less difficult. Students will be able to get the best results when they combine practise and Extramarks’ advice. The Extramarks website offers a wide range of top-notch educational courses for students to take in order to learn, practice, and succeed.

Students are guided through the chapter summary in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. These NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are presented in step-by-step detail and organised by Extramarks’ subject matter experts. These processes allow students to understand the concepts of the chapter and the logic underlying the questions with ease. In order to lay the groundwork for their core ideas, students must completely know the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. After practising the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, they will be able to tackle any difficult problems they encounter during the examination with ease. Along with the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, Extramarks also provides its students with sample papers, past years’ papers, revision notes, and important questions about all the topics and classes. Sample papers and past years’ papers are two of the best resources for students to study for examinations. By reviewing sample papers and past years’ papers, students can increase their confidence and perform well on their board examinations.

Human logic and thought heavily rely on Mathematics. It is a significant technique for strengthening mental strength and logical reasoning. Furthermore, a solid foundation in Mathematics is required to fully understand the principles of other subjects, including Science, Social Science etc. Mathematics is used in many different industries and professions. The ideas and methods of Mathematics are used to answer questions in Engineering, Science, and Economics. Mathematics is one of the most crucial subjects in a student’s academic career. Students must consistently practice the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 if they want to succeed in the topic. To perform well on their Class 10 board exams, students should also have a solid understanding of Mathematics because it is a conceptual subject.

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 will not be challenging for students. With regular practise and the right help from Extramarks, students can easily learn the answers to the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 from the Extramarks website and do well on their exams. To aid students in understanding the content, Extramarks offers in-depth NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. If students are hesitant to express questions in front of their peers at school, they can also participate in the doubt sessions provided on the Extramarks’ website.

Students should be familiar with the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 in order to get an idea of the types of questions that might be asked in the examination. Therefore, before taking their exam, Extramarks encourages students to review a number of sample papers and past years’ papers. Students can advance in their preparation by comprehending the ideas offered in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. The chapter’s content may occasionally be challenging, but with the aid of NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4, students may easily comprehend it.

**Q.1 **

$\begin{array}{l}\text{Determine the ratio in which the line\hspace{0.33em}}2\mathrm{x}+\mathrm{y}-4=0\text{\hspace{0.33em}divides the line segment joining the points A(}2,\text{}-2\text{)}\\ \text{and B(3, 7).\hspace{0.17em}}\end{array}$

**Ans**

$\begin{array}{l}\text{Let the given line divide the line segment joining the points}\\ \text{A(2,}-2)\text{and B(3, 7) in the ratio}\mathrm{k}:1.\\ \text{Coordinates of the point of division}=(\frac{3\mathrm{k}+2}{\mathrm{k}+1},\text{\hspace{0.17em}\hspace{0.17em}}\frac{7\mathrm{k}-2}{\mathrm{k}+1})\\ \text{This point of division lies on the line 2}\mathrm{x}+\mathrm{y}-4=0.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\frac{3\mathrm{k}+2}{\mathrm{k}+1}+\frac{7\mathrm{k}-2}{\mathrm{k}+1}-4=0\\ \text{or 6}\mathrm{k}+4+7\mathrm{k}-2-4\mathrm{k}-4=0\\ \text{or 9}\mathrm{k}-2=0\\ \text{or}\mathrm{k}=\frac{2}{9}\\ \text{Therefore, the required ratio is 2:9}\end{array}$

**Q.2 **

$\text{Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.}$

**Ans**

$\begin{array}{l}\text{Points (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{) are collinear if}\\ {\mathrm{x}}_{1}({\mathrm{y}}_{2}-{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}-{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}-{\mathrm{y}}_{2})=0\\ \text{Therefore,}\\ \text{Points (x,}\mathrm{y}\text{)},\text{(1, 2) and (7, 0) are collinear if}\\ \text{}\mathrm{x}(2-0)+1(0-\mathrm{y})+7(\mathrm{y}-2)=0\\ \text{or}2\mathrm{x}-\mathrm{y}+7\mathrm{y}-14=0\\ \text{or}2\mathrm{x}+6\mathrm{y}-14=0\\ \text{or}\mathrm{x}+3\mathrm{y}-7=0\text{}\end{array}$

**Q.3 ****Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).
**

**Ans**

$\begin{array}{l}\text{Let O(x, y) be the centRE of the circle and A(6,}-6),\text{B(3,}-7)\\ \text{and (3, 3) are points on the circle.}\\ \text{Then OA, OB and OC are radii of the circle and}\\ \text{OA}=\text{OB}=\text{OC.}\\ \text{Therefore, by distance formula, we have}\\ \text{}\sqrt{{(\mathrm{x}-6)}^{2}+{(\mathrm{y}+6)}^{2}}=\sqrt{{(\mathrm{x}-3)}^{2}+{(\mathrm{y}+7)}^{2}}\\ \text{or}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}+12\mathrm{y}+36={\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}+14\mathrm{y}+49\\ \text{or}-6\mathrm{x}-2\mathrm{y}+14=0\\ \text{or}-3\mathrm{x}-\mathrm{y}+7=0\\ \text{or}\mathrm{y}=-3\mathrm{x}+7\text{}...\text{(1)}\\ \text{Also,}\\ \text{OA}=\text{OC}\\ \text{or}\sqrt{{(\mathrm{x}-6)}^{2}+{(\mathrm{y}+6)}^{2}}=\sqrt{{(\mathrm{x}-3)}^{2}+{(\mathrm{y}-3)}^{2}}\\ \text{or}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}+12\mathrm{y}+36={\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-6\mathrm{y}+9\\ \text{or}-6\mathrm{x}+18\mathrm{y}+54=0\\ \text{or}-\mathrm{x}+3\mathrm{y}+9=0\\ \text{or}-\mathrm{x}+3(-3\mathrm{x}+7)+9=0\text{[From (1),}\mathrm{y}=-3\mathrm{x}+7\text{]}\\ \text{or}-\mathrm{x}-9\mathrm{x}+21+9=0\\ \text{or}-\text{10}\mathrm{x}+30=0\\ \text{or}-\mathrm{x}+3=0\\ \text{or}\mathrm{x}=3\\ \text{On putting this value of x in (1), we get}\\ \\ \text{}\mathrm{y}=-3\times \left(3\right)+7=-2\\ \text{Therefore, coordinates of the centre are (}3,\text{}-2).\text{}\end{array}$

**Q.4 ****The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
**

**Ans**

$\begin{array}{l}\text{Let ABCD be a square having A(}-1,\text{}2\left)\text{and C(3, 2}\right)\text{as opposite}\\ {\text{vertices. Let other two opposite vertices are B(x, y) and D(x}}_{1},{\text{y}}_{1}).\\ \text{We know that sides of a square are equal to each other.}\\ \therefore \text{AB}=\text{BC}\\ \text{or}\sqrt{{(\mathrm{x}+1)}^{2}+{(\mathrm{y}-2)}^{2}}=\sqrt{{(\mathrm{x}-3)}^{2}+{(\mathrm{y}-2)}^{2}}\\ \text{or}{\mathrm{x}}^{2}+2\mathrm{x}+1+{(\mathrm{y}-2)}^{2}={\mathrm{x}}^{2}-6\mathrm{x}+9+{(\mathrm{y}-2)}^{2}\\ \text{or}8\mathrm{x}=8\\ \text{or}\mathrm{x}=1\\ \text{We know that all interior angles are of 90\xb0 in a square.}\\ \text{Therefore, using Pythagoras theorem in}\mathrm{\Delta}\text{ABC, we have}\\ {\text{AB}}^{2}+{\text{BC}}^{2}={\text{AC}}^{2}\\ \text{or}{(\mathrm{x}+1)}^{2}+{(\mathrm{y}-2)}^{2}+{(\mathrm{x}-3)}^{2}+{(\mathrm{y}-2)}^{2}={(3+1)}^{2}+{(2-2)}^{2}\\ {\text{or x}}^{2}+2\mathrm{x}+1+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-4\mathrm{y}+4=16\\ {\text{or 2x}}^{2}+2{\mathrm{y}}^{2}-4\mathrm{x}+2-8\mathrm{y}=0\\ \text{or 2}+2{\mathrm{y}}^{2}-4+2-8\mathrm{y}=0\text{[}\mathrm{x}=1]\\ {\text{or 2y}}^{2}-8\mathrm{y}=0\\ {\text{or y}}^{2}-4\mathrm{y}=0\\ \text{or y(y}-4)=0\\ \text{i.e., y}=0\text{or y}=4\\ \text{We know that in a square, the diagonals are of equal length}\\ \text{and bisect each other at 90\xb0. Let O be the mid-point of AC.}\\ \text{Therefore, it is also the mid-point of BD.}\\ \text{Coordinates of point O}=(\frac{-1+3}{2},\text{}\frac{2+2}{2})=(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2})\\ \text{i.e.,}(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2})=(1,\text{}2)\\ \text{i.e.,}\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2}=1\text{and}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}=2\\ \text{i.e.,}\mathrm{x}+{\mathrm{x}}_{1}=2\text{and}\mathrm{y}+{\mathrm{y}}_{1}=4\\ \text{Putting the values of x and y, we get}{\mathrm{x}}_{1}=1\text{and}{\mathrm{y}}_{1}=0\text{or}{\mathrm{y}}_{1}=4.\\ \text{Therefore, the required coordinates of other two opposite vertices}\\ \text{of the given square are (1, 0) and (1, 4).}\end{array}$

**Q.5 ****The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown
in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
**

**Ans**

$\begin{array}{l}\text{(i)}\\ \text{We take A as origin and AD as x-axis and AB as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta}\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_{1}({\mathrm{y}}_{2}-{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}-{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}-{\mathrm{y}}_{2})\right]\\ \text{}=\frac{1}{2}\left[4(2-5)+3(5-6)+6(6-2)\right]\\ \text{}=\frac{1}{2}[-12-3+24]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\text{square units}\\ \text{(ii)}\\ \text{We take C as origin and CB as x-axis and CD as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (12, 2), (13, 6) and (10, 3) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta}\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_{1}({\mathrm{y}}_{2}-{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}-{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}-{\mathrm{y}}_{2})\right]\\ \text{}=\frac{1}{2}\left[12(6-3)+13(3-2)+10(2-6)\right]\\ \text{}=\frac{1}{2}[36+13-40]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\text{square units}\\ \text{We observe that area of the triangle in both cases is same.}\end{array}$

**Q.6 **

$\begin{array}{l}\text{The vertices of a}\mathrm{\Delta}\text{\hspace{0.17em}ABC are A(4, 6), B(1, 5) and C(7, 2).}\\ \text{A line is drawn to intersect sides AB and AC at D and E}\\ \text{respectively, such that}\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{1}}{4}.\text{Calculate the area}\\ \text{of the}\mathrm{\Delta}\text{\hspace{0.17em}ADE and compare it with the area of}\mathrm{\Delta}\text{\hspace{0.17em}ABC.}\\ \text{(Recall converse of basic proportionality theorem and}\\ \text{Theorem 6.6 related to the ratio of the areas of two}\\ \text{similar triangles).}\end{array}$

**Ans**

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}=4\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}-1=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}-1=4-1\\ \text{or}\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}}=3\\ \text{or}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1}{3}\\ \text{Therefore, D and E are two points on side AB and AC respectively}\\ \text{such that they divide side AB and AC in the ratio 1:3.}\\ \text{Coordinates of point D}=(\frac{1\times 1+3\times 4}{1+3},\text{}\frac{1\times 5+3\times 6}{1+3})\\ \text{}=(\frac{13}{4},\text{}\frac{23}{4})\\ \text{Coordinates of point E}=(\frac{1\times 7+3\times 4}{1+3},\text{}\frac{1\times 2+3\times 6}{1+3})\\ \text{}=(\frac{19}{4},\text{}\frac{20}{4})\\ \text{Area of a triangle}=\frac{1}{2}\left[{\mathrm{x}}_{1}({\mathrm{y}}_{2}-{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}-{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}-{\mathrm{y}}_{2})\right]\\ \text{Area of}\mathrm{\Delta}\text{ADE\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[4(\frac{23}{4}-\frac{20}{4})+\frac{13}{4}(\frac{20}{4}-6)+\frac{19}{4}(6-\frac{23}{4})]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[3-\frac{13}{4}+\frac{19}{16}]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\frac{48-52+19}{16}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{32}\text{square units}\\ \text{Area of}\mathrm{\Delta}\text{ABC\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[12-4+7]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{2}\text{\hspace{0.17em}\hspace{0.17em}square units}\\ \text{Ratio between the areas of}\mathrm{\Delta}\text{ADE and}\mathrm{\Delta}\text{ABC is 1:16.}\end{array}$

**Q.7 **

$\begin{array}{l}\text{Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of}\mathrm{\Delta}\text{ABC.}\\ \text{(i) The median from A meets BC at D. Find the}\\ \text{coordinates of the point D.}\\ \text{(ii) Find the coordinates of the point P on AD such that}\\ \text{AP}:\text{PD}=\text{2}:\text{1}\\ \text{(iii) Find the coordinates of points Q and R on medians}\\ \text{BE and CF respectively such that BQ}:\text{QE}=\text{2}:\text{1}\\ \text{and CR}:\text{RF}=\text{2}:\text{1.}\\ \text{(iv) What do yo observe?}\\ \text{[Note: The point which is common to all the three}\\ \text{medians is called the centroid and this point divides}\\ \text{each median in the ratio 2}:\text{1.]}\\ \text{(v) If A(}{\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{) are the vertices of}\\ \text{}\mathrm{\Delta}\text{ABC, find the coordinates of the centroid of the}\\ \text{triangle.}\end{array}$

**Ans**

$\begin{array}{l}\text{(i)}\\ \text{Median AD of}\mathrm{\Delta}\text{ABC bisects BC. So, D is the mid-point of}\\ \text{side BC.}\\ \therefore \text{Coordinates of D}=(\frac{6+1}{2},\text{}\frac{5+4}{2})=(\frac{7}{2},\text{}\frac{9}{2})\\ \text{(ii)}\\ \text{Point P divides the side AD in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of P}=(\frac{2\times \frac{7}{2}+1\times 4}{2+1},\text{}\frac{2\times \frac{9}{2}+1\times 2}{2+1})=(\frac{11}{3},\text{}\frac{11}{3})\\ \text{(iii)}\\ \text{Median BE of the triangle will divide the side AC in two equal parts.}\\ \text{Therefore, E is the mid-point of the side AC.}\\ \therefore \text{Coordinates of E}=(\frac{4+1}{2},\text{}\frac{2+4}{2})=(\frac{5}{2},\text{}3)\\ \text{Point Q divides the side BE in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of Q}=(\frac{2\times \frac{5}{2}+1\times 6}{2+1},\text{}\frac{2\times 3+1\times 5}{2+1})=(\frac{11}{3},\text{}\frac{11}{3})\\ \text{Median CF of the triangle will divide the side AB in two}\\ \text{equal parts.}\\ \text{Therefore, F is the mid-point of the side AB.}\\ \therefore \text{Coordinates of F}=(\frac{4+6}{2},\text{}\frac{2+5}{2})=(5,\text{}\frac{7}{2})\\ \text{Point R divides the side CF in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of R}=(\frac{2\times 5+1\times 1}{2+1},\text{}\frac{2\times \frac{7}{2}+1\times 4}{2+1})=(\frac{11}{3},\text{}\frac{11}{3})\\ \text{(iv)}\\ \text{We observe that coordinates of point P, Q and R are same.}\\ \text{Therefore, all these are representing the same point on}\\ \text{the plane i.e., the centroid of the triangle.}\\ \text{(v)}\\ \text{We consider}\mathrm{\Delta}\text{ABC having vertices A(}{\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and}\\ \text{C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{). Let AD is median and so D bisects side BC.}\\ \therefore \text{Coordinates of D}=(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2},\text{}\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2})\\ \text{Let O is the centroid of}\mathrm{\Delta}\text{ABC}.\text{So, O divides median AD in}\\ \text{the ratio 2:1.}\\ \therefore \text{Coordinates of centroid}\mathrm{O}=(\frac{{\mathrm{x}}_{1}+2\left(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2}\right)}{2+1},\text{}\frac{{\mathrm{y}}_{1}+2\left(\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2}\right)}{2+1})\\ \text{}=(\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{3},\frac{{\mathrm{y}}_{1}+{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{3})\end{array}$

**Q.8 ****ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
**

**Ans**

$\begin{array}{l}\text{P is the mid-point of side AB.}\\ \therefore \text{Coordinates of P}=(\frac{-1-1}{2},\text{}\frac{-1+4}{2})=(-1,\text{}\frac{3}{2})\\ \text{Similarly, coordinates of Q, R and S are}(2,\text{}4),\text{}(5,\text{}\frac{3}{2})\\ \text{and (2,}-1)\text{respectively.}\\ \text{Now, we find lengths of each side and each diagonal of}\\ \text{quadrilateral PQRS by using distance formula.}\\ \text{PQ}=\sqrt{{(-1-2)}^{2}+{(\frac{3}{2}-4)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{QR}=\sqrt{{(2-5)}^{2}+{(4-\frac{3}{2})}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{RS}=\sqrt{{(5-2)}^{2}+{(\frac{3}{2}+1)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{SP}=\sqrt{{(2+1)}^{2}+{(-1-\frac{3}{2})}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{PR}=\sqrt{{(-1-5)}^{2}+{(\frac{3}{2}-\frac{3}{2})}^{2}}=\sqrt{36+0}=6\\ \text{QS}=\sqrt{{(2-2)}^{2}+{(4+1)}^{2}}=\sqrt{0+25}=5\\ \text{We find that all sides of the quadrilateral PQRS are of equal}\\ \text{length, but diagonals are of different length. Therefore,}\\ \text{quadrilateral PQRS is a rhombus.}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. For students to study for the chapter, are the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 adequate?

The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 are sufficient for students to effectively prepare for the chapter. Students can build upon their foundational knowledge and prepare for any chapter-related problems by using the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4. For better preparation, students can also practice past years’ papers and sample papers. Students who fully comprehend the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.4 score higher on the examination.

### 2. Why are students encouraged to use Extramarks to study for examinations?

Students are encouraged to use Extramarks to study for examinations to ensure that all the chapter’s ideas and formulas are fully comprehended by the time they take the examination. Extramarks offers worksheets based on chapters, interactive exercises, an endless supply of practice questions, and more. The Extramarks website offers adaptive quizzes, MCQs, and mock examinations to help students who have enrolled track their progress. The Extramarks experts produce in-depth reports and analysis to help students get the most out of their time learning on the platform. Students’ strengths and weaknesses are highlighted in this performance analysis, allowing them to focus and improve on the concepts they find challenging to do well in the examination.