# NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

Class 10 is undoubtedly one of the most important years during the education of a student. It is the year when students have to appear for the board examinations that hold remarkable significance in the career of a student. Students in Class 10 experience challenges that they have not faced before, as the board examinations are a new experience for them. The lessons that students are taught in Class 10 influence them throughout their life. The knowledge gained in Class 10 is applied in a variety of ways throughout a person’s lives.. Each topic learnt in Class 10 can play a significant role for students in students’ success.. It requires tremendous effort for students in Class 10 to achieve success in Class 10 board examinations. Being successful in the Class 10 board examinations can help students throughout their careers. The results of the Class 10 board examinations are an important factor for admissions into various institutes of higher education. Being a significant indicator of a student’s academic performance, the results of Class 10 are also taken into consideration for admissions to colleges and universities as well as for securing employment opportunities in the future. Performing well in the board examinations for Class 10 will also boost the confidence of the students and will make them not fear other examinations in the future. They can consequently perform better in any examination in the future. Students are required to study a variety of subjects in Class 10. As a result, Class 10 serves as an important year for students to make important career decisions. Following the completion of Class 10, students are required to make decisions about choosing a stream based on their interests and abilities. They can identify their subjects of interest and choose a suitable career for themselves subsequently.

For students of Class 10, Mathematics is one of the most significant and challenging subjects. Mathematics requires a tremendous amount of practisein problem-solving in addition to a deep comprehension of the principles. It is necessary to practise a lot of questions if one wants to perform well on the board examinations in Mathematics. Noticeably, most board examination questions are derived from NCERT textbooks. The NCERT book is therefore the most crucial tool for practising problems in Mathematics. However, Class 10 students may find it difficult to complete the NCERT exercises on their own. NCERT Solutions, such as the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1, might be useful in these situations. To complete the chapter and receive higher grades, students must be able to solve every problem in the Class 10 Maths Chapter 7 Exercise 7.1. Although these problems might initially seem challenging, with practise they will become simpler to resolve. The Extramarks learning platform can assist students who have difficulty understanding questions by helping them comprehend them better and solve the questions more quickly.Students can use the Extramarks learning platform to access a wide range of instructional modules and a wealth of study resources, including the NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1, to meet their demands during board examination preparation.The NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 on the Extramarks website concentrate on a single chapter. These solutions serve as a useful tool for learning, practising, and answering questions. The fact that the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 from Extramarks have been created and compiled by experienced subject specialists ensures that all students will have a great learning experience.

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.1) Exercise 7.1

The most crucial method for getting ready for board examinations is solving NCERT exercises. Hence, it is typically advised that students practise the NCERT textbook questions. In order to verify their answers, they also need the right answers to the questions. Through NCERT Solutions, students may be able to receive the help they need when preparing for board examinations. They can apply the concepts from the lesson in a more simple and in-depth manner by using the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1. Thorough and error-free NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 are authored by highly qualified teachers. As a result, these solutions have been developed with consideration for the requirements and demands of Class 10 students. Solutions to the problems in the NCERT textbook have been divided into smaller, more manageable steps in order to make writing solutions easier. Teachers emphasise how crucial it is to give thorough, step-by-step answers in the examinations. In order to achieve the best results, students should carefully follow the instructions in the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1.

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### Chapter 7 Exercise 7.1: Topics Covered

Chapter 7 is based on a crucial concept in Mathematics called Coordinate Geometry. This chapter includes concepts that provide more details on some concepts that students have learned in Class 9. By studying this chapter, students will be able to find the distance between two points whose coordinates are known using the distance formula. Class 10 Maths Ch 7 Ex 7.1 covers the topic of Distance Formula. There are 10 questions in this exercise that involve finding the distance between two points. Students can refer to the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 while solving this exercise to learn how to approach the questions in this exercise. Students also learn the section formula and area of a triangle in this chapter. They can learn the method of finding the area of a triangle formed by three given points. They can also learn to find the coordinates of the point that divides a line segment joining two points in a given ratio. Students should thoroughly read all these concepts and practise questions pertaining to them. This will not only help them score better marks in the board examinations but also help them in their future studies as well.

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### Access NCERT solution for Class 10 Maths Chapter 7 – Coordinate Geometry

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### NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

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Along with the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1, Class 10 students studying for their board examinations can also acquire the NCERT Solutions Class 10 for all the other subjects. The NCERT Solutions Class 10 are available in both Hindi and English to better serve the needs of students who study these two languages. NCERT Solutions for Class 12 are also available for Mathematics and all other subjects for students in Class 12 preparing for the Class 12 board examinations.Extramarks provides NCERT Solutions for Class 11 students in Mathematics and other disciplines.NCERT Solutions are available in Hindi and English for all subjects for students in other classes as well. Along with NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 9, and NCERT Solutions Class 6, students can find NCERT Solutions Class 10 on the Extramarks website and smartphone application. For students in primary schools, Extramarks also provides NCERT Solutions, including NCERT Solutions Class 5, NCERT Solutions Class 2, NCERT Solutions Class 4, NCERT Solutions Class 3 and NCERT Solutions Class 1 for the NCERT textbook series. Students can obtain the PDF of the Class 10 Maths Ex 7.1 Solutions from the Extramarks website and Learning App.

### NCERT Solutions for Class 10 Maths Chapter 7 All Other Exercises

The chapter on Coordinate Geometry includes a total of four exercises. Exercise 7.1 includes questions based on the Distance Formula. There are 10 questions in this exercise. Students can refer to Extramarks’ NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 while solving the questions in this exercise. Exercise 7.2 is based on the Section Formula. There are 10 questions in this exercise. Exercise 7.3 includes 5 questions that are based on the Area of a Triangle. Students should remember all the formulas in this chapter to solve the questions. Exercise 7.4 is an optional exercise for students and includes eight questions. Students can refer to the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 and the solutions to other exercises to get assistance in solving the problems in these exercises.

**Q.1** Find the distance between following pairs of points:

(i) (2, 3), (4, 1)

(ii) (–5, 7), (–1, 3)

(iii) (a, b), (–a, –b)

**Ans.**

$\begin{array}{l}\text{(i)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}){\text{and (x}}_{2},{\text{y}}_{2}\text{) is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (}2,\text{}3)\text{and (}4,\text{}1)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(4-2\right)}^{2}+{\left(1-3\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}\text{units}\\ \text{(ii)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}){\text{and (x}}_{2},{\text{y}}_{2})\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (}-5,\text{}7)\text{and (}-1,\text{}3)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(-5-(-1)\right)}^{2}+{\left(7-3\right)}^{2}}=\sqrt{16+16}=4\sqrt{2}\text{units}\\ \text{(iii)}\\ {\text{Distance between the two points (x}}_{1},{\text{y}}_{1}){\text{and (x}}_{2},{\text{y}}_{2})\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}}.\\ \text{Therefore, distance between points (a},\text{}\mathrm{b})\text{and (}-\mathrm{a},\text{}-\mathrm{b})\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\mathrm{a}-(-\mathrm{a})\right)}^{2}+{\left(\mathrm{b}-(-\mathrm{b})\right)}^{2}}=\sqrt{4{\mathrm{a}}^{2}+4{\mathrm{b}}^{2}}=2\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\text{}\end{array}$

**Q.2** Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

**Ans.**

$\begin{array}{l}\text{Distance between points (}0,\text{}0)\text{and (}36,\text{}15)\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{(0-36)}^{2}+{(0-15)}^{2}}=\sqrt{1296+225}=\sqrt{1521}=39\text{\hspace{0.17em}\hspace{0.17em}units}\\ \text{Yes, we can find the distance between the given towns A and B.}\\ \text{Let town A be at (0, 0) and town B be at (36, 15). Then, as per}\\ \text{above calculation, distance between the two towns is}39\text{\hspace{0.17em}\hspace{0.17em}km}.\end{array}$

**Q.3** Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.

**Ans.**

$\begin{array}{l}\text{Let the given points are P(1, 5), Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(1, 5) and Q(2, 3)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PQ}=\sqrt{{(1-2)}^{2}+{(5-3)}^{2}}=\sqrt{1+4}=\sqrt{5}\\ \text{Distance between points Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{QR}=\sqrt{{(2+2)}^{2}+{(3+11)}^{2}}=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}\\ \text{Distance between points P(1, 5) and R(}-2,\text{\hspace{0.17em}}-11)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PR}=\sqrt{{(1+2)}^{2}+{(5+11)}^{2}}=\sqrt{9+256}=\sqrt{265}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{5}+2\sqrt{53}\ne \sqrt{265}=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(1, 5), Q(2, 3) and R(}-2,\text{\hspace{0.17em}}-11)\text{are}\\ \text{not collinear.}\end{array}$

**Q.4** Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

**Ans.**

$\begin{array}{l}\text{Let the given points are P(5,}-2\text{), Q(6, 4) and R(7},\text{\hspace{0.17em}}-2).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(5,}-2\text{) and Q(6, 4)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PQ}=\sqrt{{(5-6)}^{2}+{(-2-4)}^{2}}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points Q(6, 4) and R(7},\text{\hspace{0.17em}}-2)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{QR}=\sqrt{{(6-7)}^{2}+{(4+2)}^{2}}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points P(5,}-2\text{) and R(7},\text{\hspace{0.17em}}-2)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{PR}=\sqrt{{(5-7)}^{2}+{(-2+2)}^{2}}=\sqrt{4+0}=2\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{37}+\sqrt{37}\ne 2=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(5,}-2\text{), Q(6, 4) and R(7},\text{\hspace{0.17em}}-2)\text{are}\\ \text{not collinear. They form an isosceles triangle as lengths of}\\ \text{sides PQ and QR are equal.}\end{array}$

**Q.5** In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

**Ans.**

$\begin{array}{l}\text{It is given that 4 friends are seated at the points A(3, 4),}\\ \text{B(6, 7), C(9, 4) and D(6, 1).}\\ \text{Now, using distance formula we find length of each side}\\ \text{of the quadrilatral ABCD and diagonals AC and BD.}\\ \text{AB}=\sqrt{{\left(6-3\right)}^{2}+{\left(7-4\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{BC}=\sqrt{{\left(6-9\right)}^{2}+{\left(7-4\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{CD}=\sqrt{{\left(9-6\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{DA}=\sqrt{{\left(6-3\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{AC}=\sqrt{{\left(9-3\right)}^{2}+{\left(4-4\right)}^{2}}=\sqrt{36+0}=6,\\ \text{BD}=\sqrt{{\left(6-6\right)}^{2}+{\left(7-1\right)}^{2}}=\sqrt{0+36}=6.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, quadrilateral ABCD is a square. Hence, Champa}\\ \text{is correct.}\end{array}$

**Q.6** Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)

(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

**Ans.**

$\begin{array}{l}\text{(i)}\\ \text{Let the given points are A}(-1,\text{}-2),\text{B}(1,\text{}0),\text{C}(-1,\text{}2)\text{and}\\ \text{D}(-3,\text{}0).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(-1-1)}^{2}+{(-2-0)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{BC}=\sqrt{{(1+1)}^{2}+{(0-2)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{CD}=\sqrt{{(-1+3)}^{2}+{(2-0)}^{2}}=\sqrt{4+4}=2\sqrt{2},\\ \text{DA}=\sqrt{{(-1+3)}^{2}+{(-2-0)}^{2}}=\sqrt{4+4}=2\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{(-1+1)}^{2}+{(-2-2)}^{2}}=\sqrt{0+16}=4,\\ \text{BD}=\sqrt{{(1+3)}^{2}+{(0-0)}^{2}}=\sqrt{16+0}=4.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, the given points form a square.}\\ \text{(ii)}\\ \text{Let the given points are A}(-3,\text{5}),\text{B}(3,\text{1}),\text{C}(0,\text{3})\text{and}\\ \text{D}(-1,\text{}-4).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(-3-3)}^{2}+{(5-1)}^{2}}=\sqrt{36+16}=\sqrt{52},\\ \text{BC}=\sqrt{{(3-0)}^{2}+{(1-3)}^{2}}=\sqrt{9+4}=\sqrt{13},\\ \text{CD}=\sqrt{{(0+1)}^{2}+{(3+4)}^{2}}=\sqrt{1+49}=5\sqrt{2},\\ \text{DA}=\sqrt{{(-1+3)}^{2}+{(-4-5)}^{2}}=\sqrt{4+81}=\sqrt{85}\\ \text{We see that sides AB, BC, CD and DA are not equal.}\\ \text{Therefore, the given points do not form a special}\\ \text{quadrilateral. These points form a general quadrilateral.}\\ \text{(iii)}\\ \text{Let the given points are A}(4,\text{}5),\text{B}(7,\text{}6),\text{C}(4,\text{}3)\text{and D}(1,\text{}2).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(4-7)}^{2}+{(5-6)}^{2}}=\sqrt{9+1}=\sqrt{10},\\ \text{BC}=\sqrt{{(7-4)}^{2}+{(6-3)}^{2}}=\sqrt{9+9}=3\sqrt{2},\\ \text{CD}=\sqrt{{(4-1)}^{2}+{(3-2)}^{2}}=\sqrt{9+1}=\sqrt{10},\\ \text{DA}=\sqrt{{(4-1)}^{2}+{(5-2)}^{2}}=\sqrt{9+9}=3\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{(4-4)}^{2}+{(5-3)}^{2}}=\sqrt{0+4}=2,\\ \text{BD}=\sqrt{{(7-1)}^{2}+{(6-2)}^{2}}=\sqrt{36+16}=\sqrt{52}.\\ \text{We see that opposite sides of quadrilateral ABCD are equal.}\\ \text{But, diagonals AC and BD are not equal.}\\ \text{Therefore, the given points form a parallelogram.}\end{array}$

**Q.7** Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

**Ans.**

$\begin{array}{l}\text{Let T(x, 0) be the point on the x-axis which is equidistant from}\\ \text{A(2,}-\text{5) and B(}-\text{2, 9)}.\\ \text{Now, using distance formula we have}\\ \text{TA}=\sqrt{{(\mathrm{x}-2)}^{2}+{(0+5)}^{2}}=\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+4+25}=\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+29}\\ \text{and}\\ \text{TB}=\sqrt{{(\mathrm{x}+2)}^{2}+{(0-9)}^{2}}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+4+81}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+85}\\ \text{Now,}\\ \text{TA}=\text{TB}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\mathrm{x}}^{2}-4\mathrm{x}+29}=\sqrt{{\mathrm{x}}^{2}+4\mathrm{x}+85}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-4\mathrm{x}+29={\mathrm{x}}^{2}+4\mathrm{x}+85\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\mathrm{x}=29-85=-56\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-7\\ \text{Hence, the required point is (}-7,\text{0).}\end{array}$

**Q.8** Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

**Ans.**

$\begin{array}{l}\text{It is given that the distance between the points P(2,}-3)\text{and}\\ \text{(10, y) is 10 units.}\\ \text{Therefore, using distance formula we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{(2-10)}^{2}+{(-3-\mathrm{y})}^{2}}=10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64+9+6\mathrm{y}+{\mathrm{y}}^{2}=100\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^{2}+6\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^{2}+9\mathrm{y}-3\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}(\mathrm{y}+9)-3(\mathrm{y}+9)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(\mathrm{y}-3)(\mathrm{y}+9)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=3\text{or \hspace{0.17em}}\mathrm{y}=-9\end{array}$

**Q.9** If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

**Ans.**

$\begin{array}{l}\text{It is given that Q(0, 1) is equidistant from P(5,}-3)\text{and R(x, 6).}\\ \text{Therefore, using distance formula, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{(0-5)}^{2}+{(1+3)}^{2}}=\sqrt{{(0-\mathrm{x})}^{2}+{(1-6)}^{2}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25+16={\mathrm{x}}^{2}+25\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}=16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\pm 4\\ \text{When}\mathrm{x}=4,\text{then}\\ \text{QR}=\sqrt{{(0-4)}^{2}+{(1-6)}^{2}}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{(5-4)}^{2}+{(-3-6)}^{2}}=\sqrt{82}\text{units}\\ \text{When}\mathrm{x}=-4,\text{\hspace{0.17em}\hspace{0.17em}then}\\ \text{QR}=\sqrt{{(0+4)}^{2}+{(1-6)}^{2}}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{(5+4)}^{2}+{(-3-6)}^{2}}=9\sqrt{2}\text{units}\end{array}$

**Q.10** Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

**Ans.**

$\begin{array}{l}\text{It is given that the point (x, y) is equidistant from the}\\ \text{point (3,}6)\text{and R(}-3\text{, 4).}\\ \text{Therefore, using distance formula, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{(\mathrm{x}-3)}^{2}+{(\mathrm{y}-6)}^{2}}=\sqrt{{(\mathrm{x}+3)}^{2}+{(\mathrm{y}-4)}^{2}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-12\mathrm{y}+36={\mathrm{x}}^{2}+6\mathrm{x}+9+{\mathrm{y}}^{2}-8\mathrm{y}+16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-6\mathrm{x}-6\mathrm{x}-12\mathrm{y}+8\mathrm{y}+45-25=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-12\mathrm{x}-4\mathrm{y}+20=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+\mathrm{y}-5=0\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Is it possible to find NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 on the Extramarks website?

The Extramarks website and mobile application both provide quick access to the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 in PDF format.

### 2. Why should Class 10 students refer to Extramarks while preparing for the board examinations?

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### 3. Do the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 help in understanding the concepts covered in this chapter?

Students in Class 10 who are preparing for their board exams need to use NCERT solutions. To prepare for the board examination, students can use learning resources like the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1. Only giving answers will not be enough for students. They must familiarise themselves with the procedures used to get the result. Students may improve their potential to respond to each similar question that appears in the examination by becoming more proficient at following clear tactics and procedures.

### 4. Which chapter is covered in the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1?

The NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1 cover the chapter on Coordinate Geometry. The questions in this exercise are mainly based on the Distance Formula that is used to find the distance between two given points in the Cartesian plane.

### 5. What advantages do students gain from using the NCERT Solutions Class 10 Maths Chapter 7 Exercise 7.1?

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