NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry – Here are all the NCERT solutions for Class 10 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete Chapter 7 titled Coordinate Geometry of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 7 Coordinate Geometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry in one place.
NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry
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For a better understanding of this chapter, you should also see summary of Chapter 7 Coordinate Geometry , Maths, Class 10.
| Class | 10 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 7 |
| Chapter Name |
Coordinate Geometry |
NCERT Solutions Class 10 Maths chapter 7 Coordinate Geometry
Class 10, Maths chapter 7, Coordinate Geometry solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)
Ans : (i) distance between the two points is given by \( \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\) Therefore. distance between (2, 3) and (4, 1) is given by \( l=\sqrt{(2-4)^{2}+(3-1)^{2}}=\sqrt{(-2)^{2}+(2)^{2}}\) \( =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) (ii) Distance between (-S,7) and (-1,3) is given by \( l=\sqrt{(-5-(-1))^{2}+(7-3)^{2}}=\sqrt{(-4)^{2}+(4)^{2}}\) \( =\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) (iii) Distance between (a, b) and (-a -b) is given by \( l=\sqrt{(a-(-a))^{2}+(b-(-b))^{2}}\) \( =\sqrt{\left((2 a)^{2}+(2 b)^{2}\right.}=\sqrt{4 a^{2}+4 b^{2}}=2 \sqrt{a^{2}+b^{2}}\)
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 : Distance between points (0.0) and (36,15) \( =\sqrt{(36-0)^{2}+(15-0)^{2}}=\sqrt{36^{2}+15^{2}}\) \( =\sqrt{1296+225}=\sqrt{1521}=39\) Yes, we can find the distance between the given towns A and B Assume town A at origin point (O, O), Therefore, town a Rill be at point (36, 15) with respect to town A. And hence, as calculated above, the distance between town A and B Bill be 39 km
Q.3: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Ans : Let the points (1, 5), (2, 3), and (-2, -11) be representing the vertices A, B, and C of the given triangle respectively. Let A=(1,5) , B =(2,3) , C=(-2,-11) \( \therefore A B=\sqrt{(1-2)^{2}+(5-3)^{2}}=\sqrt{5}\) \(BC=\sqrt{(2-(-2))^{2}+(3-(-11))^{2}}=\sqrt{4^{2}+14^{2}}=\sqrt{16+196}=\sqrt{212}\) \( CA=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}}=\sqrt{3^{2}+16^{2}}=\sqrt{9+256}=\sqrt{265}\) Since \( \mathrm{AB}+\mathrm{BC} \neq \mathrm{CA}\) Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear,
Q.4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Ans : Let the points (5, -2), (6, 4), and (7, -2) are representing the vertices A, B, and C of the given triangle respectively. \( A B=\sqrt{(5-6)^{2}+(-2-4)^{2}}=\sqrt{(-1)^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}\) \( B C=\sqrt{(6-7)^{2}+(4-(-2))^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{1+36}=\sqrt{37}\) \( \mathrm{CA}=\sqrt{(5-7)^{2}+(-2-(-2))^{2}}=\sqrt{(-2)^{2}+0^{2}}=2\) Therefore AB=BC As two sides are equal in length, therefore, ABC is an isosceles triangle.
Q.5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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 : \( A B=\sqrt{(3-6)^{2}+(4-7)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( B C=\sqrt{(6-9)^{2}+(7-4)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( C B=\sqrt{(9-6)^{2}+(4-1)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( A D=\sqrt{(3-6)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) \( A C=\sqrt{(3-9)^{2}+(4-4)^{2}}=\sqrt{(-6)^{2}+0^{2}}=6\) \( \mathrm{BD}=\sqrt{(6-6)^{2}+(7-1)^{2}}=\sqrt{0^{2}+(6)^{2}}=6\)It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are Of the sarne length, Therefore, ABCD is a square and hence, Champa correct
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