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

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

Edited By Ramraj Saini | Updated on Nov 27, 2023 08:34 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1

NCERT Solutions for Exercise 7.1 Class 10 Maths Chapter 7 Coordinate Geometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 7.1 introduces us to the concept of a distance formula which is used to find out distance between coordinates of two points. This exercise also gives an idea of how to apply this distance formula in real life problems.

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  1. NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 7 exercise 7.1
  3. Assess NCERT Solutions for Class 10 Maths chapter 7 exercise 7.1
  4. More About NCERT Solutions for Class 10 Maths Exercise 7.1
  5. NCERT Solutions Subject Wise
  6. Subject Wise NCERT Exemplar Solutions
  7. NCERT Exemplar Class 10 Maths
NCERT Solutions for Exercise 7.1 Class 10 Maths Chapter 7 - Coordinate Geometry
NCERT Solutions for Exercise 7.1 Class 10 Maths Chapter 7 - Coordinate Geometry

NCERT solutions for exercise 7.1 Class 10 Maths chapter 7 Coordinate Geometry focus is on how to apply distance formula between two coordinates and we also see this by an analogy of right-angle triangle formation or we also know it as Pythagoras theorem. 10th class Maths exercise 7.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 7 exercise 7.1

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Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1

Q1 (i) Find the distance between the following pairs of points : (2, 3), (4, 1)

Answer:

Given points: (2, 3), (4, 1)

Distance between the points will be: (x_{1},y_{1})\ and\ (x_{2},y_{2})

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

D= \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = 2\sqrt{2}

Q1 (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)

Answer:

Given points: (– 5, 7), (– 1, 3)

Distance between the points will be: (x_{1},y_{1})\ and\ (x_{2},y_{2})

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

D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}

Q1 (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)

Answer:

Given points: (a, b), (– a, – b)

Distance between the points will be: (x_{1},y_{1})\ and\ (x_{2},y_{2})

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

D= \sqrt{(-a-a)^2+(-b-b)^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2+b^2}

Q2 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?

Answer:

Given points: (0, 0) and (36, 15)

Distance between the points will be: (x_{1},y_{1})\ and\ (x_{2},y_{2})

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

D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39

The distance between the two towns A and B is, thus 39 km for given towns location

(0,0) and (36,15) .

Q3 Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Answer:

Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle respectively.

A = (1,5),\ B = (2,3),\ C = (-2,-11)

Therefore,

AB = \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 these are not satisfied.

AB+BC \neq CA

BA+AC \neq BC

BC+CA \neq BA

As these cases are not satisfied.

Hence the points are not collinear.

Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Answer:

The distance between two points A(x_{1},y_{1})\ and\ B(x_{2},y_{2}) is given by:

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

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.

AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}

BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}

CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Q5 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.

Answer:

The coordinates of the points:

A(3,4),\ B(6,7),\ C(9,4),\ and\ D(6,1) are the positions of 4 friends.

The distance between two points A(x_{1},y_{1}), \ and\ B(x_{2},y_{2}) is given by:

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

Hence,

AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2

BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2

CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2

AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2

And the lengths of diagonals:

AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6

BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6

So, here it can be seen that all sides of quadrilateral ABCD are of the same lengths and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square and Champa is saying right.

Q6 (i) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Answer:

Let the given points (-1,-2),\ (1,0),\ (-1,2),\ and\ (-3,0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2

BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2

CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2

AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2

Finding the length of the diagonals:

AC= \sqrt{(-1+1)^2+(-2-2)^2} =\sqrt{0+16} = 4

BD= \sqrt{(1+3)^2+(0-0)^2} =\sqrt{16+0} = 4

It is clear that all sides are of the same lengths and also the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Q6 (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)

Answer:

Let the given points (-3,5),\ (3,1),\ (0,3),\ and\ (-1,-4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}

BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}

CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2

AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like square, rectangle, etc.

Q6 (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

Let the given points (4,5),\ (7,6),\ (4,3),\ (1,2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}

BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}

CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}

AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}

And the diagonals:

AC =\sqrt{(4-4)^2+(5-3)^2} = \sqrt{0+4} = 2

BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Q7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Answer:

Let the point which is equidistant from A(2,-5)\ and \ B(-2,9) be X(x,0) as it lies on X-axis.

Then, we have

Distance AX: = \sqrt{(x-2)^2+(0+5)^2}

and Distance BX = \sqrt{(x+2)^2+(0+9)^2}

According to the question, these distances are equal length.

Hence we have,

\sqrt{(x-2)^2+(0+5)^2} = \sqrt{(x+2)^2+(0+9)^2}

Solving this to get the required coordinates.

Squaring both sides we get,

(x-2)^2+25 = (x+2)^2+81

\Rightarrow (x-2+x+2)(x-2-x-2)= 81 - 25 = 56

\Rightarrow (-8x)= 56

Or, \Rightarrow x = -7

Hence the point is X(-7,0) .

Q8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Answer:

Given the distance between the points P(2,-3) and Q(10,y) is 10 units.

The distance formula :

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

So, given PQ = 10\ units

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

After squaring both sides

\Rightarrow (10-2)^2+(y-(-3))^2 = 100

\Rightarrow (y+3)^2 = 100 - 64

\Rightarrow y+3 = \pm 6

\Rightarrow y = 6 - 3\ or\ y = -6-3

Therefore, the values are y = 3\ or -9 .

Q9 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.

Answer:

Given Q(0,1) is equidistant from P(5,-3) and R(x,6) .

Then, the distances PQ = RQ .

Distance PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}

Distance RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}

\Rightarrow \sqrt{x^2+25} = \sqrt{41}

Squaring both sides, we get

\Rightarrow x^2 = 16

\Rightarrow x = \pm 4

The points are: R(4,6)\ or\ R(-4,6.)

CASE I: when R is (4,6)

The distances QR and PR.

QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}

PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}

CASE II: when R is (-4,6)

The distances QR and PR.

QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}

PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}

Q10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Answer:

Let the point P(x,y ) is equidistant from A(3,6) and B(-3,4) .

Then, the distances AP =BP

AP = \sqrt{(x-3)^2+(y-6)^2} and BP = \sqrt{(x-(-3))^2+(y-4)^2}

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

Squaring both sides: we obtain

\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2

\Rightarrow (2x)(-6)+(2y-10)(-2)= 0 \left [\because a^2-b^2 = (a+b)(a-b) \right ]

\Rightarrow -12x-4y+20 = 0

\Rightarrow 3x+y-5 = 0

Thus, the relation is 3x+y-5 = 0 between x and y.

More About NCERT Solutions for Class 10 Maths Exercise 7.1

The questions in exercise 7.1 Class 10 Maths, broadly consist of all the basic topics which we have learned till now about coordinate geometry such as isosceles triangle, collinear points and it also has many questions about equidistant. NCERT solutions for Class 10 Maths exercise 7.1 also consist of questions in which we are given a real-life scenario and we have to use a distance formula. Exercise 7.1 Class 10 Maths covers all types of basic questions that can be formed on the distance formula. Students can find Coordinate Geometry Class 10 Notes here which can be used to quick revision of the concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 7.1

  • Class 10 Maths chapter 7 exercise 7.1 broadly covers a basic knowledge which we would be required to proceed in Class 10 Maths chapter 7 exercise 7.2
  • NCERT Class 10 Maths chapter 7 exercise 7.1, will be helpful in Class 11 Maths chapter 10 straight line and Class 11 Maths chapter 11 conic section.
  • Exercise 7.1 Class 10 Maths, is also a base builder concept for many topics of Joint Entrance Exam (JEE) Main.

Also, see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions


Frequently Asked Question (FAQs)

1. What is analogy of deriving distance formula by pythagoras theorem according to NCERT solutions for Class 10 Maths exercise 7.1?

We assume that the line between two coordinates is hypotenuse so with the help x axis and y axis we will find perpendicular and base and by applying pythagoras theorem we will find hypotenuse .

2. What is the formula of pythagoras theorem ?

H =square root of [(P)²+(B)²]

In the above formula, H is the hypotenuse, P is perpendicular and B is base

3. What do you mean by collinear points as referred in NCERT solutions for Class 10 Maths exercise 7.1?

Collinear points are set of three or more points which is on exact same line

4. According to NCERT solutions for Class 10 Maths exercise 7.1 where can we use distance formula.

The distance formula is used in NCERT solutions for Class 10 Maths exercise 7.1  to find area of triangle length of vertices of different types of triangles such as an isosceles triangle.

5. According to NCERT solutions for Class 10 Maths exercise 7.1 what is equidistant points.

Set of points all at equal distance from each other.

6. What is the total number of solved examples prior to the Class 10 Maths Exercise 7.1?

Before the Class 10 Mathematics chapter 7 activity 7.1, there are five key questions that must be answered.

7. In the NCERT answers for Class 10 Maths chapter 7 exercise, how many questions are covered?

There are 10 questions in Class 10 Maths chapter exercise 7.1. Question 1 consists of three subparts.

8. What are the types of questions covered in the NCERT solutions for Class 10 Maths chapter 7 exercise?

Question 1 and question 2 is very basic coordinates are given and we have to find the distance then we have question 3 based on collinear point question 4 on finding vertices of an isosceles triangle then we have question 5 based on real-life scenario and then we have some question of  equidistant point

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