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.

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.

## Assess NCERT Solutions for Class 10 Maths chapter 7 exercise 7.1

Coordinate Geometry Class 10 Chapter 7 Exercise: 7.1

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

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

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

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

Given points: (– 5, 7), (– 1, 3)

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

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

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

Given points: (a, b), (– a, – b)

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

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

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

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 $(36,15)$ .

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.

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.

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.

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.

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.

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.

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 $X(-7,0)$ .

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 $y = 3\ or -9$ .

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.

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

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

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-

## Subject Wise NCERT Exemplar Solutions

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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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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