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Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 NCERT Solutions

Get NCERT Solutions for Exercise 7.1 with clear step-by-step answers based on the distance formula. It will help you calculate distances between two points accurately in coordinate geometry.

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 focus on the distance formula in coordinate geometry. This exercise will help you calculate the distance between two points on a coordinate plane using a standard formula derived from the Pythagorean theorem.

This exercise builds a foundation for advanced concepts by teaching how to plot points using ordered pairs and interpret their positions on the x–y plane. These  NCERT Solutions for Class 10 Maths are presented in a clear step-by-step manner. 

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

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)

Answer:

(i) Distance between the points is given by
 
Therefore the distance between (2,3) and (4,1) is given by l = chapter 7-Coordinate Geometry Exercise 7.1/image002.png
=chapter 7-Coordinate Geometry Exercise 7.1/image002.png
= √4+4 = √8 = 2√2

(ii)Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), 
we get l =chapter 7-Coordinate Geometry Exercise 7.1/image002.png
=chapter 7-Coordinate Geometry Exercise 7.1/image002.png
= √16+16
= √32
= 4√2

(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b),
we get l = chapter 7-Coordinate Geometry Exercise 7.1/image002.png=chapter 7-Coordinate Geometry Exercise 7.1/image002.png=chapter 7-Coordinate Geometry Exercise 7.1/image002.png 

2. Find the distance between the points (0, 0) and (36, 15). Also, find the distance between towns A and B if town B is located at 36 km east and15 km north of town A.

Answer:

Applying the Distance Formula to find the distance between points (0, 0) and (36, 15), 

we get = chapter 7-Coordinate Geometry Exercise 7.1/image004.png
= √1296 + 225
= √1521
= 39.
Yes, we can find the distance between the given towns A and B.
Assume town A at the origin point (0, 0). Therefore, town B will be at point (36, 15) with respect to town A. And hence, as calculated above, the distance between town A and B will be 39km. 

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

Answer:
Let A = (1, 5), B = (2, 3) and C = (–2, –11). Using the Distance Formula to find the distance AB, BC, and CA.
 chapter 7-Coordinate Geometry Exercise 7.1/image007.png
BC = chapter 7-Coordinate Geometry Exercise 7.1/image008.png
CA = chapter 7-Coordinate Geometry Exercise 7.1/image009.png

Since AB+BC ≠ CA, therefore, the points (1, 5), (2, 3), and (−2, 11) are not collinear. 

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

Answer:
Let A = (5, –2), B = (6, 4), and C = (7, –2). Using the Distance Formula to find distances AB, BC, and CA.
AB = chapter 7-Coordinate Geometry Exercise 7.1/image011.png
BC = chapter 7-Coordinate Geometry Exercise 7.1/image012.png
CA = chapter 7-Coordinate Geometry Exercise 7.1/image013.png
Since AB = BC. Therefore, A, B, and C are the vertices of an isosceles triangle.  

5. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4), and D (6, 1). 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 the distance formula, find which of them is correct.
 
NCERT solutions for class 10 maths/image014.jpg

Answer:

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1) Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = NCERT solutions for class 10 maths/image015.png
BC = NCERT solutions for class 10 maths/image016.png
CD = NCERT solutions for class 10 maths/image017.png
AD = NCERT solutions for class 10 maths/image018.png
Therefore, All the sides of ABCD are equal here Now, we will check the length of its diagonals
NCERT solutions for class 10 maths/image019.png
AC = NCERT solutions for class 10 maths/image019.png
BD =NCERT solutions for class 10 maths/image020.png
So, the Diagonals of ABCD are also equal. We can definitely say that ABCD is a square. Therefore, Champa is correct. 

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)

Answer:

(i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0) Using Distance Formula to find distances AB, BC, CD and DA,
we get AB = NCERT solutions for class 10 maths/image021.png
BC = NCERT solutions for class 10 maths/image022.png
CD = NCERT solutions for class 10 maths/image023.png
AD = NCERT solutions for class 10 maths/image024.png
Therefore, all four sides of quadrilateral are equal. Now, we will check the length of diagonals.
AC =NCERT solutions for class 10 maths/image025.png
BD = NCERT solutions for class 10 maths/image026.png
Therefore, diagonals of quadrilateral ABCD are also equal. we can say that ABCD is a square.

(ii)Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA,
we get AB = NCERT solutions for class 10 maths/image027.png
BC = NCERT solutions for class 10 maths/image028.png
CD = NCERT solutions for class 10 maths/image029.png
DA = NCERT solutions for class 10 maths/image030.png
We cannot find any relation between the lengths of different sides. Therefore, we cannot give any name to the quadrilateral ABCD.

(iii)Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2) Using the Distance Formula to find distances AB, BC, CD, and DA,
we get AB =NCERT solutions for class 10 maths/image031.png
BC = NCERT solutions for class 10 maths/image032.png
CD =NCERT solutions for class 10 maths/image033.png
DA = NCERT solutions for class 10 maths/image034.png
Here opposite sides of quadrilateral ABCD are equal. We can now find out the lengths of diagonals.
AC =NCERT solutions for class 10 maths/image035.png
BD = NCERT solutions for class 10 maths/image036.png
Here diagonals of ABCD are not equal. We can say that ABCD is not a rectangle, therefore it is a parallelogram. 

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

Answer:
Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).
Using Distance Formula and according to given conditions
we have:NCERT solutions for class 10 maths/image037.pngNCERT solutions for class 10 maths/image038.png
Squaring both sides, we get ⇒ NCERT solutions for class 10 maths/image039.png
(x-2)² + 25 = (x+2)² + 81 x² + 4 - 4x + 25 = x² + 4 + 4x + 81 8x = - 25 - 81 8x = -56 x = - 7
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0) 

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

Answer:
Using the Distance formula,
we haveNCERT solutions for class 10 maths/image041.png
NCERT solutions for class 10 maths/image042.png
⇒ 64 + (y +3)² = 100
⇒ (y+3)² = 100-64 = 36
⇒ y+3 = ± 6
⇒ y+3=6 or y+3 = - 6
Therefore, y = 3 or -9

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. 

Answer:
It is given that Q is equidistant from P and R. Using the Distance Formula,
 
we get PQ = RQ NCERT solutions for class 10 maths/image047.pngNCERT solutions for class 10 maths/image047.png
⇒√25+16 = √x² + 25
⇒41 = x² + 25 16 = x² x = ± 4.
Thus, R is (4, 6) or (–4, 6). 
When point R is (4,6), PR = NCERT solutions for class 10 maths/image054.png
QR =NCERT solutions for class 10 maths/image055.png
When point R is (- 4,6) PR = NCERT solutions for class 10 maths/image056.png
QR =NCERT solutions for class 10 maths/image057.png

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

Answer:
It is given that (x, y) is equidistant from (3, 6) and (–3, 4). Using Distance formula,
we can write NCERT solutions for class 10 maths/image060.pngNCERT solutions for class 10 maths/image061.png
⇒ (x-3)² + (y-6)² = (x+3)² + (y-4)² ⇒ x² + 9 -6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y ⇒36- 16 = 6x + 6x + 12y  - 8y ⇒20 = 12x + 4y ⇒3x + y = 5 ⇒3x + y - 5 = 0

⇒ (x-3)² + (y-6)² = (x+3)² + (y-4)² 

⇒ x² + 9 -6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y 

⇒36- 16 = 6x + 6x + 12y  - 8y 

⇒20 = 12x + 4y 

⇒3x + y = 5 

⇒3x + y - 5 = 0

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practise, and a focus on accuracy and answer presentation. To score better, you should:

  • Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

  • Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

  • Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

  • Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

  • Solve Previous Year Papers:

Practising CBSE Class 10 Previous Year Question Papers(PYQs) helps you understand question patterns and important topics.

 

 

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 FAQs

What does Exercise 7.1 of Class 10 Coordinate Geometry cover?

It covers plotting points on the Cartesian plane, distance formula, and midpoint formula.

How does this exercise help in exams?

It builds a strong foundation for coordinate geometry, which is essential for scoring well in Class 10 Maths.

What is the importance of the midpoint formula in coordinate geometry?

The midpoint formula is crucial for finding the center point between two given points, which is often needed in problems related to geometry, symmetry, and constructing shapes.

Are there any tricky questions in this exercise?

The exercise is primarily focused on the application of the distance and midpoint formulas, but some problems may involve multi-step solutions or require careful attention to coordinate signs (positive or negative). However, with practice, these problems become easier to tackle.
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