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

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

The academic team at Neutronclasses has meticulously crafted NCERT solutions for Chapter 7 Exercise 7.1 of Class 10 Mathematics, which focuses on Coordinate Geometry. These solutions comprehensively cover all the exercises within this chapter. Presented below are the step-by-step solutions to all the questions found in the NCERT textbook for 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

D =

=

= √4+4 = √8 = 2√2 units

(ii)Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get

D =

=

= √16+16 = √32 = 4√2 units

(iii)Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get

D =

=

= units

**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 Distance Formula to find distance between points (0, 0) and (36, 15), we get

=

= √1296 + 225 = √1521 = 39

Yes, we can find the distance between the given towns A and B.

Assume town A at 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 Distance Formula to find distance AB, BC and CA.

BC =

CA =

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 Distance Formula to find distances AB, BC and CA.

AB =

BC =

CA =

Since AB = BC.

Therefore, A, B and C are 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 distance formula, find which of them is correct.**

**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 =

BC =

CD =

AD =

Therefore, All the sides of ABCD are equal here

Now, we will check the length of its diagonals.

AC =

BD =

So, 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 =

BC =

CD =

AD =

Therefore, all four sides of quadrilateral are equal.

Now, we will check the length of diagonals.

AC =

BD =

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 =

BC =

CD =

DA =

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 Distance Formula to find distances AB, BC, CD and DA, we get

AB =

BC =

CD =

DA =

Here opposite sides of quadrilateral ABCD are equal.

We can now find out the lengths of diagonals.

AC =

BD =

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:

⇒

Squaring both sides, we get

⇒

(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 Distance formula, we have

⇒

⇒ 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 Distance Formula, we get

PQ = RQ

⇒√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 =

QR =

When point R is (- 4,6)

PR =

QR =

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

⇒

⇒ (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