Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 – Coordinate Geometry. This Exercise contains 10 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

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

**NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 Coordinate Geometry**

**1. Find the distance between the following pair of point**

**i. (2, 3), (4, 1)**

Using distance formula

Distance =

=

=

=

=

=

**ii. (-5, 7), (-1, 3)**

Using distance formula

Distance =

=

=

=

=

=

**iii. (a, b), (-a, -b)**

Using distance formula

Distance =

=

=

=

=

**2. Find the distance between the two points (0, 0) and (36, 15).**

Using distance formula

Distance =

=

=

=

=

= 39

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

Let **A**, **B** and **C** denote the points (1, 5), (2, 3) (-2, -11) respectively

Using distance formula

AB = = = =

BC = = = =

CA = = =

∵ AB + BC ≠ CA

BC + CA ≠ BA

CA + AB ≠ CB

∴ A, B and C are non-collinear

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

Let** A**, **B** and **C** denote the points (5, -2), (6, 4) (7, -2) respectively

Using distance formula

AB = = = =

BC = = = =

CA = = = = 2

∵ AB = BC

∴ Two sides of triangle are equal in length. Therefore, ABC is an isosceles triangle

**5. In a classroom, 4 friends are seated at the points A, B, C and D as shown. 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.**

Points **A**, **B**, **C** and **D** are respectively (3, 4), (6, 7), (9, 4) and (6, 1)

Using distance formula

Length of sides

AB = = = =

BC = = = =

CD = = = =

DA = = = =

∵ AB = BC = CD = DA

⇒ ABCD is a rhombus

Length of diagonals

AC = = = = 6

BD = = = = 6

∵ AC = BD

∴ Diagonals are equal

A rhombus having diagonals of equal length is a square

⇒ ABCD is a square

Hence, 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)**

Let **A**, **B**, **C** and **D** denote the points (–1, –2), (1, 0), (–1, 2) and (–3, 0) respectively

Using distance formula

Length of sides

AB = = = =

BC = = = =

CD = = = =

DA = = = =

Length of diagonals

AC = = = = 4

BD = = = = 4

∵ All sides are equal (AB = BC = CD = DA) and both the diagonals are of same length (AC = BD)

∴ The quadrilateral formed by these points is a square

**ii. (-3, 5), (3, 1), (0, 3), (-1, -4)**

Let **A**, **B**, **C** and **D** denote the points (-3, 5), (3, 1), (0, 3) and (-1, -4) respectively

Using distance formula

Length of sides

AB = = = = =

BC = = = =

CD = = = = =

DA = = = =

Length of diagonals

AC = = = =

BD = = = =

∵ AC + CB = + = = AB

∴ ABC lie on the same line

⇒ ABCD is not a quadrilateral

**iii. (4, 5), (7, 6), (4, 3), (1, 2)**

Let **A**, **B**, **C** and **D** denote the points (4, 5), (7, 6), (4, 3) and (1, 2) respectively

Using distance formula

Length of sides

AB = = = =

BC = = = =

CD = = = =

DA = = = =

Length of diagonals

AC = = = = 2

BD = = = = =

∵ Opposite sides are equal (AB = CD and BC = DA) and the diagonals are of unequal lengths

∴ ABCD is a parallelogram

**7. Find the point on the x-axis which is equidistant form (2, -5) and (-2, 9).**

Let** A** and **B **denote the points (2, -5) and (-2, 9) respectively

Let the point **C(x, 0)** on x-axis be equidistance from **A** and **B **∴ AC = BC

=

=

On squaring both sides

2

^{2}+ x

^{2}– 4x + 25 = 2

^{2}+ x

^{2}+ 4x + 81

-4x + 25 = 4x + 81

8x = -81 + 25

8x = -56

x = -7

Therefore, the point (-7, 0) is the point on x-axis that is equidistant from A and B

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

According to the question

PQ = 10

= 10

= 10

= 10

On squaring both sides

y^{2} + 6y + 73 = 100

y^{2} + 6y – 27 = 0

By Splitting the middle term

y^{2} + 9y – 3y – 27 = 0

y(y + 9) – 3(y + 9) = 0

(y – 3)(y + 9) = 0

⇒ y = 3, -9

Therefore, for y = 3 and y = -9, the distance between the points P and Q is 10 units.

**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.**

According to the question

PQ = QR

On squaring both sides

25 + 16 = x^{2} + 25

x^{2} = 16

⇒ x = ±4

If x = +4

PR = = = =

QR = = = =

If x = -4

PR = = = =

QR = = = =

**10. Find a relation between x and y such that the point (x, y) is equidistant form the point (3, 6) and (-3, 4).**

Let **A**, **B** and **C** denote the points (x, y), (3, 6) and (-3, 4) respectively

According to the question

AB = AC

On squaring both sides

x^{2} + 9 – 6x + y^{2} + 36 – 12y = x^{2} + 9 + 6x + y^{2} + 16 – 8y

12x + 4y – 20 = 0

3x + y – 5 = 0

**NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1 – Coordinate Geometry, has been designed by the NCERT to test the knowledge of the student on the topic – Distance Formula**

**Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.1 – Coordinate Geometry**