**Download 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

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

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