NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 1

Question 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)

Solution:

Formula used in the above question is : √(x₂ – x₁)² + (y₂ – y₁)² (i.e., Distance Formula)

(i) Here, x₁ = 2, y₁= 3, x₂ = 4, y₂ = 1

Now, applying the distance formula :

= √(4-2)² + (1-3)²

=√(2)² + (-2)²

= √8

= 2√2 units

(ii) Here, x₁= -5, y₁= 7, x₂ = -1, y₂ = 3

Now, applying the distance formula :

= √(-1 – (-5))² + (3 – 7)²

= √(4)² + (-4)²

= 4√2

(iii) Here, x₁ = a, y₁ = b . x₂ = -a, y₂ = -b

Now, applying the distance formula :

= √(-a – a)² +(-b – b)²

= √(-2a)² + (-2b)²

= √4a² + 4b²

= 2√a² + b²

Question 2) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in section 7.2.

Solution:

Formula used in the above question is : √(x₂ – x₁)² + (y₂ – y₁)² (i.e, Distance Formula)

Considering, Point A as (0, 0) and Point B as (36, 15) and applying the distance formula we get :

Distance between the two points : √(36 – 0)² + (15 – 0)²

= √(36)² + (15)²

= √1296 + 225

= √1521

= 39 units

Hence, the distance between two towns A and B is 39 units.

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

Solution:

Let (1, 5), (2, 3) and (-2, -11) be points A, B and C respectively.

Collinear term means that these 3 points lie in the same line. So, to we’ ll check it.

Using distance formula we will find the distance between these points.

AB = √(2 – 1)² + (3 – 5)²

=√(1)² + (-2)² =√1 + 4 =√5

BC = √(-2 – 2)² + (-11 – 3)²

= √(-4)² + (-14)² = √16 + 196 = √212

CA = √(-2 – 1)² + (-11 – 5)²

= √(-3)² + (-16)² = √9 + 256 =√265

As, AB + BC ≠ AC (Since, one distance is not equal to sum of other two distances, we can say that they do not lie in the same line.)

Hence, points A, B and C are not collinear.

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

Solution:

Let (5, – 2), (6, 4) and (7, – 2) be the points A, B and C respectively.

Using distance formula :

AB = √(6 – 5)² + (4 – (-2))²

= √(1 + 36) = √37

BC = √(7 – 6)² + (-2 – 4)²

= √(1 + 36) = √37

AC = √(7 – 5)² + (-2 – (-2))²

= √(4 + 0) = 2

As, AB = BC ≠ AC (Two distances equal and one distance is not equal to sum of other two)

So, we can say that they are vertices of an isosceles triangle.

Question 5) In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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.

Solution:

From the given fig, find the coordinates of the points

AB = √(6 – 3) + (7 – 4)

= √9+9 = √18 = 3√2

BC = √(9 – 6) + (4 – 7)

= √9+9 = √18 = 3√2

CD = √(6 – 9) + (1 – 4)

= √9 + 9 = √18 = 3√2

DA = √(6 – 3) + (1 – 4)

= √9+9 =√18 =3√2

AB = BC = CD = DA = 3√2

All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.

Question 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)

Solution:

(i) Here, let the given points are P(-1, -2), Q(1, 0), R(-1, 2) and S(-3, 0) respectively.

PQ = √(1 – (-1))² + (0 – (-2))²

= √(1+1)²+(0+2)²

= √8 = 2 √2

QR = √(−1−1)²+(2−0)²

​ = √(−2)²+(2)²

= √8 = 2 √2

RS = √(−3−(−1))²+(0−2)²

= √8 = 2 √2

PS = √((−3−(−1))²+(0−(−2))²

= √8 = 2 √2

Here, we found that the length of all the sides are equal.

Diagonal PR = √(−1−(−1))²+(2−(−2))²

= √ 0+16

​ = 4

Diagonal QS = √(−3−1)²+(0−0)²

= √ 16 = 4

​ Finally, we also found that the length of diagonal are also same.

Here, PQ = QR = RS = PS = 2√2

and QS = PR = 4

This is the property of SQUARE. Hence, the given figure is SQUARE.

(ii) Let the points be P(-3, 5), Q(3, 1), R(0, 3) and S(-1, -4)

PQ = √(3−(−3))²+(1−5)²

= √(3+3)²+(−4)²

= √36+16 = √52 = 2 √13

QR = √(0−3)²+(3−1)²

= √(−3)²+(2)²

= √9 + 4 = √13

RS = √(−1−0)²+(−4−3)²

​ = √(−1)²+(−7)²

= √1+49 = √50 = 5 √2

PS = √(−1−(−3))²+(−4−5)²

= √(−1+3)²+(−9)²

= √4+81 = √85

Here, All the lengths of sides are unequal.

So, The given points will not create any quadrilateral.

(iii) Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)

PQ = √(7−4)²+(6−5)²

= √(3)²+(1)² = √9+1 = √10

QR = √(4−7)²+(3−6)²

= √(−3)²+(−3)² = √9+9 =3 √2

RS = √(1−4)²+(2−3)²

= √(−3)²+(−1)² = √9+1 = √10

PS = √(1−4)²+(2−5)²

= √(−3)²+(−3)² = √9+9 =3 √2

We see that the opposite sides are equal. Lets find the diagonal now.

Diagonal PR = √(4−4)²+(3−5)²

= √0+4 = 2

Diagonal QS = √(1−7)²+(2−6)²

= √36+16

= √52

Here, PQ = RS = √10

and QR = PS = 3√2

We see that the diagonals are not equal.

Hence, the formed quadrilateral is a PARALLELOGRAM.

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

Solution:

Let the point on the X axis be (x, 0)

Given, distance between the points (2, -5), (x, 0) = distance between points (-2, 9), (x, 0)

[ Applying Distance Formula ]

⇒ √(x – 2)²+(0 – (-5))² = √(x – (-2))²+(0 – 9)²

On squaring both the sides, we get

⇒ (x – 2)² + 5² = (x + 2)² + 9²

⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81

⇒ -4x -4x = 85 – 29

⇒ -8x = 56

⇒ x = -7

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

Solution:

It is given that, the distance b/w two points is 10 units.

So, we’ll find the distance and equate

PQ = √ (10 – 2)² + (y – (-3))²

= √ (8)² + (y +3)²

On squaring both the sides, we get :

64 +(y+3)² = (10)²

(y+3)² = 36

y + 3 = ±6

y + 3 = +6 or y + 3 = −6

Hence, y = 3 or -9.

Question 9) If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Solution:

Given, PQ = QR

We will apply distance formula and find the distance between them,

PQ = √(5 – 0)² + (-3 – 1)²

= √ (- 5)² + (-4)²

= √ 25 + 16 = √41

QR = √ (0 – x)² + (1 – 6)²

= √ (-x)² + (-5)²

= √ x² + 25

As they both are equal so on equating them, x² + 25 = 41

x² =16, x = ± 4

So, putting the value of x and obtaining the value of QR and PR through distance formula

For x = +4, PR = √ (4 – 5)² + (6 – (-3))²

= √ (-1) ²+ (9)²

= √ 82

QR = √ (0 – 4)² + (1 – 6)²

= √ 41

For x = -4, QR = √ (0 – (-4))² + (1 – 6)²

= √ 16 + 25 = √ 41

PR = √ (5 + 4)² + (-3 -6)²

= √ 81 + 81 = 9 √ 2

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

Solution:

Let, (x, y) be point P and (3, 6), (-3, 4) be pt A and B respectively.

It is given that their distance is equal, so we will equate the equations.

PA = PB (given)

⇒ √(x – 3)² +(y – 6)² = √(x-(-3))²+ (y – 4)² [ By applying distance formula ]

On squaring both sides,

(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

NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry- The team of subject matter experts at GFG have made detailed NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry to make sure that every student can understand how to solve Coordinate Geometry problems in a stepwise manner.

Coordination of the point of a line divided into two halves, section formula, the distance between two points, and areas of the triangle are some examples of problems covered in Coordinate Geometry.

This article provides solutions to all the problems asked in Class 10 Maths Chapter 7 Coordinate Geometry of your NCERT textbook in a step-by-step manner. They are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

All the exercises in Class 10 Maths Chapter 7 Coordinate Geometry of your NCERT textbook have been properly covered in NCERT Solutions for Class 10 Maths.