Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

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:

(i) According to the question, the points are : (2, 3); (4, 1)

Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)

= \(\sqrt{4+4}\)

= \(\sqrt{8}\) = \(\sqrt{2 \times 2 \times 2}\)

= 2\(\sqrt{2}\) units

(ii) According to the question the points are : (- 5, 7); (- 1, 3)

Required distance

= \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)

= \(\sqrt{16+16}\) = \(\sqrt{32}\) = \(\sqrt{4 \times 4 \times 2}\)

= 4\(\sqrt{2}\) units

(iii) According to the question the points are : (a, b), (- a, – b)

Required distance

= \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)

= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)

= \(\sqrt{4 a^{2}+4 b^{2}}\)

= 2 \(\sqrt{a^{2}+b^{2}}\) units

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:

According to the question, the points are : A (0, 0) and B (36, 15)

distance AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)

= \(\sqrt{1296+225}\) = \(\sqrt{1521}\)

= 39 units

According to Section 7.2, a town B is located 36 km east and 15 km north of the town A. Now we find distance between A and B. Coordinate of the point A as the origin (0, 0) and the coordinate of B are (36, 15), then

= 39 units

Hence the required distance between the two cities A and B is 39 km.

Question 3.

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

Solution:

According to the question the points are :

A(1, 5); B(2, 3) and (- 2, -11)

Here AB + BC ≠ CA; BC + CA ≠ AB; CA + AB ≠ BC

From the above, it is clear that sum of any two is not equal to the third. So, the given points are not collinear.

Question 4.

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

Solution:

According to the question the points are :

A(5, – 2); B(6, 4) and C(7, – 2)

From the above distances it is clear that AB= BC = \(\sqrt {37}\) units

∴ The given points are the 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 Figure 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:

In the given figures the coordinates of the given points are: A(3, 4); B(6, 7); C(9, 4); and D(6, 1)

From the above distances it is clear that

AB = BC = CD = DA = 3\(\sqrt {2}\) units.

and AC = BD = 6 units

∴ ABCD forms a square and Champa’s statement is 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) According to the question the points are :

A(- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0)

From above it is clear that :

AB = BC = CD = DA = \(\sqrt {8}\) = 2\(\sqrt {2}\) units and AC= BD = 4 units

So, the given quadrilateral ABCD is a square.

(ii) According to the question the points are : A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4)

∴ A, B and C are collinear. So A, B, C and D do not form a quadrilateral.

(iii) According to the question the points are :

A(4, 5), B(7, 6), C(4, 3) and D(1, 2)

From above it is clear that

AB = CD

and BC = DA

and AC ≠ BD

i.e. opposite sides are equal, but their diagonals are not equal. Hence the given quadrilateral is parallelogram.

Question 7.

Find the point on the jr-axis which is equidistant from (2, – 5) and (- 2, 9)

Solution:

Let the required point on x- axis be P(x, 0) and the given points are :

A(2, – 5) and B(-2, 9)

According to the question,

PA = PB

or (PA)^{2} = (PB)^{2}

or (2 – x)^{2} + (- 5 – 0)^{2} = (- 2 – x)^{2} + (9 – 0)^{2}

or 4 + x^{2} – 4x + 25 = 4 + x^{2} + 4.x + 81

or – 8x = 56

or x = \(-\frac{56}{8}\) = -7

Hence the required point on the x-axis is (- 7, 0)

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:

According to the question the points are :

P(2, – 3) and Q(10, y)

PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)

= \(\sqrt{64+y^{2}+9+6 y}\)

= \(\sqrt{y^{2}+6 y+73}\)

Again according to the question,

PQ = 10

or \(\sqrt{y^{2}+6 y+73}\) = 10

squaring both sides or y^{2} + 6y + 73 = 100

or y^{2} + 6y – 27 = 0

or y^{2} + 9y – 3y – 27 = 0

or y (y + 9) – 3 (y + 9) = 0

or (y + 9) (y – 3) = 0

y + 9 = 0 or y – 3 = 0

y = – 9 or y = 3

So y = – 9 and 3

Question 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.

Solution:

According to the question the points are :

Q(0, 1), P(5, – 3) and R(x, 6)

QP= \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)

= \(\sqrt{25+16}\) = \(\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)

= \(\sqrt{x^{2}+25}\)

again according to the question :

QP = QR

or \(\sqrt{41}\) = \(\sqrt{x^{2}+25}\)

squaring both sides

or 41 = x^{2} + 25 or x^{2} = 16

or x,= ±\(\sqrt{16}\) = ±4

Thus the coordinates of R will be R(4, 6) or R(- 4, 6)

When x = 4 then R(4, 6)

QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)

= \(\sqrt{16+25}\) = \(\sqrt{41}\)

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 the required point be P(x, y)

The given points are : A(3, 6) and B(- 3, 4)

squaring both sides,

or x^{2} + y^{2} – 6x – 12y + 45

= x^{2} + y^{2} + 6x – 8y + 25

or -12x – 4y + 20 = 0

[dividing both sides by (-4)]

or 3x + y – 5 = 0

This is the required relation.

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