RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

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 + x² – 4x + 25 = 4 + x² + 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 y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 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 = x² + 25 or x² = 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 x² + y² – 6x – 12y + 45
= x² + y² + 6x – 8y + 25
or -12x – 4y + 20 = 0
[dividing both sides by (-4)]
or 3x + y – 5 = 0
This is the required relation.