HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 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, 6), (- a, – b)
Solution :
(i) Let the given points be A(2, 3) and B(4, 1), then
AB = \(\sqrt{(4-2)^2+(1-3)^2}\)
⇒ AB = \(\sqrt{(2)^2+(-2)^2}\)
⇒ AB = \(\sqrt{4+4}\)
⇒ AB = √8
⇒ AB = \(\sqrt{2 \times 2 \times 2}\)
⇒ AB = 2√2
Hence, distance (AB) = 2√2.

(ii) Let the given points be P(- 5, 7) and Q(- 1, 3), then
PQ = \(\sqrt{(-1+5)^2+(3-7)^2}\)
PQ = \(\sqrt{(4)^2+(-4)^2}\)
⇒ PQ = \(\sqrt{16+16}\)
⇒ PQ = √32
⇒ PQ = \(\sqrt{4 \times 4 \times 2}\)
⇒ PQ = 4√2
Hence, distance (PQ) = 4√2

(iii) Let the given points be P(a, b) and Q(- a, – 6), then
PQ = \(\sqrt{(-a-a)^2+(-b-b)^2}\)
⇒ PQ = \(\sqrt{(-2 a)^2+(-2 b)^2}\)
⇒ PQ = \(\sqrt{4 a^2+4 b^2}\)
⇒ PQ = \(\sqrt{4\left(a^2+b^2\right)}\)
⇒ PQ = 2\(\sqrt{a^2+b^2}\)
Hence, the distance (PQ) = 2\(\sqrt{a^2+b^2}\).

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.

Solution:
Let the given points be A(0,0) and B(36, 15) then
AB = \(\sqrt{(36-0)^2+(15-0)^2}\)
⇒ AB = \(\sqrt{(36)^2+(15)^2}\)
⇒ AB = \(\sqrt{1296+225}\)
⇒ AB = \(\sqrt{1521}\)
⇒ AB = 39
Hence, distance between two towns = 39 km.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Let the given points be A(1, 5), B(2, 3) and C(- 2, – 11). Then,

⇒ AC = √256 = 16.28 (approx)
Now,AB + BC = 2.24 + 14.56 = 16.8
AC = 16.28
∴ AB + BC ≠ AC
Hence, the given points (1, 5), (2, 3) and (- 2, – 11) are not collinear.

Question 4.
Check whether (5, – 2), (6,4) and (7, – 2) are the vertices of an isosceles triangle.
Solution :
Let the given points be A(5, – 2), B(6, 4) and C(7, – 2), then
AB = \(\sqrt{(6-5)^2+(4+2)^2}\)
= \(\sqrt{(1)^2+(6)^2}\)
= \(\sqrt{37}\)

BC = \(\sqrt{(7-6)^2+(-2-4)^2}\)
= \(\sqrt{(1)^2+(-6)^2}\)
= \(\sqrt{37}\)
Since, AB = BC
Therefore, the given joints are the vertices of an isosceles triangle.

Question 5.
In a class room, 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:
From the figure, coordinates of A, B, C, D, are respectively (3, 4), (6, 7), (9, 4), (6, 1)

⇒ BD = √36
⇒ BD = 6.
Since, AB = BC = CD = DA and diagonal AC = diagonal BD.
Therefore, ABCD is a square.
Hence, Champa 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, 6), (7, 6), (4, 3), (1, 2)
Solution :
(i) Let the given points of the quadrilateral be A(- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0).
Join AC and BD

Since, AB = BC = CA = DA and diagonal AC = diagonal BD
Therefore, given points form a square.

(ii) Let the given points of the quadrilateral be A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4).
Join AC and BD

Now,
AC + BC = √13 + √13 = 2√13
AB = 2√13
Since, AC + BC = AB.
Therefore, A, B, C are collinear. So, the given points do not form any quadrilateral.

(iii) The given points of a quadrilateral be A(4, 5), B(7, 6), C(4, 3) and D(l, 2). Join AC and

Diagonal AC ≠ diagonal BD.
Since, opposite sides are equal but diagonals are not equal.
Therefore, the given points form a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution :
Let the given points be A(2, – 5) and B(- 2, 9) and let the required point be POc, 0), then
PA = PB
⇒ PA² = PB²
⇒ (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)²
⇒ x² + 4 – 4x + 25 = x² + 4 + 4x + 81
⇒ x² – 4x + 29 = x² + 4x + 85
⇒ x² – 4x – x² – 4x = 85 – 29
⇒ – 8x = 56.
⇒ x = \(\frac{56}{-8}\)
x = – 7
Hence, the required point = (- 7, 0)

Question 8.
Find the values ofy for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution :
The given points are P(2, – 3) and Q(dfgvbtyh10, y)
and PQ = 10 units
∴ PQ = \(\sqrt{(10-2)^2+(y+3)^2}\)
10 = \(\sqrt{(8)^2+y^2+9+6 y}\)
10 = \(\sqrt{64+y^2+9+6 y}\)
10 = \(\sqrt{73+y^2+6 y}\)
100 = 73 + y² + 6y (squaring both sides)
y² + 6y + 73 – 100 = 0
y² + 6y – 27 = 0
y² + (9 – 3)y – 27 = 0
y² + 9y – 3y – 27 = 0
[∵ 1 × – 27 = – 27
9 × – 3 = – 27
9 – 3 = 6]
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9) (y – 3) = 0
⇒ y + 9 = 0 or y – 3 = 0
⇒ y = – 9 or y = 3.
Hence, y = – 9 or 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:
The given points are P(5, -3), Q(0, 1) and R(x, 6)
∵ Q is the equidistant from P and R.
∴ PQ = RQ
⇒ PQ² = RQ²
⇒ (0 – 5)² + (1 + 3)² = (0 – x)² + (1 – 6)²
⇒ (- 5)² + (4)² = (- x)² + (- 5)²
⇒ 25 + 16 = x² + 25
⇒ 25 – 25 + 16 = x²
⇒ x² = 16
⇒ x = ± 4

Case II: x = 4

PR = \(\sqrt{(-1)^2+81}\)
PR = \(\sqrt{1+81}=\sqrt{82}\)

Case II: If x = – 4

Hence, x = ± 4, QR = √41 and PR = √82 or 9√2.

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-. 3, 4).
Solution:
Let P(x, y) be the equidistant from the points M(3, 6) and B(- 3, 4), then
PA = PB
PA² = PB² .
⇒ (3 – x)² + (6 – y)²
⇒ (- 3 – x)² + (4 – y)²
⇒ 9 + x² – 6x + 36 + y² – 12y = 9 + x² + 6x + 16 + y² – 8y
⇒ x² + y² + 6x – 8y + 6x + 12y – x² -y² = 45 – 25
⇒ 12x + 4y = 20
⇒ 3x + y = 5
⇒ 3x + y – 5 = 0
Hence, the required relation is 3x + y – 5 = 0