Jharkhand Board JAC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

Question 1.

Find the distance between the following pairs of points :

1. (2, 3), (4, 1)

2. (- 5, 7), (1, 3)

3. (a, b), (- a, – b)

Answer:

1. Let A (2, 3) and B (4, 1) be the given points. Then,

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

= \(\sqrt{4+4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)

Thus, the distance between the given points is 2\(\sqrt{2}\).

2. Let A (- 5, 7) and B (- 1, 3) be the given points. Then,

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

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

Thus, the distance between the given points is 4\(\sqrt{2}\).

3. Let P (a, b) and Q(- a, – b) be the given points. Then,

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

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

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

Thus, the distance between the given points is 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 discussed in section 7.2.

Answer:

Let A (0, 0) and B (36, 15) be the given points.

Then,

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

= \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39

Thus, the distance between the given points is 39.

Yes, now we can find the distance between the two towns A and B discussed in section 7.2 in the textbook.

Town B is located 36 km east and 15 km north of the town B. So, if we take town A to be situated at the origin with coordinates (0, 0) then the coordinates of town B become (36, 15). Now, as calculated above, the distance between A(0, 0) and B (36, 15) is 39. Hence, the distance between town A and town B is 39 km.

Question 3.

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

Answer:

Let A(1, 5), B(2, 3) and C (- 2, – 11) be the given points. Then,

Now, 14.56 + 2.24 = 16.80 ≠ 16.28

Thus, AB + BC ≠ AC

Moreover, BC + AC ≠ AB and AC + AB ≠ BC are obvious.

Hence, the given points are not collinear.

Question 4.

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

Answer:

Let the given points A (5, – 2), B (6, 4) and C(7, – 2) be the vertices of ΔABC. Then,

Here, in ΔABC, AB = BC ≠ AC.

Hence, ABC is an isosceles triangle.

Thus, the given points (5, – 2), (6, 4) and (7, – 2) 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 the given 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.

Answer:

Here A(3, 4), B(6, 7), C (9, 4) and D(6, 1) give the position of points where those four friends are seated. Then,

Thus, in quadrilateral ABCD,

AB = BC = CD = DA and AC = BD.

In other words, all the sides of quadrilateral ABCD are equal and its diagonals are also equal.

Hence, ABCD is a square.

So. Champa is correct stating that ABCD is a square.

Question 6.

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

1. (- 1, – 2), (1, 0), (1, 2), (- 3, 0)

2. (- 3, 5), (3, 1), (0, 3), (- 1, – 4)

3. (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

1. If possible, let A (- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) be the vertices of quadrilateral ABCD. Then,

Thus, in quadrilateral ABCD,

AB = BC = CD = DA and AC = BD.

Hence, ABCD is a square.

2. If possible, let A (- 3, 5), B(3, 1), C(0, 3) and D(- 1, – 4) be the vertices of quadrilateral ABCD. Then,

Hence, A, B and C are collinear points in which C lies between A and B.

Hence, A, B, C and D do not form a quadrilateral.

3. If possible, let A (4, 5), B(7, 6), C(4, 3) and D (1, 2) be the vertices of quadrilateral ABCD. Then,

Thus, in quadrilateral ABCD, AB = CD, BC = DA, i.e., both the pairs of opposite sides are equal, but AC ≠ BD. i.e.. diagonals are not equal.

Hence, ABCD is a parallelogram.

Question 7.

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

Answer:

Here, A (2, – 5) and B(- 2, 9) are two given points.

Coordinates of any point of the x-axis are of the form (x, 0).

Let P(x, 0) be the required point on the x-axis which is equidistant from A and B.

So, PA = PB

∴ PA² = PB²

∴ (x – 2)² + (0 + 5)² = (x + 2)² +(0 – 9)²

∴ x² – 4x + 4 + 25 = x² + 4x + 4 + 81

∴ -8x=56

∴ x = – 7

Thus, the required point on the x-axis which is equidistant from (2, – 5) and (- 2, 9) 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.

Answer:

The distance between P(2, – 3) and 9(10, y) is given to be 10.

∴ PQ = 10

∴ PQ² = 100

∴ (2 – 10)² + (- 3 – y)² = 100

∴ 64 + 9 + 6y + y² = 100

∴ y² + 6y – 27 = 0

∴ (y + 9) (y – 3) = 0

∴ y + 9 = 0 or y – 3 = 0

∴ y = – 9 or y = 3

Thus, the required values of y are – 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.

Answer:

Q(0, 1) is equidistant from P(5, -3) and R (x, 6).

∴ PQ = RQ

∴ PQ² = RQ²

∴ (5 – 0)² + (- 3 – 1)² = (x – 0)² + (6 – 1)²

∴ 25 + 16 = x² + 25

∴ x² = 16

∴ x = ± 4

∴ x = 4 or x = – 4

Thus, x = ± 4, QR = \(\sqrt{41}\) and PR = \(\sqrt{82}\) or 9\(\sqrt{2}\).

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).

Answer:

Here, point P(x, y) is given to be equidistant from points A (3, 6) and B (- 3, 4).

∴ PA = PB

∴ PA² = PB²

∴ (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²

∴ x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16

∴ – 12x – 4y + 20 = 0

∴ 3x + y – 5 = 0 (Dividing by – 4)

Thus, 3x + y – 5 = 0 is the required relation between x and y.