NCERT Solutions Maths Class 10 Exercise 7.1

# NCERT Solutions Maths Class 10 Exercise 7.1

## Maths Class 10 Exercise 7.1

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) Using the distance formula to find the distance between points (2, 3) and (4, 1), we have

(ii) Using the distance formula to find the distance between points (–5, 7) and (–1, 3), we have

(iii) Using the distance formula to find the distance between points (a, b) and (–a, –b), we have

2. Find the distance between the points (0, 0) and (36, 15). Also, find the distance between towns A and B if the town B is located at 36 km east and 15 km north of the town A.

Solution: Using the distance formula to find the distance between points (0, 0) and (36, 15), we have

Town B is located at 36 km east and 15 km north of town A. So, the location of town A and B can be shown as:

Clearly, the coordinates of point A are (0, 0) and coordinates of point B are (36, 15).

To find the distance between them, we use distance formula:

Thus, d = 39 km

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

Solution: Let A(1, 5), B(2, 3) and C(–2, –11) are three points.

Let’s use the distance formula to find the distances AB, BC and CA.

Since AB + BC ≠ AC

Therefore, the points A(1, 5), B(2, 3) and C(–2, –11) are not collinear.

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

Solution: Let A(5, –2), B(6, 4) and C(7, –2) are three points.

Let’s use the distance formula to find the distances AB, BC and CA.

Since AB = BC.

Therefore, A(5, –2), B(6, 4) and C(7, –2) are vertices of an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the 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: We have A(3, 4), B(6, 7), C(9, 4) and D(6, 1) are the four points where four friends are seated.

Let’s use the distance formula to find the distances AB, BC, CD and DA.

Therefore, all the sides of ABCD are equal here.

Now, we will check the length of its diagonals.

Let’s use the distance formula to find the distances AC and BD.

So, the diagonals of ABCD are also equal.

Since all the four sides and two diagonals of quadrilateral ABCD are equal.

We can definitely say that ABCD is a square.

Therefore, Champa is correct.

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) Let A(–1, –2), B(1, 0), C(–1, 2) and D(–3, 0) are the three points.

Let’s use the distance formula to find the distances AB, BC, CD and DA.

Therefore, all four sides of the quadrilateral ABCD are equal.

Now, we will check the lengths of diagonals.

Let’s use the distance formula to find the distances AC and BD.

Therefore, the diagonals of the quadrilateral ABCD are also equal.

Since all the four sides and two diagonals of quadrilateral ABCD are equal.

Therefore, the quadrilateral ABCD is a square.

(ii) Let A(–3, 5), B(3, 1), C(0, 3) and D(–1, –4) are the four points.

Let’s use the distance formula to find the distances AB, BC, CD and DA.

We could not find any relation between the lengths of the different sides.

Therefore, we cannot give any name to the quadrilateral ABCD.

(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) are the four points.

Let’s use the distance formula to find the distances AB, BC, CD and DA.

Here the opposite sides of the quadrilateral ABCD are equal.

Now, we will check the lengths of diagonals.

Let’s use the distance formula to find the distances AC and BD.

Here the diagonals of the quadrilateral ABCD are not equal.

In quadrilateral ABCD, the opposite sides are equal but the diagonals are not equal.

Therefore, the quadrilateral is a parallelogram.

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

Solution: Let the point be (x, 0) on x-axis which is equidistant from (2, –5) and (–2, 9).

Using distance formula according to the given conditions, we have

The distance between (x, 0) and (2, –5) = The distance between (x, 0) and (–2, 9)

Squaring both sides, we get

x2 + 4 − 4x + 25 = x2 + 4 + 4x +81

−4x + 29 = 4x + 85

8x = −56

x = −7

Therefore, the point on the x-axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0).

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

Solution: Using the distance formula, we have

Squaring both sides, we get

100 = 73 + y2 + 6y

y2 + 6y – 27 = 0

Let’s solve this quadratic equation by factorization.

y2 + 9y – 3y – 27 = 0

y(y + 9) – 3(y + 9) = 0

(y + 9) (y − 3) = 0

y = 3, −9

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: It is given that Q(0, 1) is equidistant from P(5, –3) and R(x, 6).

Since PQ = RQ

Using the distance formula, we get

Squaring both sides, we get

25 + 16 = x2 + 25

x2 = 16

x = 4, −4

Using the distance formula to find QR, we have

Therefore, QR = √41

Using the distance formula to find PR, we have

Therefore, x = 4, –4

QR =√41 and PR = √82, 9√2

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: According to question, the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

Using the distance formula, we have

Squaring both sides, we get

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

−6x − 12y + 45 = 6x − 8y + 25

12x + 4y = 20

3x + y = 5

This is the required relation between x and y.