## Solution of Exercise 7.1 (NCERT)

Question: 1 - Find the distance between the following pairs of points:

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:

In particular, the distance of a point (x, y) from the origin is given by

Now, as discussed in Section 7.2 in the book, town B is 36 east and 15 north from the town A. Thus, distance between town A and B will be equal to 39 unit. If distance is measured in km, then distance will be equal to 39 km.

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

Solution:

Thus, distance between AB by distance formula

Now, distance between BC by distance formula

And distance between AC by distance formula

Thus, given points are non-collinear.

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

Thus, given triangle is an isosceles triangle.

Question: 5 – In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa as Chameli, “Don’t you think ABCD is square?” Chameli disagrees. Using distance formula find which of them is correct.

Solution:

Here, coordirantes are A (3, 4), B (6, 7), C (9, 4) and D (6, 1)

Since, all four sides are equal and diagonals are also equal thus, it is clear that ABCD is a square and hence, Champa is right.

Question: 6 – Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

Thus, ABCD is a square.

Since, sides are not equal, thus only a general and not a specific quadrilateral such as square or parallelogram is formed.

Here opposite sides are equal but diagonals are not equal, thus with the given points a parallelogram is formed.

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

Solution:

Let given coordinates are A(2, –5), BC (–2, 9) and P(x, 0) is on the x-axis.

Since, the third coordinate is at equidistance from the given coordinates, thus AP = BP

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

Solution:

Given, P(2, –3) and Q(10, y) and QP = 10 unit

Thus, required value = 3 or – 9

Question: 9 – If Q (0, 1) is equidistant from P(5, – 3) and R (x, 6), find the value of x. Also find the distances QR and PR.

Solution:

Given, Q(0, 1), P(5, –3) and R(x, 6) and QP = QR

Question: 10 – Find the relation between x and y such that the point (x, y) is equidistance from the point (3, 6) and (–3, 4).

Solution:

Let P(x, y), A (3, 6) and B (– 3, 4)

Here AP = BP

## Post a Comment