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.
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.
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.
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).
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.
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.
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).
Let P(x, y), A (3, 6) and B (– 3, 4)
Here AP = BP