Exercise 7.1

1. Find the distance between the following pairs of points :

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

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

Hence, x1 = , y1 = , x2 = , y2 =

We know that the distance between the two points is given by the Distance Formula,

= x2−x12+y2−y12....(1)

= 36−02+15−02

=1296+225

= 1521

=

YesNo, it is possible to find the distance between the given towns A and B.

The positions of towns A & B are given by (0, 0) and (36, 15), hence, as calculated above, the distance between town A and B will be 39 km.

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

Let the points (1, 5), (2, 3), and (- 2, - 11) be represented as A, B, and C.

For A, B, and C to be collinear, they must lie on the same line.

Hence, we will have to check if AB + BC = AC or BC + AC = AB or AB + AC = BC.

We know that the distance between any two points is given by,

Distance Formula = x2−x12+y2−y12 ....(1)

To find AB, the Distance between the Points A (1, 5) and B (2, 3), let x1 = , y1 = , x2 = , y2 =

∴ AB = 2−12+3−52 (By Substituting in (1))

= 5

To find BC Distance between Points B (2, 3) and C (- 2, - 11), let x1 = , y1 = , x2 = , y2 =

Therefore, BC = −2−22+−11−32

= −42+−142 (By Substituting in the Equation (1))

= 16 +

= 212

To find AC Distance between Points A (1, 5) and C (-2, -11), let x1 = , y1 = , x2 = , y2 =

Therefore, CA = −2−12+−11−52

= −32+−162 (By Substituting in the Equation (1))

= 9 +

= 265

AB = 5, BC = 212, CA = 265

Since AB + AC ≠ BC and BC + AC ≠ AB and AB + BC ≠ AC, therefore, the points (1, 5), (2, 3), and (- 2, - 11) are not collinear.

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

Distance Formula = x2−x12+y2−y12

To find AB, that is distance between points A (5, - 2) and B (6, 4), let x1 = , y1 = , x2 = , y2 =

AB = 6−52+4−−22

= 12+62

= 1 +

= 37

To find BC, distance between Points B (6, 4) and C (7, - 2), let x1 = , y1 = , x2 = , y2 =

BC = 7−62+−2−42

= 12+−62

= 1 +

= 37

To find AC, that is distance between Points A (5, - 2) and C (7, - 2), let x1 = , y1 = , x2 = , y2 =

AC = 7−52+−2−−22

= 22+02

=

From the above values of AB, BC and AC we can conclude that AB = BC. As the two sides are equal in length, therefore, ABC is an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 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

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

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

Distance Formula = x2−x12+y2−y12

Let's assume a point P on the x-axis which is of the form P(x, 0).

We have to find a point on the x-axis which is equidistant from A (2, - 5) and B (- 2, 9).

To find the distance between P and A, substitute the values of P (x, 0) and A (2, - 5) in the distance formula.

PA = x−22+0−−52

= x−22+52 --------- (1)

To find the distance between P and B, substitute the values of P (x, 0) and B (- 2, 9) in the distance formula.

PB = x−−22+0−92

= x+22+−92 ---------- (2)

By the given condition, these distances are equal in measure.

Hence, PA = PB

x−22+52 = x+22+−92 [From equation (1) and (2)]

Squaring on both sides, we get

x−22 + = x−22 +

x2 + - 4x + = x2 + + 4x +

x = 25 -

8x = -

x = -

Therefore, the point equidistant from the given points on the x-axis 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.

Distance Formula = x2−x12+y2−y12

By substituting the values of points P (2, - 3) and Q (10, y) in the distance formula, we get

PQ = 2−102+−3−y2 =

PQ = −82+3+y2 =

Squaring on both sides, we get

64 + y+32 =

y+32 =

y + 3 = 36

y + 3 = ±

y + 3 = 6 or y + 3 = -

Therefore, y = 3 or - 9 are the possible values for y.

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.

Distance Formula = x2−x12+y2−y12

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

So, PQ = QR

Hence by applying the distance formula we get,

5−02+−3−12= 0−x2+1−62

52+−42 = −x2+−52

By squaring both the sides,

25 + 16 = x2 +

16 = x2

x = ±

Therefore, point R is (4, 6) or (- 4, 6).

Case (1): When point R is (4, 6),

Distance between P (5, - 3) and R (4, 6) can be calculated using the Distance Formula as,

PR = 5−42+−3−62

= 12+−92

= 1 +

= 82

Distance between Q (0, 1) and R (4, 6) can be calculated using the distance formula as,

QR = 0−42+1−62

= −42+−52

= 16 +

= 41

Case (2): When point R is (- 4, 6)

Distance between P (5, - 3) and R (- 4, 6) can be calculated using the distance formula as,

PR = 5−−42+−3−62

= 92+−92

= 81 +

= 92

Distance between Q (0, 1) and R (- 4, 6) can be calculated using the distance formula as,

QR = 0−−42+1−62

= 42+−52

= 16 +

= 41

Thus, we see that using R (- 4, 6) we get PR = QR. Thus, the point is R (- 4, 6). Hence, x = - 4.