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Question-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) (2, 3) (4, 1)
Distance =
             =
             = = 2

(ii) (-5, 7) (-1, 3)
Distance =
             = = 4

(iii) (a, b) (-a, -b)
Distance =
             = .

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.

Solution:
Let A (0, 0), B(36, 15)
AB =
    =
    =
In section 7.2, A is (36, 0)
And B is (0, 15)
Distance is = .

Question-3

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 (7,-2) be the vertics of triangle.

then we have


Here AB=BC
triangle ABC is  an isosceles triangle.

Question-4

In a classroom, 4 friends are seated at the points A, B, C and D as shown in this 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:
In the figure A is (3, 4)
B is (6. 7)
C is (9, 4)
D is (6, 1)
AB = = = 3
BC = = 3
CD = = 3
DA = = 3
Since AB = BC = CD = DA
ABCD is a square or rhombus
BD = =
AC = =
Diagonal, BD = AC hence it is a square.

Question-5

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

AB = = =
BC = =
CD = = =
DA = =
AB = BC = CD = DA
AC = = = 4
BD = = 4
Diagonals AC = BD

ABCD is a square

(ii) (-3, 5) (3 , 1) (0, 3) (-1, -4)
A = (-3, 5)
B = (3 , 1)
C = (0, 3)
D = (-1, -4)
AB = = =
BC = = =
CD = = =
DA = = =
AB BC CD DA

 It is a quadrilateral

(iii) (4, 5), (7, 6), (4, 3), (1, 2)
A = (4 , 5)
B = (7 , 6)
C = (4 , 3)
D = (1 , 2)
AB = =
BC = =
CD = =
DA = =
AB = CD
BC = DA.
Opposite sides are equal so it is a parallelogram.

Question-6

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

Solution:
Let the point on x axis be X(x, 0)
Let P(2, - 5), Q(-2, 9)
PX =
    =
QX =

Given PX = QX PX2 = QX2

  (x - 2)2 + 25 = (x + 2)2 + 92   
Þ x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
8x = 56
x = 7
The required point on x - axis is (7, 0).

Question-7

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

Solution:
Given PQ = 10
PQ =
     = = 10
y2 + 6y + 73 = 100
y2 + 6y - 27 = 0
y2 + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0

⇒ y = 3, y = -9
∴ Co-ordinate of y are -9 or 3.


Question-8

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:
Q = (0, 1)
P = (5, -3)
R = (x, 6)
PQ = QR (given)
PQ = = =
QR = =
Given PQ2 = QR2
41 = x2 + 25

 x2 = 16
⇒ x = ± 4
QR =
=
PR = or =.

Question-9

Find the coordinates of the point which divides the join of (–1, 7) and (4, -3) in the ratio 2 : 3.

Solution:
Given, A(-1, 7) and B(4, -3) divides internally in the ratio 2 : 3.
Let C be the required point, using section formula, we get
 

Question-10

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Solution:


Let C and D be the point of trisection C divides AB in the ratio 1 : 2 and D divides AB in the ratio 2 : 1.
∴ The coordinates of C are

                                      
Now, D divides AB internally in the ratio, 2 : 1.
∴ The coordinates of D are

Question-11

 To conduct Sports Day activities, in your rectangular shaped school ground ABCD lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs
1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?.

 


Solution:
Now let us find out the distance AD. Since 100 flower pots have been arranged at a distance of 1 m between each pot, the total distance of AD is (100 × 1 m) = 100 m.
To find out the location of the green flag taken by Niharika please observe below)
Distance run by Niharika = = 25 m
As Niharika starts from the 2nd line the position of green flag P is (2, 25) (To find out the location of the blue flag taken by Preet please observe below)
Distance run by preet = = 20 m.
As Preet starts from 8th line location of red flag Q is (8, 20) (To find the distance between the two flags, the green at P and red at Q we find the same using distance formula)
Distance between P(2, 25) and Q(8, 20) using the formula

                                           
                                                                                 =
                                                                                 =
                                                                                 =
                                                                                 = 7.81 m
(To find the position where Rashmi can post her blue flag exactly between the other two flags we can find the same using the mid-point formula)
Mid point of P(2, 25) and Q(8, 20) using the formula
                                                                       =
                                                                       =
Rashmi has to travel a distance of 22.5 starting from the fifth line.

Question-12

Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Solution:

Let the ratio be(m1, m2)

∴ The coordinates of C(-1, 6) = 
= -1
6m1 – 3m2 = -m1 – m2
6m1 + m1 = -m2 + 3m2
7m1 = 2m2

.
∴ The ratio at which the point (-1, 6) divides the line segments joining the points (-3, 10) and (6, -8) is 2 : 7.

Question-13

Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:
A(1, -5), B(-4, 5)
Let the co-ordinate of point of division be (x, 0)
Let the ratio be m1 : m2

The coordinate of C= 
Given the y co-ordinate = 0 (since it cuts the x axis)
m1 : m2 = 1 : 1 = 0
5m = 5m2 :1

The co-ordinates of point of division = ( - 3/2, 0)

Question-14

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

Ler A (1, 2), B (4, y), C (x, 6), D(3, 5) are the vertices of parallelogram ABCD. 
 
We know that diagonals of a parallelogram bisect each other
A, C are divided at O in the ratio 1 : 1

∴ Coordinate of O =  ……………………………. (1)
Also B, D are divided at O in the ratio 1 : 1

∴ Coordinate of O =  …………………………. (2)
since both the coordinates in (1) and (2) are equal.

x = 6

y = 8 – 5 = 3


 x = 6, y = 3.

Question-15

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Solution:
Let the coordinate of A be (x, y), O(2, -3) is the mid point of AB.
Coordinate of O dividing AB in ratio 1 : 1 is

 Given the centre of the circle is O(2, -3)
1 + x = 4 = x = 3
Þ y = -6 – 4 = -10
∴ The coordinates of A (x, y) is (3, -10).

Question-16

If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such
that AP =
AB and P lies on the line segment AB.

Solution:

Given AP = AB

∴ PB =
(ie) P divides AB in the ratio 3 : 4

Also A(-2, -2) = A(x1, y1) and B(2, -4) = B(x2, y2)
Co-ordinates of P is


Question-17

Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Solution:

Let P, Q, R be the points which divide the line into 4 equal parts
To find P, AP : PB is 1 : 3

A(-2, 2) = A(x1, y1),  B(2, 8) = B(x2, y2)  
Co-ordinates of P
= =
To find Q, AQ : QB = 2 : 2 (ie) 1 : 1
Coordinates of Q =
                       =  = (0, 5)
To find R, AR : RB is 3 : 1
Co-ordinates of R  = (1, )
Required Co-ordinates (-1, 7/2), (0, 5), (1, 
) = (1, ).

Question-18

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1)
taken in order. [Hint: Area of a rhombus =(product of its diagonals)]
.

Solution:

Area of rhombus = 1/2 (product of its diagonals)
Let A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) be the vertices of the rhombus.
BD = = =
AC = = =
Area of the rhombus =
                               = = 24 sq. unit.

Question-19

Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
.

Solution:
(i)
A = (x1, y1) = (2, 3)
B = (x2, y2) = (-1, 0)
C = (x3, y3) = (2, -4)
Area of triangle = 
 
                      = [2(0 + 4) -1(-4 - 3) + 2(3 - 0)]
                      =
[8 + 7 + 6] = sq.units.
(ii)
A = (x1, y1) = (-5, -1)
B = (x2, y2) = (3, -5)
C = (x3, y3) = (5, 2)
Area of triangle =
                     

Question-20

In each of the following find the value of "k", for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5).

Solution:
(i)
Given the points are collinear

⇒ Area of triangle = 0
A (7, -2)
B (5, 1)
C (3, k)
Area of triangle =
(i.e.,) = 0
-2k + 8 = 0
k = 4

(ii)

A(8, 1)
B(k, -4)
C(2, -5)

 Area of Δ ABC = 0
  = 0
           = 18 – 6k = 0

 k = 3.

Question-21

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution:

Let A(0, -1), B(2, 1) and C(0, 3) be vertices of the DABC.
Area of triangle ABC =  [0(1 - 3) + 2 (3 + 1) +0 ( -1 -1)
                                    = -1 = | -1 | = 1   (∵ Area of triangle is non negative)
Let P, Q, R be the midpoints of the sides
Co-ordinates of P are = (1, 0)
Co-ordinates of Q are = (0, 1)
Co-ordinates of R are = (1, 2)
 Area of DPQR = [1(1 - 2) + 0(2 - 0) + 1(0 - 1)] 
                           =
                           = -1 = | -1 | = 1   (Area of triangle is non negative)
 Area of DPQR = 1 unit.
Hence the ratio of area of triangle ABC and area of triangle PQR  = 1 unit.
The ratio of area of triangle ABC and area of triangle PQR = 1 : 1

Question-22

Find the area of the quadrilateral whose vertices are (-4, -2), (-3, -5) and (3, -2), (2, 3).

Solution:

Area of quadrilateral = Area of Δ ABC + Area of Δ ADC
Area of Δ ABC =
                    = = Area of Δ ADC =
                       = =
                       =
Area of quadrilateral = + == 28 sq.units.

Question-23

The median of a triangle divides it into two triangles of equal areas . Verify this result for Δ ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:

If AD is the median of ΔABC. D is the midpoint of BC given by (ie) (4, 0)
Hence Co-ordinates of D is (4, 0)
Area of ΔABD =
= = = 3

Area of ΔADC =[4(0 + 2) + 4(-2 + 6) + 5(-6 - 0)]  
=
[8 + 16 - 30] 
 Area of Δ ABD = Area of Δ ADC = .




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