# 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).

(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, â€“ 2), (6, 4) and (7, â€“ 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, "Donâ€™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) (â€“ 1, â€“ 2), (1, 0), (â€“ 1, 2), (â€“ 3, 0)

(ii) (â€“3, 5), (3, 1), (0, 3), (â€“1, â€“ 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2).(i) (â€“ 1, â€“ 2), (1, 0), (â€“ 1, 2), (â€“ 3, 0)

(ii) (â€“3, 5), (3, 1), (0, 3), (â€“1, â€“ 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2).

**Solution:**

**(i)**

**(â€“ 1, â€“ 2), (1, 0), (â€“ 1, 2), (â€“ 3, 0)**AB = = =

BC = =

CD = = =

DA = =

AB = BC = CD = DA

AC = = = 4

BD = = 4

Diagonals AC = BD

âˆ´ 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 â‰ BC â‰ CD â‰ DA

**âˆ´**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, â€“5) and (â€“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 â‡’ PX

^{2}= QX

^{2}

â‡’ (x - 2)

^{2}+ 25 = (x + 2)

^{2}+ 9

^{2}

Ãž x

^{2}- 4x + 4 + 25 = x

^{2}+ 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, â€“ 3) and Q(10, y) is 10 units.****Solution:**

Given PQ = 10

PQ =

= = 10

y

^{2}+ 6y + 73 = 100

y

^{2}+ 6y - 27 = 0

y

^{2}+ 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

â‡’ y = 3, y = -9

âˆ´ 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 PQ

^{2}= QR

^{2}

41 = x

^{2}+ 25

**âˆ´**x

^{2}= 16

â‡’ x = Â± 4

QR = =

PR = or =.

# Question-9

**Find the coordinates of the point which divides the join of (â€“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, â€“1) and (â€“2, â€“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.

âˆ´ The coordinates of C are

Now, D divides AB internally in the ratio, 2 : 1.

âˆ´ 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 2**

^{nd}line and posts a green flag. Preet runs 1/5

^{th}**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 Ã— 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 2

^{nd}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 8

^{th}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 (â€“ 3, 10) and (6, â€“ 8) is divided by (â€“ 1, 6).****Solution:**

Let the ratio be(m

_{1}, m

_{2})

âˆ´ The coordinates of C(-1, 6) =

= -1

6m

_{1}â€“ 3m

_{2}= -m

_{1}â€“ m

_{2}

6m

_{1}+ m

_{1}= -m

_{2}+ 3m

_{2}

7m

_{1}= 2m

_{2}

â‡’.

âˆ´ 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, â€“ 5) and B(â€“ 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 m

_{1 }: m

_{2}

The coordinate of C=

Given the y co-ordinate = 0 (since it cuts the x axis)

âˆ´m

_{1}: m

_{2}= 1 : 1 = 0

â‡’ 5m = 5m

_{2}â‡’ :1

**âˆ´**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

âˆ´ Coordinate of O = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Also B, D are divided at O in the ratio 1 : 1

âˆ´ Coordinate of O = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)

since both the coordinates in (1) and (2) are equal.

â‡’ x = 6

y = 8 â€“ 5 = 3

**âˆ´**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, â€“ 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)

â‡’ 1 + x = 4 = x = 3

Ãž y = -6 â€“ 4 = -10

âˆ´ The coordinates of A (x, y) is (3, -10).

# Question-16

**If A and B are (â€“ 2, â€“ 2) and (2, â€“ 4), respectively, find the coordinates of P such**

that AP =that AP =

**AB and P lies on the line segment AB.****Solution:**

Given AP = AB

**âˆ´**PB =

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

Also A(-2, -2) = A(x

_{1}, y

_{1}) and B(2, -4) = B(x

_{2}, y

_{2})

Co-ordinates of P is

# Question-17

**Find the coordinates of the points which divide the line segment joining A(â€“ 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(x

_{1}, y

_{1}), B(2, 8) = B(x

_{2}, y

_{2})

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), (â€“ 1, 4) and (â€“ 2, â€“ 1)**

taken in order. [Hint: Area of a rhombus =(product of its diagonals)]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)(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

**.****Solution:**

(i)

A = (x

_{1}, y

_{1}) = (2, 3)

B = (x

_{2}, y

_{2}) = (-1, 0)

C = (x

_{3}, y

_{3}) = (2, -4)

Area of triangle =

= [2(0 + 4) -1(-4 - 3) + 2(3 - 0)]

= [8 + 7 + 6] = sq.units.

(ii)

A = (x

_{1}, y

_{1}) = (-5, -1)

B = (x

_{2}, y

_{2}) = (3, -5)

C = (x

_{3}, y

_{3}) = (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).(i) (7, -2), (5, 1), (3, k)

(ii) (8, 1), (k, -4), (2, -5).

**Solution:**

(i)

Given the points are collinear

â‡’ 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)

**âˆµ**Area of Î” ABC = 0

**âˆ´**= 0

= 18 â€“ 6k = 0

**âˆ´**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.

**âˆ´**Area of triangle ABC = [0(1 - 3) + 2 (3 + 1) +0 ( -1 -1)

= -1 = | -1 | = 1 (

**âˆµ**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)

**âˆ´**Area of DPQR = [1(1 - 2) + 0(2 - 0) + 1(0 - 1)]

=

= -1 = | -1 | = 1 (

**âˆµ**Area of triangle is non negative)

**âˆ´**Area of DPQR = 1 unit.

Hence the ratio of area of triangle ABC and area of triangle PQR = 1 unit.

**âˆ´**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 Î” ABC + Area of Î” ADC

Area of Î” ABC =

= = âˆ´ Area of Î” 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 Î” ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).****Solution:**

If AD is the median of Î”ABC. D is the midpoint of BC given by (ie) (4, 0)

Hence Co-ordinates of D is (4, 0)

\ Area of Î”ABD =

= = = 3

Area of Î”ADC =[4(0 + 2) + 4(-2 + 6) + 5(-6 - 0)]

=[8 + 16 - 30] =

**âˆ´**Area of Î” ABD = Area of Î” ADC = .