7/4/2018 ... · topic: section formulae find the coordinates of a point , where is the diameter of...

7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=465332%2C+4653…

https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=465332%2C+465331%2C+46… 1/20

#465322

Topic: Distance Between Two Points

Find the distance between the following pairs of points :

(i)

(ii)

(iii)

Solution

Distance between two points and is

(i) Distance between and is

(ii) Distance between and is

(iii )Distance between and is

#465323


Find the distance between the points and .

Solution

Distance Between two given point=

Here

and

Distance between the points and =

#465324


Determine if the points and are collinear.

Solution

Three points , and are collinear if

Here, point and

Hence the given points are collinear.

(2, 3), (4, 1)

(−5, 7), (−1, 3)

(a, b), (−a, −b)

( , )x1 y1 ( , )x2 y2

( − + ( −x1 x2)2 y1 y2)2− −−−−−−−−−−−−−−−−−√

(2, 3) (4, 1)

= (2 − 4 + (3 − 1)2 )2− −−−−−−−−−−−−−−√

= = = = 2(−2 + (2)2 )2− −−−−−−−−−√ 4 + 4− −−−√ 8√ 2√

(−5, 7) (−1, 3)

= (−5 − (−1) + (7 − 3)2 )2− −−−−−−−−−−−−−−−−−−√

= = = = 4(−4 + (4)2 )2− −−−−−−−−−√ 16 + 16− −−−−−√ 32−−√ 2√

(a, b) (−a, −b)

= (a− (−a) + (b− (−b))2 )2− −−−−−−−−−−−−−−−−−−−√

= = = 2(2a + (2b)2 )2− −−−−−−−−−√ 4 + 4a2 b2− −−−−−−−√ +a2 b2− −−−−−√

(0, 0) (36, 15)

( − + ( −x2 x1)2 y2 y1)2− −−−−−−−−−−−−−−−−−√

= 0, = 26x1 x2

= 0, = 15y1 y2

∴ (0, 0) (36, 15) (36 − 0 + (15 − 0)2 )2− −−−−−−−−−−−−−−−√

= +362 152− −−−−−−−√

= 1296 + 225− −−−−−−−−√

= = 391521− −−−√

(1, 5), (2, 3) (−2, −11)

A B C

AB+BC = AC

A(1, 5),B(2, 3) C(−2, −11).

∴ AB = = = = = 2.23(2 − 1 + (3 − 5)2 )2− −−−−−−−−−−−−−−√ + (− )12 22− −−−−−−−−

√ 1 + 4− −−−√ 5√

BC = = = = = 14.56((−2) − (2) + ((−11) − (3))2 )2− −−−−−−−−−−−−−−−−−−−−−−−√ (−4 + (−14)2 )2− −−−−−−−−−−−√ 16 + 196− −−−−−−√ 212−−−√

AC = = − 3 + (−16 = = = 16.27((−2) − (1) + ((−11) + (5))2 )2− −−−−−−−−−−−−−−−−−−−−−−−√ (√ )2 )2 9 + 256− −−−−−√ 265−−−√

AB+BC = 2.23 + 14.56 = 16.79

AB+BC ≃ AC



#465325


Check whether and are the vertices of an isosceles triangle.

Solution

Let the given points are , and

Here . Hence ABC is an isosceles.

#465327


In a classroom, friends are seated at the points , , and as shown in Fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks

Chameli, Dont you think is a square? Chameli disagrees. Using distance formula, find which of them is correct.

Solution

According to the fig. ), , and are the position of friends .

If we join all these point then

if we join the point and than

Diagonal

Diagonal

Hence all the sides of quadrilateral are same and the diagonal are also same so is a square.

Hence Champa is correct.

#465328


(5, −2), (6, 4) (7, −2)

A = (−5, 2) B = (6, 4) C = (−7, 2)

∴ AB = = = =(6 − 5 + (4 − (−2)2)2 )2− −−−−−−−−−−−−−−−−−√ +12 62− −−−−−√ 1 + 36− −−−−√ 37−−√

BC = = = =(7 − 6 + ((−2) − 4)2 )2− −−−−−−−−−−−−−−−−√ +12 62− −−−−−√ 1 + 36− −−−−√ 37−−√

AC = = = = 2(7 − 5 + ((−2) − (−2))2 )2− −−−−−−−−−−−−−−−−−−−√ +22 02− −−−−−√ 4√

AB = BC △

4 A B C D

ABCD

A(3, 4 B(6, 7) C(9, 4) D(6, 1) 4

AB = = = = = 3(3 − 6 + (4 − 7)2 )2− −−−−−−−−−−−−−−√ (−3 + (−3)2 )2− −−−−−−−−−−√ 9 + 9− −−−√ 18−−√ 2√

BC = = = = = 3(6 − 9 + (7 − 4)2 )2− −−−−−−−−−−−−−−√ (−3 + (3)2 )2− −−−−−−−−−√ 9 + 9− −−−√ 18−−√ 2√

CB = = = = = 3(9 − 6 + (4 − 1)2 )2− −−−−−−−−−−−−−−√ (3 + ( ))2 32− −−−−−−−−

√ 9 + 9− −−−√ 18−−√ 2√

AD = = = = = 3(3 − 6 + (4 − 1)3 )2− −−−−−−−−−−−−−−√ −3 + (3)2 )2− −−−−−−−−√ 9 + 9− −−−√ 18−−√ 2√

AC BD

AC = = = = 6(3 − 9 + (4 − 4)2 )2− −−−−−−−−−−−−−−√ (−6 + (0)2 )2− −−−−−−−−−√ 36−−√

BD = = = = 6(6 − 6 + (7 − 1)2 )2− −−−−−−−−−−−−−−√ (0 + (6)2 )2− −−−−−−−−√ 36−−√

ABCD



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

(i)

(ii)

(iii)

Solution

(i) Let the given points are , , and Then,

Since the four sides and are equal and the diagonals and are equal .

Quadrilateral is a square.

(ii)Let the given points are , , and Then

Here

it is not a quadrilateral.

(iii)Let the given points are , , and Then

Here . But

Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.

#465329


Find the point on the X-axis which is equidistant from and .

Solution

(−1, −2), (1, 0), (−1, 2), (−3, 0)

(−3, 5), (3, 1), (0, 3), (−1, −4)

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

A(−1, −2) B(1, 0) C(−1, 2) D(−3, 0)

AB = = = =(1 + 1 + (0 + 2)2 )2− −−−−−−−−−−−−−−√ +22 22− −−−−−√ 4 + 4− −−−√ 8√

BC = = = =(−1 − 1 + (2 − 0)2 )2− −−−−−−−−−−−−−−−√ ( + )22 22− −−−−−−

√ 4 + 4− −−−√ 8√

CD = = = =((−3) − (−1) + (0 − 2)2 )2− −−−−−−−−−−−−−−−−−−−√ + (−222 )2− −−−−−−−−

√ 4 + 4− −−−√ 8√

DA = = = =(−3) − (−1) + (0 − (−2))2 )2− −−−−−−−−−−−−−−−−−−−−−√ (−2 +)2 22− −−−−−−−−

√ 4 + 4− −−−√ 8√

AC = = = 4((−1) − (−1) + (2 − (−2))2 )2− −−−−−−−−−−−−−−−−−−−−−−√ 0 + 42− −−−−√ 16−−√

BD = = = = 4(−3 − 1 + (0 − 0)2 )2− −−−−−−−−−−−−−−−√ −42− −−√ 16−−√

AB,BC,CD DA AC BD

∴ ABCD

A(−3, 5) B(3, 1) C(0, 3) D(−1, −4)

AB = = = =(−3 − 3 + (5 − 1)2 )2− −−−−−−−−−−−−−−−√ (−6 +)2 42− −−−−−−−−

√ 36 + 16− −−−−−√ 52−−√

BC = = = =(3 − 0 + (1 − 3)2 )2− −−−−−−−−−−−−−−√ ( + (−2) )32 22− −−−−−−−−−−

√ 9 + 4− −−−√ 11−−√

CD = = = =(0 − (−1) + (3 − (−4))2 )2− −−−−−−−−−−−−−−−−−−−√ + (712 )2− −−−−−−

√ 1 + 49− −−−−√ 50−−√

DA = = = =(−1) − (−3) + ((−4) − 5))2 )2− −−−−−−−−−−−−−−−−−−−−−−√ (2 + (−9)2 )2− −−−−−−−−−√ 4 + 81− −−−−√ 85−−√

AB ≠ BC ≠ CD ≠ DA

∴

A(4, 5) B(7, 6) C(4, 3) D(1, 2)

AB = = = =(7 − 4 + (6 − 5)2 )2− −−−−−−−−−−−−−−√ +32 12− −−−−−√ 9 + 1− −−−√ 10−−√

BC = = = =(4 − 7 + (3 − 6)2 )2− −−−−−−−−−−−−−−√ ((−3 + (−3 ))2 )2− −−−−−−−−−−−−√ 9 + 9− −−−√ 18−−√

CD = = = =(1 − 4 + (2 − 3)2 )2− −−−−−−−−−−−−−−√ (−3 + (−1)2 )2− −−−−−−−−−−√ 9 + 1− −−−√ 10−−√

DA = = = =(1 − 4 + (2 − 5)2 )2− −−−−−−−−−−−−−−√ (−3 + (−3)2 )2− −−−−−−−−−−√ 9 + 9− −−−√ 18−−√

AC = = = 2(4 − 4 + (3 − 5)2 )2− −−−−−−−−−−−−−−√ 0 + (−2)2− −−−−−−−√ 4√

BD = = = =(1 − 7 + (2 − 6)2 )2− −−−−−−−−−−−−−−√ (−6 + (−4)2 )2− −−−−−−−−−−√ 36 + 16− −−−−−√ 52−−√

AB = CD,BC = DA AC ≠ BD

(2, −5) (−2, 9)



Let the required point on the X-axis be

As given is equidistant from and

Required point on X-axis

#465330


Find the values of y for which the distance between the points and is units.

Solution

As given distance between point and

here

Take the square of both side

the possible values of are .

#465331


If is equidistant from and , find the values of . Also find the distances and .

Solution

As given is equidistant from and

Distance between and

Distance between and

P (x, 0)

P A(2, −5) B(−2, 9)

∴ PA = PB

⇒ =(x− 2 + (0 − (−5))2 )2− −−−−−−−−−−−−−−−−√ (x+ 2 + (0 − 9)2 )2− −−−−−−−−−−−−−−√

⇒ (x− 2 + 25 = (x+ 2 + 81)2 )2

⇒ − 4x+ 4 + 25 = + 4x+ 4 + 81x2 x2

⇒ −8x = 85 − 29

⇒ −8x = 56

⇒ x = = −756

−8∴ = (−7, 0)

P (2, −3) Q(10, y) 10

P Q = 10

∴ PQ = ( − + ( −x2 x1)2 y2 y1)2− −−−−−−−−−−−−−−−−−√

= 2, = 10, = −3, = yx1 x2 y1 y2

⇒ 10 = (10 − 2 + (y − (−3))2 )2− −−−−−−−−−−−−−−−−−√

⇒ 100 = (8 + (y + 3)2 )2

⇒ 100 = 64 + + 6y + 9y2

⇒ 100 = + 6y + 73y2

⇒ + 6y − 27 = 0y2

⇒ + 9y − 3y − 27 = 0y2

⇒ y(y + 9) − 3(y + 9) = 0

⇒ (y + 9)(y − 3) = 0

⇒ y = −9, 3

∴ y (−9, 3)

Q(0, 1) P (5, −3) R(x, 6) x QR PR

Q(0, 1) P (5, −3) R(x, 6)

∴ PQ = QR

⇒ =(0 − 5 + (1 − (−3))2 )2− −−−−−−−−−−−−−−−−√ (x− 0 + (1 − 6)2 )2− −−−−−−−−−−−−−−√

⇒ 25 + 16 = + 25x2

⇒ 41 = + 25x2

⇒ = 16x2

⇒ x = 4

Q(0, 1) R(4, 6)

QR = = = =(4 − 0 + (6 − 1)2 )2− −−−−−−−−−−−−−−√ +42 52− −−−−−√ 16 + 25− −−−−−√ 41−−√

P (5, −3) R(4, 6)

PR = = =(4 − 5 + (6 + 3)2 )2− −−−−−−−−−−−−−−√ 1 + 81− −−−−√ 82−−√



#465332


Find a relation between and such that the point is equidistant from the point and .

Solution

Let the point is equidistant from the points and

Hence this is a relation between and .

#465333

Topic: Section Formulae

Find the coordinates of the point which divides the join of and in the ratio .

Solution

Let the co-ordinates of be

Here

and

Here

#465334


Find the coordinates of the points of trisection of the line segment joining and .

Solution

x y (x, y) (3, 6) (−3, 4)

P (x, y) A(3, 6) B(−3, 4)

∴ PA = PB

⇒ =(x− 3 + (y − 6)2 )2− −−−−−−−−−−−−−−√ (x+ 3 + (y − 4)2 )2− −−−−−−−−−−−−−−√

⇒ (x− 3 + (y − 6 = (x+ 3 + (y − 4)2 )2 )2 )2

⇒ − 6x+ 9 + − 12y + 36 = + 6x+ 9 + − 8y + 16x2 y2 x2 y2

⇒ −6x− 6x− 12y + 8y + 36 − 16 = 0

⇒ −12x− 4y + 20 = 0

⇒ 3x+ y − 5 = 0

x y

(−1, 7) (4, −3) 2 : 3

P (x, y)

∴ x =+m1x2 m2x1

+m1 m2

= 2, = 3, = −1, = 4m1 m2 x1 x2

⇒ x =2 × 4 + 3 × −1

2 + 3

⇒ x = = = 18 − 3

5

5

5

y =+m1y2 m2y1

+m1 m2

= 7, = −3y1 y2

⇒ y =2 × −3 + 3 × 7

2 + 3

⇒ y = = = 3−6 + 21

5

15

5∴ P = (1, 3)

(4, −1) (−2, −3)



Let and be the points of trisection of and

so

For :

and

For :

and

#465335


To conduct Sports Day activities, in your rectangular shaped school ground , lines have been drawn with chalk powder at a distance of m each. flower pots have

been placed at a distance of m from each other along , as shown in Fig. Niharika runs th the distance on the line and posts a green flag. Preet runs th the

distance 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

A(4, −1) B(−2, −3) P Q

AP = PQ = QB

P

: = AP : PB = 1 : 2;m1 m2

( , ) = (4, −1)x1 y1

( , ) = (−2, −3)x2 y2

∴ x =+m1x2 m2x1

+m1 m2

⇒ x = = = = 21 × −2 + 2 × 4

1 + 2

−2 + 8

3

6

3

∴ y = = = =+m1y2 m2y1

+m1 m2

1 × −3 + 2 × −1

1 + 2

−3 − 2

3

−5

3

∴ P = (2, )−5

3Q

: = AQ : QB = 2 : 1;m1 m2

( , ) = (4, −1)x1 y1

( , ) = (−2, −3)x2 y2

∴ x =+m1x2 m2x1

+m1 m2

⇒ x = = = = 02 × −2 + 1 × 4

1 + 2

−4 + 4

3

0

3

∴ y = = = =+m1y2 m2y1

+m1 m2

2 × −3 + 1 × −1

1 + 2

−6 − 1

3

−7

3

∴ Q = (0, )−7

3

ABCD 1 100

1 AD1

4AD 2nd

1

5AD



m and m

The coordinates of Niharika's flag and the coordinates of Preet's flag .

and

AB =

Distance between both flags is m.

Rashmi has to place her flag at the midpoint of .

MIdpoint of AB = =

Rashmi must place her flag along the fifth line at m from .

#465336


Find the ratio in which the line segment joining the points and is divided by .

Solution

Let the required ratio be

Take and

The required ratio is .

#465337


Find the ratio in which the line segment joining and is divided by the X-axis. Also find the coordinates of the point of division.

Solution

Let the required ratio be and the point on the X-axis be

Let

Required ratio

The ratio and the required point of intersection

AD = 100 AB = 10

= (2, 25) = (8, 20)

A(2, 25) B(8, 20)

+(20 − 25)2 (8 − 2)2− −−−−−−−−−−−−−−−

√

= (−5 +)2 62− −−−−−−−−

√

= 25 + 36− −−−−−√

= 61−−√

61−−√

AB

( , )2 + 8

2

25 + 20

2(5, 22.5)

22.5 A

(−3, 10) (6, −8) (−1, 6)

:m1 m2

( , ) = (−3, 10); ( , ) = (6, −8)x1 y1 x2 y2 (x, y) = (−1, 6)

∴ x =+m1x2 m2x1

+m1 m2

⇒ −1 =× 6 + × −3m1 m2

+m1 m2

⇒ − − = 6m− 3m1 m2 m2

⇒ − − 6 = −3 +m1 m1 m2 m2

⇒ −7 = −2m1 m2

⇒ =m1

m2

2

7

∴ 2 : 7

A(1, −5) B(−4, 5)

k : 1 (x, 0)

( , ) = (1, −5), ( , ) = (−4, 5)x1 y1 x2 y2

⇒ y =k +y2 y1

k+ 1

⇒ 0 =k× 5 + (−5)

k+ 1

⇒ 0 =5k− 5

k+ 1

⇒ 0 = 5k− 5

⇒ −5k = −5

⇒ k =−5

−5∴ ( : ) = 1 : 1m1 m2

∴ x =+m1x2 m2x1

+m1 m2

⇒ x =1 × −4 + 1 × 1

1 + 1

⇒ x = = =−4 + 1

2

−3

2

−3

2

∴ = 1 : 1 = ( , 0)−3

2



#465338


If and are the vertices of a parallelogram taken in order, find and .

Solution

Let , , and are the vertices of a parallelogram

and are the diagonals .

is the midpoint of and

If is the mid-point of ,then the coordinates of are

If is the mid-point of BD then coordinates of O are

Since both coordinates are of the same point

Hence and .

#465339


Find the coordinates of a point , where is the diameter of a circle whose centre is and is .

Solution

Let the coordinates of be

is a diameter of circle then Midpoint of is center of circle which is )

Mid point of A and B

$

Coordinates of is .

#465340


If and are and , respectively, find the coordinates of such that and lies on the line segment .

Solution

(1, 2), (4, y), (x, 6) (3, 5) x y

A(1, 2) B(4, y) C(x, 6) D(3, 5) ABCD

.AC BD

O AC BD.

O AC O = ( , ) = ( , 4)1 + x

2

2 + 6

2

x+ 1

2

O ( , ) = ( , )4 + 3

2

5 + y

2

7

2

5 + y

2

O

∴ =1 + x

2

7

2⇒ 1 + x = 7

⇒ x = 7 − 1 = 6

∴ = 45 + y

2⇒ 5 + y = 8

⇒ y = 8 − 5 = 3

x = 6 y = 3

A AB (2, −3) B (1, 4)

A (x, y)

AB AB (2, −3

= ( , )+x1 x2

2

+y1 y2

2

∴ (2, −3) = ( , )x+ 1

2

y + 4

2

⇒ = 2x+ 1

2⇒ x+ 1 = 4 ⇒ x = 3

⇒ = −3y + 4

2⇒ y + 4 = −6

⇒ y = −10

∴ A (3, −10)

A B (−2, −2) (2, −4) P AP = AB3

7P AB



As given the coordinates of and and is a point lies on .

And

Then, ratio of and

Let the coordinates of be .

And

Coordinates of

#465341


Find the coordinates of the points which divide the line segment joining and into four equal parts.

Solution

Let , , be the points which divide the line into equal parts

Then the ratio of and

Here

Co-ordinates of P

The ratio of and =

Here

Co-ordinates of Q

Then the ratio of and =

Here

Co-ordinates of R

Coordinates are

#465343

Topic: Triangles and Quadrilateral

Find the area of a rhombus, if its vertices are , , and taken in order.

Solution

A(−2, −2) B(2, −4) P AB

AP = AB3

7

∴ BP =4

7AP PB = : = 3 : 4m1 m2

P (x, y)

∴ x =+m1x2 m2x1

+m1 m2

⇒ x = = =3 × 2 + 4 × (−2)

3 + 4

6 − 8

7

−2

7

y =+m1y2 m2y1

+m1 m2

⇒ y = = =3 × (−4) + 4 × (−2)

3 + 4

−12 − 8

7

−20

7

∴ P = ,−2

7

−20

7

A(−2, 2) B(2, 8)

P QR 4

AP PB = , = 1 : 3m1 m2

A( , ) = A(−2, 2),B( , ) = B(2, 8)x1 y1 x2 y2

∴ = ( , )+m1x2 m2x1

+m1 m2

+m1y2 m2y1

+m1 m2

⇒ ( , )1 × 2 − 2 × 3

1 + 3

1 × 8 + 3 × 2

1 + 3

⇒ ( , ) = (−1, )2 − 6

4

8 + 6

4

7

2

AQ QB , = 2 : 2m1 m2

A( , ) = A(−2, 2),B( , ) = B(2, 8)x1 y1 x2 y2

∴ = ( , )+m1x2 m2x1

+m1 m2

+m1y2 m2y1

+m1 m2

⇒ ( , )2 × −2 + 2 × 2

1 + 3

2 × 8 + 2 × 2

1 + 3

⇒ ( , ) = (0, 5)−4 + 4

4

16 + 4

4

AR RB , = 3 : 1m1 m2

A( , ) = A(−2, 2),B( , ) = B(2, 8)x1 y1 x2 y2

∴ = ( , )+m1x2 m2x1

+m1 m2

+m1y2 m2y1

+m1 m2

⇒ ( , )3 × 2 + 1 × −2

1 + 3

3 × 8 + 1 × 2

1 + 3

⇒ ( , ) = (1, )6 − 2

4

24 + 2

4

13

2

∴ P (−1, ), (0, 5), (1, )7

2

13

2

(3, 0) (4, 5) (−1, 4) (−2, −1)



Let ), , and are the vertices of rhombus

and are the diagonals of rhombus.

Here and

Here and

Area of rhombus Product of diagonals

Sq.unit

#465344


Find the area of the triangle whose vertices are :

(i)

(ii)

Solution

(i) Let , and are the vertices of

Area of triangle=

Here

sq.unit

(i) Let , and are the vertices of

Area of triangle=

Here

sq.unit

#465345


In each of the following find the value of , for which the points are collinear.

(i)

(ii)

Solution

A(3, 0 B(4, 5) C(−1, 4) D(−2, −1) ABCD.

AC BD

∴ BD = ( − + ( −x1 x2)2 y1 y2)2− −−−−−−−−−−−−−−−−−√

= 4, = −2x1 x2 = 5, = −1y1 y2

⇒ BD = (4 − (−2) + (5 − (−1))2 )2− −−−−−−−−−−−−−−−−−−−√

⇒ BD = = = = 6+62 62− −−−−−√ 36 + 36− −−−−−√ 72−−√ 2√

∴ AC = ( − + ( −x1 x2)2 y1 y2)2− −−−−−−−−−−−−−−−−−√

= 3, = −1x1 x2 = 0, = 4y1 y2

⇒ BD = (3 − (−1) + (0 − 4)2 )2− −−−−−−−−−−−−−−−−√

⇒ BD = = = = 4+ (−442 )2− −−−−−−−−

√ 16 + 16− −−−−−√ 32−−√ 2√

= ×1

2

= × 6 × 41

22√ 2√

= = 2424 × 2

2

(2, 3), (−1, 0), (2, −4)

(−5, −1), (3, −5), (5, 2)

A(2, 3) B(−1, 0) C(2, −4) △ABC

[ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

( , ) = (2, 3)x1 y1

( , ) = (−1, 0)x2 y2

( , ) = (2, −4)x3 y3

= [2(0 + 4) − 1(−4 − 3) + 2(3 − 0)]1

2

= [8 + 7 + 6]1

2

=21

2

A(−5, 1) B(3, −5) C(5, 2) △ABC

[ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

( , ) = (−5, −1)x1 y1

= ( , ) = (3, −5)x2 y2

= ( , ) = (5, 2)x3 y3

= [−5(−5 − 2) + 3(2 + 1) + 5(−1 + 5)]1

2

= [35 + 9 + 20]1

2

= = 3264

2

k

(7, −2), (5, 1), (3, k)

(8, 1), (k, −4), (2, −5)



(i) let the given points are and

These point are collinear if area ( )=0

Here

Hence the given points are collinear for

(i) let the given points are ), and

These point are collinear if area ( )=0

Here

Hence the given points are collinear for

#465346


Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are and . Find the ratio of this area to the area of

the given triangle.

Solution

A(7, −2),B(5, 1) C(3, k)

△ABC

⇒ ( − ) + ( − ) + ( − ) = 0x1 y2 y3 x2 y3 y1 x3 y1 y2

, ) = (7. − 2)x1 y1

( , ) = (5, 1)x2 y2

( , ) = (3, k)x3 y3

⇒ 7(1 − k) + 5(k+ 2) + 3(−2 − 1) = 0

⇒ 7 − 7k+ 5k+ 10 − 9 = 0

⇒ 8 − 2k = 0

⇒ 2k = 8

⇒ k = 4

k = 4

A(8, 1 B(k, −4) C(2, −5)

△ABC

⇒ ( − ) + ( − ) + ( − ) = 0x1 y2 y3 x2 y3 y1 x3 y1 y2

, ) = (8, 1)x1 y1

( , ) = (k, −4)x2 y2

( , ) = (2, −5)x3 y3

⇒ 8(−4 + 5) + k(−5 − 1) + 2(1 + 4) = 0

⇒ 8 − 6k+ 10 = 0

⇒ −6k = −18

⇒ k = 3

⇒ k = 4

k = 3

(0, −1), (2, 1) (0, 3)



Let , and are the vertices of

Area of

Here

sq.unit

Let are the mid-point of and

then coordinates of

Coordinates of

Coordinates of

Area of

sq.unit.

Ratio of and

#465347


Find the area of the quadrilateral whose vertices taken in order, are and .

Solution

A(0, −1) B(2, 1) C(0, 3) △ABC

∴ △ABC = [ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

( , ) = (0, −1)x1 y1

( , ) = (2, 1)x2 y2

( , ) = (0, 3)x3 y3

= [0(1 − 3) + 2(3 + 1) + 0(−1 − 1)]1

2

= = 48

2P ,Q,R AB,AC BC.

P = [ , ] = [1, 0]0 + 2

2

−1 + 1

2

Q = [ , ] = [0, 1]0 + 0

2

−3 + 1

2

R = [ , ] = [1, 2]2 + 0

2

1 + 3

2

∴ △PQR = [ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

= [1(1 − 2) + 0(2 − 0) + 1(0 − 1)]1

2

= = = −1 = 1−1 − 1

2

−2

2∴ △ABC △PQR= 4 : 1

(−4, −2), (−3, −5), (3, −2) (2, 3)



Let , , and be the vertices of the quadrilateral .

Area of a quadrilateral Area of + Area of

Area of ABC=

sq.units

Area of ACD=

sq.units

Area of quadilateral= sq.unit

#465348


You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for whose vertices are and

.

Solution

A(– 4, – 2) B(– 3, – 5) C(3, – 2) D(2, 3) ABCD

ABCD = △ABC △ACD

△ [−4(−5 + 2) + −3(−2 + 2) + 3(−2 + 5)]1

2

= [12 + 9]1

2

=21

2

△ [−4(3 + 2) + −2(−2 + 2) + 3(−2 − 3)]1

2

= [−20 − 15]1

2

=35

2

∴ + = = 2821

2

35

2

56

2

ΔABC A(4, −6),B(3, −2)

C(5, 2)



Let is the median of . Then is the midpoint of .

Co-ordinates of

Median divide the in to two triangles.

Area of

Here

Area of

Area of

Here

Area of

Area of =Area of

Hence it is proved that a median of a triangle divides it into two triangles of equal areas.

#465349


Determine the ratio in which the line divides the line segment joining the points and .

Solution

Let the given line divides the line segment joining the points and in ratio

And

This point also lies on

Put the value of x and y in the given equation

The required ratio is

AD △ABC D BC

∴ D= ( , ) = ( , ) = (4, 0)+x1 x2

2

+y1 y2

2

3 + 5

2

−2 + 2

2

AD △ABC

∴ △ABD = [ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

( , ) = (4, −6)x1 y1

( , ) = (3, −2)x2 y2

( , ) = (4, 0)x3 y3

△ABD = [4(−2 − 0) + 3(0 + 6) + 4(−6 + 2)]1

2

= [−8 + 18 − 16] = = 31

2

−6

2

∴ △ADC = [ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

( , ) = (4, −6)x1 y1

( , ) = (4, 0)x2 y2

( , ) = (5, −2)x3 y3

△ABD = [4(0 + 2) + 4(−2 + 6) + 5(−6 − 0)]1

2

= [−8 + 16 − 30] = = 31

2

−6

2∴ △ABD △ADC

2x+ y − 4 = 0 A(2, −2) B(3, 7)

A(2, −2) B(3, 7) : = k : 1m1 m2

∴ x =+m1x2 m2x1

+m1 m2

⇒ x = =k× 3 + 1 × 2

k+ 1

3k+ 2

k+ 1

y =+m1y2 m2y1

+m1 m2

=k× 7 + 1 × −2

k+ 1

=7k− 2

k+ 12x+ y − 4 = 0

∴ 2 × [ ]+ [ ]− 4 = 03k+ 2

k+ 1

7k− 2

k+ 1

= = 06k+ 4 + 7k− 2 − 4k− 4

k+ 1⇒ 9k− 2 = 0

⇒ k =2

9∴ 2 : 9.



#465351


Find the centre of a circle passing through the points and .

Solution

Let is the centre of the circle and and are the points on the circumference of the circle.

Radii of the circle are equal

....................................................................(i)

Similarly,

............................................................(ii)

Adding (i) and (ii)

Substitute the value of in (i)

The center of the circle is .

#465352


The two opposite vertices of a square are and . Find the coordinates of the other two vertices.

Solution

(6, −6), (3, −7) (3, 3)

O(x, y) A(6. − 6),B(3, −7) C(3, 3)

∴ OA = ( − + ( −x1 x2)2 y1 y2)2− −−−−−−−−−−−−−−−−−√

⇒ OA = (x− 6 + (y + 6)2 )2− −−−−−−−−−−−−−−√

⇒ OB = (x− 3 + (y + 7)2 )2− −−−−−−−−−−−−−−√

⇒ OC = (x− 3 + (y − 3)3 )2− −−−−−−−−−−−−−−√

∵

∴ OA = OB

⇒ =(x− 6 + (y + 6)2 )2− −−−−−−−−−−−−−−√ (x− 3 + (y + 7)2 )2− −−−−−−−−−−−−−−√

⇒ + 36 − 12x+ + 36 + 12y = + 9 − 6x+ + 49 + 14yx2 y2 x2 y2

⇒ −6x− 2y + 14 = 0

⇒ 3x+ y = 7

OA = OC

⇒ =(x− 6 + (y + 6)2 )2− −−−−−−−−−−−−−−√ (x− 3 + (y − 3)2 )2− −−−−−−−−−−−−−−√

⇒ + 36 − y + + 36 + 12y = + 9 − 6x+ + 9 − 6yx2 y2 x2 y2

⇒ −6x+ 18y + 54 = 0

⇒ −3x+ 9y = −27

⇒ 10y = −20

⇒ y = −2

y

⇒ 3x+ −2 = 7

⇒ 3x = 9

⇒ x = 3

∴ (3, −2)

(−1, 2) (3, 2)



Let is a square where two opposite vertices are .

Let B(x,y) and D ids the other two vertices.

In Square ABCD

Hence

[by distance formula]

Squaring both sides

In [all angles of square are ]

Then according to the Pythagorean theorem

As

Put the value of

Hence or .

As diagonals of a square are equal in length and bisect each other at

Let is the midpoint of

CO-ordinates of

P is also mid point of

then co-ordinates of mid-point of =co-ordinates of

Then other two vertices of square are or and .

#465353


ABCD A(−1, 2)andC(3, 2)

( . )x1 y1

AB = BC = CD = DA

AB = BC

⇒ =(x+ 1 + (y − 2)2 )2− −−−−−−−−−−−−−−√ (3 − x + (2 − y)2 )2− −−−−−−−−−−−−−−√

⇒ (x+ 1 + (y − 2 = (3 − x + (2 − y)2 )2 )2 )2

⇒ + 2x+ 1 + + 4 − 4y = 9 + − 6x+ 4 + − 4yx2 y2 x2 y2

⇒ 2x+ 5 = 13 − 6x

⇒ 2x+ 6x = 13 − 5

⇒ 8x = 8

⇒ x = 1

△ABC, ∠B = 90∘ 90∘

A +B = AB2 C 2 C 2

AB = BC

∴ 2A = AB2 C 2

⇒ 2 =( )x+ 1 + (y − 2)2 )2− −−−−−−−−−−−−−√ 2 ( )(3 − (−1) + (2 − 2)2 )2− −−−−−−−−−−−−−−−−√ 2

⇒ 2((x+ 1 + (y − 2 ) = (3 + 1 + (2 − 2)2 )2 )2 )2

⇒ 2( + 2x+ 1 + + 4 − 4y) = (4x2 y2 )2

x = 1

⇒ 2( + 2 × 1 + 1 + + 4 − 4y) = 1611 y2

⇒ 2( − 4y + 8) = 16y2

⇒ 2 − 8y + 16 = 16y2

⇒ 2 − 8y = 0y2

⇒ 2y(y − 4) = 0

⇒ (y − 4) = 0

y = 0 4

90∘

P AC

∴ P = ( , ) = (1, 2)3 − 1

2

2 + 2

2

BD

BD P

⇒ ( , ) = 1, 2x1 y1

∴ = 1, = 2x1 y1

ABCD (1, 0) (1, 4) (1, 2)



The Class students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmoharare planted on the

boundary at a distance of m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on

the remaining area of the plot.

(i) Taking as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of if is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

Solution

(i) If we take as origin then is -axis and is the -axis Then the coordinates of the vertices of is

(ii) If we take as origin then is -axis and is the -axis Then the coordinates of the vertices of is

Area of the triangle =

First case-

Area of

Sq.unit

Second case-

Area of

Sq.unit

Hence we observed that the area of the triangle in both case is equal.

#465354


The vertices of a are and . A line is drawn to intersect sides and at and respectively, such that . Calculate the

area of the and compare it with the area of . (Recall Theorem and Theorem ).

Solution

X

1

A

ΔPQR C

A AD X AB Y △PQR

P = (4, 6),Q = (3, 2),R = (6, 5)

C CB X CD Y △PQR

P = (12, 2),Q = (13, 6),R = (10, 3)

[ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

△PQR = [4(2 − 5) + 3(5 − 6) + 6(6 − 2)]1

2

= (4 × −3 + 3 × −1 + 6 × 4)1

2

= (−12 − 3 + 24) =1

2

9

2

△PQR = [12(6 − 3) + 13(3 − 2) + 10(2 − 6)]1

2

= (12 × 3 + 13 × 1 + 10 × −4)1

2

= (36 + 13 − 40) =1

2

9

2

ΔABC A(4, 6),B(1, 5) C(7, 2) AB AC D E = =AD

AB

AE

AC

1

4

ΔADE ΔABC 6.2 6.6



In and

Area of

sq. unit

A line is drawn to intersect sides and at and

Then

The point divide the line in and .The ratio of and

Co-ordinates of =

Here, and

The point divide the line C in and . The ratio of and =

Co-ordinates of

Here, and

Now, the area of

sq.unit

Hence, Area Area ( )=

#465356


Let and be the vertices of .

(i) The median from meets at . Find the coordinates of the point .

(ii) Find the coordinates of the point on such that

(iii) Find the coordinates of points and on medians and respectively such that and .

(iv) What do yo observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio .]

(v) If and are the vertices of , find the coordinates of the centroid of the triangle.

Solution

△ABC A(4, 6),B(1, 5) C(7, 2)

△ABC = [ ( − ) + ( − ) + ( − )]1

2x1 y2 y3 x2 y3 y1 x3 y1 y2

= [4(5 − 2) + 1(2 − 6) + 7(4 − 5)]1

2

= [12 − 4 + 7] =1

2

15

2AB AC D E

=AD

AB

1

4D AB AD DB AD DB = : = 1 : 3m1 m2

∴ D ( , )+m1x2 m2x1

+m1 m2

+m1y2 m2y1

+m1 m2

( , ) = (4 : 6)x1 y1 ( , ) = (1, 5)x2 y2

= ( , )1 × 1 + 4 × 3

1 + 3

1 × 5 + 3 × 6

3 + 1

= ( , )13

4

23

4

E A AE EC AE EC : = 1 : 3m1 m2

∴ E = ( , )+m1x2 m2x1

+m1 m2

+m1y2 m2y1

+m1 m2

( , ) = (4 : 6)x1 y1 ( , ) = (7, 2)x2 y2

= ( , )1 × 7 + 3 × 4

1 + 3

1 × 2 + 3 × 6

3 + 1

= ( , 5)19

4

△ADE = [4( − 5)+ (5 − 6) + (6 − )]1

4

23

4

13

4

19

4

23

4

= [ − + ]1

2

12

4

13

4

19

16

= [ ]1

2

48 − 52 + 19

16

= × =1

2

15

16

15

32

(△ADE) : △ABC : = : = 1 : 1615

32

15

2

1

32

1

2

A(4, 2),B(6, 5) C(1, 4) ΔABC

A BC D D

P AD AP : PD = 2 : 1

Q R BE CF BQ : QE = 2 : 1 CR : RF = 2 : 1

2 : 1

A ( , ) ,B ( , )x1 y1 x2 y2 C ( , )x3 y3 ΔABC



i)

Median is the line joining the midpoint of one side of a triangle to the opposite vertex. So, the coordinates of would be

ii)

divides in the ratio .

Using section formula, we get the coordinates of .

iii)

Coordinates of will be and the coordinates of .

Coordinates of

Coordinates of

iv)

The coordinates of and are the same which is .

This point is called the centroid, denoted by .

v)

Centroid of triangle

D ( , )7

2

9

2

P AD 2 : 1

A( , ) = (4, 2), D( , ) = ( , )x1 y1 x2 y27

2

9

2

m : n = 2 : 1

P

P (x, y) = ( , )n +mx1 x2

m+ n

n +my1 y2

m+ n

= ,

⎛

⎝⎜⎜

1 ⋅ 4 + 2 ⋅7

22 + 1

1 ⋅ 2 + 2 ⋅9

22 + 1

⎞

⎠⎟⎟

= ( , )11

3

11

3

E ( , 3)5

2F = (5, )7

2

Q = ( , )n +mx1 x2

m+ n

n +my1 y2

m+ n

= ,

⎛

⎝⎜⎜

1 ⋅ 6 + 2 ⋅5

22 + 1

1 ⋅ 5 + 2 ⋅ 3

2 + 1

⎞

⎠⎟⎟

= ( , )11

3

11

3

R = ( , )n +mx1 x2

m+ n

n +my1 y2

m+ n

= ,

⎛

⎝⎜⎜

1 ⋅ 1 + 2 ⋅ 5

2 + 1

1 ⋅ 4 + 2 ⋅7

22 + 1

⎞

⎠⎟⎟

= ( , )11

3

11

3

P ,Q R ( , )11

3

11

3

G

ABC = ( , )+ +x1 x2 x3

3

+ +y1 y2 y3

3



#465359


is a rectangle formed by the points , , and . and are the mid-points of , , and respectively. Is the

quadrilateral a square? a rectangle? or a rhombus? Justify your answer.

Solution

Let and are the midpoint of and .

Co-ordinates of

Co-ordinates of

Co-ordinates of

Co-ordinates of

Now, length of

Length of

Length of

Length of

Length of diagonal

Length of diagonal QS=

Hence the all the sides of the quadrialteral are equal but the diagonals are not equal then is a rhombus.

ABCD A(−1, −1) B(−1, 4) C(5, 4) D(5, −1) P ,Q,R S AB BC CD DA

PQRS

P ,Q,R S AB,BC,CD DA

∴ P = [ , ] = [ , ] = [−1, ]+x1 x2

2

+y1 y2

2

−1 − 1

2

−1 + 4

2

3

2

∴ Q = [ , ] = [ , ] = [2, 4]+x1 x2

2

+y1 y2

2

−1 + 5

2

4 + 4

2

∴ R = [ , ] = [ , ] = [5, ]+x1 x2

2

+y1 y2

2

5 + 5

2

4 − 1

2

3

2

∴ S = [ , ] = [ , ] = [4, −1]+x1 x2

2

+y1 y2

2

5 − 1

2

−1 − 1

2

PQ = = = =(−1 − 2 + ( − 4)23

2)2

− −−−−−−−−−−−−−−−−√ (−3 + (−)2

5

2)2

− −−−−−−−−−−−√ 9 +

25

4

− −−−−−√ 61

4

−−−√

QR = = = =(2 − 5 + (4 −)23

2)2

− −−−−−−−−−−−−−−√ (−3 + ()2

5

2)2

− −−−−−−−−−−√ 9 +

25

4

− −−−−−√ 61

4

−−−√

RS = = = =(5 − 2 + ( + 1)23

2)2

− −−−−−−−−−−−−−−√ (3 + ()2

5

2)2

− −−−−−−−−√ 9 +

25

4

− −−−−−√ 61

4

−−−√

S = = = =(2 + 1 + (−1 −)23

2)2

− −−−−−−−−−−−−−−−−√ (3 + (−)2

5

2)2

− −−−−−−−−−−√ 9 +

25

4

− −−−−−√ 61

4

−−−√

PR = = = = 6(−1 − 5 + ( −)23

2

3

2)2

− −−−−−−−−−−−−−−−−√ ( )62

− −−√ 36−−√

= = = 5(2 − 2 + (4 + 1)2 )2− −−−−−−−−−−−−−−√ (5)2− −−√ 25−−√

PQRS PQRS

7/4/2018 ... · topic: section formulae find the coordinates of a point , where is the diameter of...

Documents