1. On which axis do the following points lie?(i) P(5,0)(ii) Q(0 - 2)(iii) R(—4, 0)(iv) S(0, 5)Sol:

(i) P (5,0) lies on jt - axis

(ii) Q{0,-2) lies oil y — axis

(iii) 7? (-4,0) lies on x — axis

(iv) £(0,5) lies on y - axis

Exercise 14.1: Coordinate Geometry

2. Let ABCD be a square of side 2a. Find the coordinates of the veitices of this square when(i) A coincides with the origin and AB and AB and coordinate axes are parallel to the

sides AB and AD respectively.(ii) The center of the square is at the origin and coordinate axes are parallel to the sides

.AB and .AD respectively.¥

2 a B

yA

2* 2ft

O

A O 3 , B

b

Sol:(i) Coordinate of the veitices of the square of side 2a are: A(0.0),B(2a,0),C(2a,2a) mid D(0,2a)

(ii) Coordinate of the vertices of the square of side 2a are:a.— o) and (a,— a)

3. The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the veitices R and R' of the triangles.

Sol:We have two equilateral triangle PQR and PQR' with side 2a.O is the mid-point of PQ. hi AQOR. ZQOR = 90°Hence, by Pythagoras theorem OR2 + OQ1 = QR1

OR2 = ( 2 a f - ( a ) 2

OR2 = 3 a 2

OR - y j ( 3 ) a

Coordinates of vertex R is ( -J3a. o] and coordinate of vertex R' is

1. Find the distance between the following pair o f points(i) (-6 ,7 ) and (-1 ,-5 )

(ii) ( a+b,b+c) and (a -b ,c -b )

(iii) (rtsinff,-£>cos«) and (-f/cos«,Z )sin«)

(iv) ( a.O) and (O.b)

Sol:

Exercise 14.2: Coordinate Geometry

(i) We have P(-6,7) and £>(-1,-5)Here.xl = -6, y\ = 7 and ■v3 = ~ h y 2 = —5

PQ = l / ( X2 - ) 2 + 0 ' l - J'l )"

i j e = ^ - i - ( - 6 ) ] 2 + ( - s - 7 ) 2

P{? = 1/( - l + 6)2+ (-5 -7 )2

/v =V(5)2+ H 2)2P0 = ,/25 + !44P Q = -yfl69

/ > 0 = 13

(ii) we have P{a + b,b+c) and Q (a -b ,c -h ) here, xl - a + b, v: = b + r and r3 — <:/ — Z>, y2 =c — b

PQ='J(xi - * ) 2 + {y - > i ) 2

PQ = \ j{a —~b— ~a— ~b)~ + ( c —A—Z>— c )2

PQ = j { -2 b f+ ( -2 b f

PQ- 44b2 +4*2

PQ = 4

PQ = -j4x2b2 PQ = 2j2b(iii) wehave P(asina,-icoscr) and £)(-<7Coscr,frsinar) here xi = a sin ay }\ = -b cos a andx2 = -a cos a , y2 = b sin a

PQ = yj{X2~ Xl )2 + (>2 - >1 f

PQ = J(-a cos or - a sin a+ [-6 sin a - (—b cos J"

PQ = yj(—flcoscr)' + (— a sinor)* + 2(—ocosa')^asincr)+(&sinGr)2 + (—bcosa}' — 2(2>sincir)(—bcosa)

PQ = yfc,7 cos2 a + a2 sin2 a +2a2 cos a sin a + b2 sin2 a +b2 cos2 a + 2b2 sin a cos a

PQ = yja2 (cos2 a + sin2 « ) + 2«2 cos a-sin cr+Zr (sin2 a + cos2 a-) + 2Z>2 sincrcosa:

PQ = a 2 x 1 + 2f/2 cos a sin a + b2 x 1 + lb2 sin a cos a [v sin2 a +cos2 a - 1]

PQ=yja2 +b2 +2cc cos a sin a + 2b1 sincrcosa

p Q = § P 7 iP

P Q = J ( 7 P b 2

(iv) We have / ’(a?0) and Q(0,b)

Here.xl = a,yl =0,x2=0?y2=b,

pQ=V(*2 - ■xi f + ( ^ 2 - yi f P Q -y j (^ ~ a)2 + (b -0 )2

PQ = <J(-af+(b)2

P Q -y ja 2 +b2

) + 2 cos ashlar [a2 +b2)

V l + 2 co sa s in a )

2 . Find the value of a when the distance between the points (3, a) and (4, 1) is VlO. Sol:We have P(3,a) and Q(4, l)

Here,

*!=3 >yx=a

xi =4,y2 =1 pg=JiopQ = Vta - ■*i )2 + (y2 - y, 7=>VT0 = (4-3)2+(l-<7)2

^ 7 io = ^ (i)2+ (i-« )2

= > J l O = J l + l + a 2 - 2 a

= ^ 2 + a 2 - 2 a

Squaring both sides

*.*( a - b y —a2 +b2 - lab

= > ( V T o f = ( ^ + a 2- 2 a j

!0 = 2 + a 2 — 2 a

=>a~ — 2 a + 2 —10 = 0 => a 2 - 2 a - 8 = 0

Splitting the middle team.=>a2 - 4 a + 2 a - 8 = 0 = > a { # -4 ) + 2 ( a - 4 ) = 0

=>(a-4)(a + 2) = 0 a = 4 , a = -2

3. If the points (2,1) and (1,-2) are equidistant from the point (x, y) from (-3, 0) as well as from (3, 0) are 4.Sol:

We have P (2 ,l) and 0 (1 ,-2 ) and R { X ,Y )

Also. PR = QR

P R = J ( x - 2 ) 2 + ( y - \ ) ‘

=> PR = J x 2 + (2 )J — 2xxx2 + y 1 - 2 x y x l

=> PR = ^ x 2 + 4 — 4x + y 2 +1 — 2 y

=> PR = ^jx2 + 5 - 4y + y 2 - 2y

QR = ( x - l ) 2 + (y+2 )2

=> PR = *Jx2 +1 - 2.v + y2 + 4+4 y

=> P£ = «Jy2 + 5 - 2y + y 2 + 4y

V P P = g P

= > y y 2 + 5 - 4y + y 2 - 2 v = ^ y2 + 5 - 2y + y 2 + 4y => y 2 + 5 - 4y + y2 - 2 y = x2 + 4 y=> y 2 + 5 - 4y + y2 - 2y = x2 + 5 - 2x + y 2 + 4y = > -4 y + 2 y - 2 v - 4 v = 0

~> —2 y — 6 y = 0

=> -2 (y + 3 y) = 0

, 0= > y + 3 y = —

=>x + 3y = 0 Hence proved.

Find the values of x. y if The distances of The point (x. y) horn (—3.0) as well as from (3.0) are 4.Sol:We have / J(x ,y ).0 (-3 ,O ) and *(3 ,0)

PQ = J { x + r f + ( y - o f

=> 4 = ^ x 2 +9+6 x + y 2

Squaring both sides

=>(4)‘ = i[yj+2 + 9 + 6x + y 2 j

=> 16 = x2 + 9 + 6x+ y1

=> x 2 + y2 = 1 6 - 9 — 6x

=X>r2 + v2 = 7 - 6 x ................(1)

PJ? = ( ^ ( . v - 3 ) 2 + ( v - 0 ) 2 )

= > 4 = *Jx2 + 9 — 6.y + y 2

Squaring both sides

(2)

( 4 ) 2 = U . v 2 + 9 - 6 . t + v 2 ) 2

x=>16 = x 2 +9-6x + y2 => x2 + y 2 = 16-9 + 6x =2> x2 + y2 = 7 + 6xEquating (1) and (2)7 — 6x = 7 + 6x=> 7 - 7 = 6x + 6x = > 0 = 1 2 r

->x = l2Equating (1) and (2)1 — 6x = ! + 6.x =2> 7 — 7 = 6x + 6.x=>0 = 12x => x = 12Substituting the value of x = 0 in (2)

7 2x ' + v = 7 + 6x

0 + / = 7 + 6x0

y 2 = 7

y = ±V7

Exercise 14.3: Coordinate Geometry1. Find tlie coordinates of tlie point which divides the line segment joining (— 1. 3) and (4. —

7) internally in the ratio 3:4.Sol:Let P(x, v) be the requited point.

m x2 + m \

m + n

nrv0 +n\\ y = ——m + n

Here. = -1

m: n = 3:4

3x4 + 4x(-i)x = ------------- 1— -3 + 4

12-4x = ---------7

8x = —

1

3x(-7) + 4x3y = ---------------------3 + 4

- 21+12

-97

/.The coordinates of P are

2. Find the points of trisection of the line segment joining the points:(i) (5, — 6 ) and (— 7, 5).(ii) (3, — 2) and (— 3, — 4)(iii) (2, — 2) and (— 7, 4).Sol:(i) Let P and Q be the point of trisection of AB i.e.. A P = P Q = Q B

(5,-6) (-7,5)

Therefore. P divides AB internally in the ratio of 1:2. thereby applying section formula, the coordinates of P will be r l(-7 ) + 2 (5)'U l(5) + 2 (-6 )V f. -7

v 1 + 2 7 V 1 + 21,

V

Now. Q also divides AB internally in the ratio of 2:1 there its coordinates are2(-7) + l(5)l 2(5) + l ( - 6 ) ._

f - + )l 2 + 1 J * 2 + 1 l 3 J

(ii)Let P. Q be the point of tri section of AB i.e.. A P = P Q = Q B

hA ? t H6

(-3 .-4 )(3,-2)

Therefore. P divides AB internally in the ratio of 1:2 Hence by applying section formula. Coordinates of P are

l( -3 ) + 2 (3 )l l( -4 ) + l ( - 2 )1 + 2 1 + 2

\ u A 1 -8

Now. Q also divides as internally in the ratio of 2:1 So. the coordinates of Q are

rf2 ( -3 )+ l(3 )^ 2 ( -4 ) + l( -2 )

VV 2+1 2 + 1i .e .\ —1

J

-10

Let P and Q be the points of bisection of AB i.e.. A P - P Q = O Q

A _P_______Q_ B(2,-2) ( - 7 , 4 )

Therefore. P divides AB internally in the ratio 1 :2 . Therefore, the coordinates of P. by applying the section formula, are

0)|Yl(-7) + 2(2)j { H4) + 2(-2 )jil l (1 + 2) J l O + d JJ

Now. Q also divides AB internally in the ration 2 : 1. So, the coordinates of Q are f 2 (-7 ) + l(2) 2(4) + l(2)> .

2 + 1 5 2 + 1 r '

3. Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (— 2, — 1), (1, 0), (4, 3) and (1, 2) meet.Sol:

Let P(.r. v) be the given points.

W e know that diagonals of a parallelogram bisect each other. - 2 + 4

Y = ------------------0

x =

y-1 + 3

2

Coordinates of P are (1,1)

4. Prove that the points (3, — 2), (4, 0), (6 . — 3) and (5. — 5) are the vertices of a parallelogram.Sol:

Let P ( .r. v) be the point of intersection of diagonals A C and 80 o f A B C D .

_ 3 + 6 _ 92

- 2 - 3 -5

Mid - point of A C —

Again.5 + 4 9

2 ~ 2-5 + 0 -5v — ■

2 2

Mid - point of B D =

Here mid-point of AC

9 _ 5_2‘_ 2;- Mid - point of BD i.e. diagonals AC and BD bisect each other.

We know that diagonals of a parallelogram bisect each other A B C D is a parallelogram.

5. Three consecutive vertices of a parallelogram are (—2, — 1), (1. 0) and (4, 3). Find the fourth vertex.Sol:

Let j4 (-2 ,-1 ),B (1 ,0 ),C (4 ,3 ) and D[x ,y ) be the vertices of a parallelogram ABCD taken

in order.

Since the diagonals of a parallelogram bisect each other.

Coordinates of the mid - point o f AC = Coordinates of the mid-point of BD.-2 + 4 \ + x=>-------= ------

2 22 x + 1

2

=>JC+1 = 2

x — 1

And,

2

i

v + 02

=> v = 2

Hence, fourth vertex of the parallelogram is (l,2j

Exercise 14.4: Coordinate Geometry

1. Find the centroid of the triangle whose vertices are:

(i) ( l ,4 ) , ( - l , - l ) a m /( 3 , -2 )

Sol:We know that the coordinates of the centroid of a triangle whose vertices are

(W i) * (W 2 M*3 > ) arexl + x 2 + x i vt + y 2 + y 3

3 ' 3 )

So, the coordinates of the centroid of a triangle whose vertices are

(1 ,4 ),(-1 ,-1 ) a n d (3 ,-2 ) are

2. Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the coordinates of the tlmd vertex.Sol:

1 -1 + 3 4 - 1 - 2 3 ’ 3 J

Let the coordinates of the third vertex bee ( r, y ). Then

Coordinates of centroid of triangle are

x +1 + 3 j/ + 2 + 5 |3 * 3 J

W e have centroid is at origin (0.0)

.■.JC± 1± 3= ( W ^ ± 2± 5 = o 3 3

=>x+4 = 0 =>y + 7 = 0=> x = -4 y = -7

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.Sol:

Let ABC be a triangle such that BC is along x-axis Coordinates o f A, B and C are (w_y)n((f 0) and ( jc Vj)

D and E are the mid-points o f AB and AC respectively

Coordinates o f D are % + 1 + ^

f x f

Coordinates o f E are V + V l'2 J

Length of BC = yfx[+yf

4. Prove That the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.Sol:

v

Let O B C D b e the quadrilateral P. O . R. S be the midpoint off O B . C D . O D and B C.

Let the coordinates of O . B . C . D are (0,0),(.Y,0),(.r,j) and (0,>>)

Coordinates of P are

Coordinates of Q are f x \ —, v

V 2 ' J

v \0 .4 -

2Coordinates of R are

Coordinates o f S are x ,—l 2 )

Coordinates o f midpoint o f PQ are

x x

Coordinates o f midpoint o f RS are2 ’ 2 2 ’ 2

Since, tlie coordinates o f the mid-point o f PQ = coordinates o f mid-point o f RS

PQ and RS bisect each other

5. If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA2+ PB2 -f PC2 = GA2 + GB2 + GC2 + 3GP2.Sol:Let ^ (0 .0 ) ,5 (o ,0 ) , and C(c, d ) are the o-ordinates of triangle ABC

let P (x ,y )

To prove:PA2 + PB2 + PC2 = GA2 + GB2 + GC2 + 3 GP2Or, PA2 + PB2 + PC2 = GA2 + GB2 + GC2 + GP2 + GP2 + GP2Or. PA2 - GP2 + PB2 - GP2 + PC2 + GP2 = GA2 + GB2 + GC2Proof:

Hence. Gc + 0 + a d

3 * 3

a + c d

PB2 = { x - a ) + y 2

P C 2 = ( x - c ) 2+ ( y - d f

L.H.S

2 | a + cx + 2xfl+c'l 2 d 2 2 vd

r

+ (x - c ) + { y d )

x +h “ J - ^ r r r

d 2 2 yd H-----------9 3

= .y + y + x + x + a — 2ax + y z + x~ + c —2xc + y 2 + d~ — 2 yd — 3

-2x|\2

2 I a + c |x +

3 J KtK d 2 2ydH---------9 3

= + y f ' + a2 + c 2 + d 1 -2 a x — 2 x c —2yd — — --------— + 2x (a + c ) — 2 y ^ -------v2 vd3 ' 3

= a2 + c2 +d2 -^ d x +c^ + a c + dx + dx - ^ - + 2Mf

3a + 3 c+ 3 d - a - c - lac - d 2a + 2c + 2d - lac = L.H.S

Solving R.H.SGA2 + GB1 +GC2

GA2 =

GC1 =

a t c lac d---------------- 1----9 9

d_)3

2 /' ^2 c — Za d \

3

a2 + 4c2 -4 ca 4d2

c2 + 4a1 - 4ac d2---------------------------------------------------+ —

„ o a + c +2ac d a + 4 c —4ac 4d c + 4 a — 4ac dGA +GB +G C = ----------------- + — + -------------------+ ------ h---------------------+ —

9 9 9 9 9 9a2 + c2 +2 ac + d2 + o 2 +4c2-4 a c + 4d2 + c 2 +4a2- 4ac + d2

6a2+6c2 + 6d2+6ac la 1 + l c 2 + l d 2 +2ac

.L.H.S = R.H.S

Exercise 14.5: Coordinate Geometry

1. Find the area o f a triangle whose vertices are(i) (6 ,3 ) ,( -3 ,5 ) and (4 ,-2 )

(ii)

(hi)

(at?, 2atx) , ( at\, 2 at2) [at3, 2at3)]

{a,c + a ),{a ,c) and (—a ,c —a)

Sol:(i) Area o f a triangle is given by

| h ( v 2 - v3)+ ^ (y 3 -3 i)+ ^ (> i +v2)]

Here, x, = 6. yL = 3, = -3 , y 2 ~ 5 7x^ = 4, y3 = -2

Let A (6 ,3 ), B ( -3 ,5 ) and C (4, - 2 ) be the given points

Area o f A 4£C = ^ [6 (5 + 2) + ( - 3 ) ( - 2 - 3 ) + 4 ( 3 - 5 ) ]

= I [ 6 x 7 - 3 x ( - 5 ) + 4 ( - 2 ) ]

= i[4 2 + l5 -8 ]

4 9- — sq.umts

(ii) Let A = (xx, yx) = {atf, 2atx)

B = {x1,y2) = {atl,lat2)

= (x3,y3) = (atl, 2at3) be the given points.The area of AABC

= [ at2 {2at2 — 2atl ) + at2 (2at2 - 2atl) + at2 (2atr — 2at2) J

= 2 a 2tlt2 -2 a 1t\tl +2a1t\tl -2 a lt\tl +2a1t\tl -2 a 2t\t2J

= “ x 2 \_a ~ri ( r2 ~ ri ) + a 2 f 2 ( r3 ~ h ) + a ~ + 3 Oi ~ h )_

= [a2 (h - h)+tl {h-h )+ti (h-h)](iii) Let A = (.Tj, yx) = (a, c + a )

■S = (-r 2>>’2) = ( n -<:')

C = (x3, _>’3) = a ) be the given pointsThe area of AABC

= — \a (c — {c - tt}) + a (c — a — (c + 17)) + (—f7)(c + a — a) J2

A a (c -c + a )+ a (

= i| £z x a + ax ( —2a ) — a x a J

c —a —c —a ) — a(c + a —

2. Find the area of the quadrilaterals, the coordinates of whose vertices are(i) (— 3, 2), (5, 4), (7, — 6 ) and (— 5, — 4)(ii) (1, 2), (6 , 2), (5, 3) and (3, 4)(hi) (--1 , — 2), ( - 3 , - 5), (3. - 2). (2 , 3 )

Let r i( -3 ,2 ) ,5 (5 .4 ) ,C (7 .-6 ) and D ( -5 , - 4 ) be the given points.

Area of AA B C

= — [—30 — 40 —14]

= -42But area caimot be negative

Area of & .4 D C = 42 square units

Area of &ADC

= \ [~ 3(— 6 + 4) + 7 (-4 -2 )+ (-5 )(2 + 6)]

= ^ [-3 (-2 ) + 7{ - 6)-5 x S ]

[6 -4 2 -4 0 ]

= —x-76 2

- - 3 8

But area cannot be negative .'.Area of A.4DC = 38 square units Now, area of quadrilateral ,4BCD = Ar. o f ABC + Ar o f ADC= (42 + 38)= 80 square, units

Sol:

_ [ _ 3 ( 4 + 6 ) + 5 ( - 6 - 2 ) + 7 ( 2 - 4 ) ]

I [ _ 3 x l 0 + 5x(-S) + 7 ( -2 )]

(i)

Let J(l,2),S(6,2),C (5,3) and (3,4)be the given points Area of AABC

= | [ 1(2 -3 ) + 6 (3 -2 )+ 5 (2 -2 )]

= i [ - l + 6x ( j)+0]

4 i - i+6i_ 5_

_ 2Area of AADC

= ^ [l (3 -4 ) + 5 (4 -2 )+ 3 (2 -3 )]

= I [ - lx 5 x2 + 3 (- l) ]

Now, Area of quadrilateral A B C D

= Area of A B C + Area of A D C

5

= 3

= | — + 3 I s q . u n its

11= — sq. units

2(«)

A

Let A ( -4 , -2)■, B ( -3 , - 5 ) , C (3, -2 ) and D (2,3) be the given points

Area of A L 8 C = ^ | ( -4 ) ( -5 + 2 ) -3 ( -2 + 2 ) + 3 ( - 2 + 5)|

= -t |H ) ( - 3 ) - 3 ( 0 ) + 3 ( 3 ) |

2 |

2

Area o f A.4CD = -t|(-4)(3 + 2) + 2 ( -2 + 2 ) + 3(-2 -3 )|

-4 (5 )+ 2(0)+3(-5)| = -3522

But area can't be negative, hence area o f AADC = —2

Now. area o f quadrilateral (ABCD ) = or (AABC ) + ar(/SADC )

21 35Area (quadrilateral ABCD) = — + —

S6.Area (quadrilateral ABCD ) = —

.Area (quadrilateral ABCD ) = 28 square. Units