Math, asked by gangadharamkottunuru, 5 months ago

22. Show that the two pair of lines 3x2 + 8xy – 3y2 = 0 and
3x² + 8xy – 3y² + 2x – 4y - 1 = 0 form a square​

Answers

Answered by amitnrw
5

Given :  two pair of lines 3x² + 8xy – 3y²   = 0  and 3x² + 8xy – 3y² + 2x – 4y - 1 = 0

To Find : Show these lines form a square

Solution:

3x² + 8xy – 3y²   = 0

=> 3x² + 9xy -xy – 3y²   = 0

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

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

3x - y = 0

x + 3y = 0

3x² + 8xy – 3y² + 2x – 4y - 1 = 0

=> (3x - y  - 1) (x  + 3y  + 1)  = 0

3x - y  - 1 = 0

x  + 3y  + 1 = 0

3x - y = 0  and x + 3y = 0

=> ( 0, 0)

3x - y  - 1 = 0  x  + 3y  + 1 = 0

(0.2 , -0.4)

3x - y = 0   and x  + 3y  + 1 = 0

=> (-0.1 , -0.3)

x + 3y = 0  and  3x - y  - 1 = 0

(0.3 , -0.1)

3x - y = 0 , x + 3y = 0    Slopes are  3 ,  -1/3  ( hence perpendicular )

3x - y  - 1 = 0 , x  + 3y  + 1 = 0 Slopes are  3 ,  -1/3  ( hence perpendicular )

( 0, 0) ,  (0.3 , -0.1) , (0.2 , -0.4) ,  (-0.1 , -0.3)

Using distance formula

Each side length = √0.1

Hence it forms a square

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Attachments:
Answered by jchennaiahjchennaiah
0

Answer:

3x²+8xy-3y²=0

=3x²+9xy-xy-3y²=0

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

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

L1=x+3y =0

L2=3x-y=0

let L3=x+3y+L=0

L4=3x-y+m=0

=3x²+8xy-3y²+2x-4y-1=(x+3y+L) (3x-y+m)

=3x²-xy+xm+9xy-3y²+3ym+3xl-lm

=x+3y=1==x+3y-1=0_____L3

=3x-y=0==3x-y-1=0______L4

solving:-

y c x y

3 -1 1 3

-1 -1 3. -1

=x/3-1 , y/3+1 , c/1-9

=x/2 , y/4 , 1/-8

=x/2 , 1/-8

=2x=-8

x=-8/2

x=-4

== y/4 , 1/-8

4y =-8

y=-8/4

y=-2

let L3= x+3y+L=0 ____ x+3y-1=0

L4 =3x-y+m =0_______3x-y-1 =0

origin (0,0) to |c/√a²+b²|

L3 =

D1= |1/√(1)²+(3)²|=|1/√1+9|

D1 =|1/√10|

L4=

D2=|1/√(3)²-(1)²|= |1/9+1|

D2= |1/√10|

D1,D2,=1/√10 and L1 is perpendicular to L2

OABC is a square...

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