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