Question

The lines represented by the equation x2 + 2V3xy + by2
3x - 2fy - 4 = 0, are parallel
(A) b=3
(B) f = 3V3
(C)
33
f =
2
(D) the line mid way
between the parallel
lines is 2x + 2V3y -
3=0​

Answer 1

Step-by-step explanation:

Let m

1

,m

2

,m

3

be the slopes of the three lines represnted by the given equation such that m

1

m

2

=−1

We have m

1

m

2

m

3

=−

d

1

so that

m

3

=

d

1

Since y=m

3

x⇒x=dy satisfies the given equation, we get

d

3

−6d

2

+3d+d=0⇒d(d

2

−6d+4)=0

If d=0, the given equation represents the line x=0 and x

2

−6xy+3y

2

=0 which are not perpendicular

∴d



=0 and d

2

−6d+4=0⇒d=

2

6±

36−16

=3±

5

which gives two real values of d

hope it's helpful❤❤

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Answer 2

Answer:

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