Question

The perpendicular distance between the lines given by 8x²+2y² + 8xy +26x +13y +15 = 0 is...

Answer 1

Answer:

ANSWER

Here h

2

−ab=(4)

2

−2×8=0.

∴ the lines are parallel.

On dividing by 2,

4x

2

+4xy+y

2

+13x+

2

13

y+

2

15

=0

Suppose its factors are (2x+y+p)(2x+y+q)

On comparing the coefficients, we get

2(p+q)=13,p+q=

2

13

and pq=

2

15

,

or p(

2

13

−p)=

2

15

or 2p

2

−13p+15=0

∴(p−5)(2p−3)=0

or p=5,

2

3

;

∴q=

2

3

,5

∴2x+y+5=0, 2x+y+

2

3

=0

are the required lines.

If p

1

and p

2

be their distances from origin, then the distance between them is

p=p

1

∼p

2

=

(4+1)

5

∼

(4+)

3/2

=

2

5

7