Question

2x-ky+3=0 and 6x+12y+8=0 are pair of parallel lines. find K. ​

Answer 1

Answer:

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mysticd

Hi ,

Compare given equations

2x - ky + 3 = 0 ,

4x + 6y - 5 = 0 with

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

we get ,

a1 = 2 , b1 = -k , c1 = 3 ;

a2 = 4 , b2 = 6 , c2 = -5 ;

a1/a2 = b1/b2 ≠ c1/c2

[ Since , lines are parallel ]

a1/a2 = b1/b2

2/4 = ( -k )/6

( 2 × -6 )/4 = k

-3 = k

Therefore ,

k = -3

Answer 2

Answer:

Step-by-step explanation:

2x - ky + 3 = 0 ,

4x + 6y - 5 = 0 with

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

we get ,

a1 = 2 , b1 = -k , c1 = 3 ;

a2 = 4 , b2 = 6 , c2 = -5 ;

a1/a2 = b1/b2 ≠ c1/c2

[ Since , lines are parallel ]

a1/a2 = b1/b2

2/4 = ( -k )/6

( 2 × -6 )/4 = k

-3 = k

Therefore ,

k = -3