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