Find the value of k for which the lines (k+1)x+13ky+15=0 , 5x+ky +15=0 are coincident
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Hi ,
Compare ( k+1)x + 13ky + 15 = 0 ,
5x + ky + 15 = 0 with a1 x+b1y +c1= 0
and a2 x+b2 y + c2 = 0 we get ,
a1 = k + 1 , b1 = 13k, c1 = 15 ,
a2 = 5 , b2 = k , c2 = 15 ;
a1/a2 = b1/b2 = c1/c2
[ Since lines are coincident each
other ]
( k + 1 )/5 = 13k/k
( k + 1 ) = 13 × 5
k + 1 = 65
k = 65 - 1
k = 64
I hope this helps you.
: )
Compare ( k+1)x + 13ky + 15 = 0 ,
5x + ky + 15 = 0 with a1 x+b1y +c1= 0
and a2 x+b2 y + c2 = 0 we get ,
a1 = k + 1 , b1 = 13k, c1 = 15 ,
a2 = 5 , b2 = k , c2 = 15 ;
a1/a2 = b1/b2 = c1/c2
[ Since lines are coincident each
other ]
( k + 1 )/5 = 13k/k
( k + 1 ) = 13 × 5
k + 1 = 65
k = 65 - 1
k = 64
I hope this helps you.
: )
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