Question

Find the value of k for which
the lines ( k+1)x+3ky +15=0
and 5x ky + 5 =0 are coincident​

Answer 1

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.

: )