Question 1184011
Unless we assume K=k there is no unique solution to this problem. So let's get started...

(2,3) implies x=2, y=3

Using substitution we can say kx-2y+k=0 becomes k(2)-2(3)+k=0. Then

2k-6+k=0

2k-6+6+k=0+6

3k=6

3k/3=6/3

k=2

which leads to L: 2x-2y+2=0

and we can simplify to get L: x-y+1=0