KX – Y – 2K + 1 = 0 K2 ; 13 2 COS CAB = COS DBA 1087 K K K K2        K = 1 , AC

1.B(11; 5) AC: kx – y – 2k + 1 = 0 k2 ; 13

2

cos CAB = cos DBA 1087 k k k k

2

       k = 1 , AC : x – y – 1 = 0