K 3 2 k 1 2 July 8, 2019 16 32 64 128 K 3 )-( 2 k )-( 1 4 k 2 2k 3 x 2 2x k 1 0 K 1 )( k 2 )( 2k 3 K 3 )+ 4 k 2 )- 2 3k 1 K 1 2 3 2 k 3 45 1 7 3 k 5 2 1 k 3 2 The points k 3 2 4 and (- k 1 2 are collinear K 1 )( k 2 )( 2k 3 )/ 6 K 1 2 2k 1 3 k 3 4 Solve for k k 2 1 2 3 If a 2 k 2 1 3 k P k 1 k 2 2k 3 2k 1 k 2 k 1 3k 2 3k 3
