Your Practice Set – Applications and Interpretation for IBDP Mathematics SE Production Limited 214 ( e ) ( i ) 5 1 1 2 1 1 t t Ae Be � � � � � � � � � � � � � � � � � � � � X ( A1 ) for correct approach 0 5 ( 0 ) 1 2 1 2 5 1 1 Ae Be � � � � � � � � � � � � � � � � � � � � � � � � � � ( M1 ) for substitution 1 2 2 5 A B A B � � � � � �� � � � By solving this system , 2 A � and 3 B � . ( A1 ) for correct values 5 3 t t x e e � � � � � A1 ( ii ) 5 2 3 t t y e e � � � A1 [ 5 ] ( f ) The required coordinates 0.5 5 ( 0.5 ) 0.5 5 ( 0.5 ) ( 3 , 2 3 ) e e e e � � � � � � ( M1 ) for substitution ( 35.94095122 , 37.7605432 ) � ( 35.9 , 37.8 ) � A1 [ 2 ] ( g ) For starting at ( 2 , 5 ) A1 For positive gradient A1 [ 2 ]