19 + 5 5 P1 for appropriate process to find gradient y = −4x e.g. (=) P1 process to find coordinate 7 + × 5 (= 9) y or coordinate 9 + × 20 (= x − −1) P1 method to find gradient of line (= L −4) M1 substitution of found values for and into equation for x, y m straight line A1 4x + 5 y = – 20 45 4 B1 = 5 p M1 405 = 5q4 M1 = (=3) q A1 cao 21 5 M1 = a 3b or = (a 3b) – – M1 = b + (a 3b) or = a + (a 3b) or – – = 2a 2b – M1 two of = b + (a 3b) or = a + (a 3b) or = 2a 2b – – –
19 y = −4x + 5 5 P1 for appropriate process to find gradient e.g. (=) P1 process to find y coordinate 7 + × 5 (= 9) or x coordinate − 9 + × 20 (= −1) P1 method to find gradient of line L (= −4) M1 substitution of found values for x, y and m into equation for straight line A1 y = – 4x + 5 20 45 4 B1 p = 5 M1 405 = 5q4 M1 q = (=3) A1 cao 21 5 M1 = a – 3b or = (a – 3b) M1 = b + (a – 3b) or = a + (a – 3b) or = 2a – 2b M1 two of = b + (a – 3b) or = a + (a – 3b) or = 2a – 2b
GCSE_Mathematics_2H_MS-1.docx