(a) let A be (x1,y1) and B be (x2,y2),       Gradient = m = (y2-y1)/(x2-x1)       m = (4- -5)/(-1-2) = 9/(-3) = -3       Gradient  = -3
(b) equation of a straight line line: y = mx + c 		=>  y1 = m(x1) + c			 -5 = -3(2) + c			 c = -5 +6 = 1=> Equation of line through A and B: y = -3x +1at (-100, 301), x = -100 => y = -3(-100) + 1 = 301 (from equation of line)Therefore this point does lie on the line