Pearson Progression Step Q Scheme Marks AOs and Progress descriptor Makes an attempt to rearrangeto make the subject. For t 1a M1 2.2a 5th example,is seen. Convert between Correctly states parametric A1 1.1b equations and Makes an attempt to substituteinto cartesian forms 2.2a M1 using substitution. For example,is seen. Simplifies the expression showing all steps. A1 1.1b For example, (4) Interprets the gradient of line being asand finds 1b −1 M1 2.2a 5th Convert between Substitutes = to find =and = t x y −1 M1 1.1b parametric And substitutes = 0 to find = 1 and = 2 t x y equations and cartesian forms Makes an attempt to use Pythagoras’ Theorem to find the length M1 1.1b using substitution. of the line: Correctly finds the length of the line segment,or states A1 1.1b (4) (8 marks) Notes Pearson Progression Step Q Scheme Marks AOs and Progress descriptor Recognises that the identity can be used to find the cartesian 2.2a 6th 2 M1 equation. Convert between parametric Statesor M1 1.1b equations and Also states cartesian using Substitutesandinto M1 1.1b trigonometry. Solves to find, accept < 1 or A1 1.1b x (4 marks)
Q Scheme Marks AOs Pearson Progression Step and Progress descriptor 1a Makes an attempt to rearrangeto make t the subject. For example,is seen. M1 2.2a 5th Convert between parametric equations and cartesian forms using substitution. Correctly states A1 1.1b Makes an attempt to substituteinto For example,is seen. M1 2.2a Simplifies the expression showing all steps. For example, A1 1.1b (4) 1b Interprets the gradient of line being −1 asand finds M1 2.2a 5th Convert between parametric equations and cartesian forms using substitution. Substitutes t = −1 to find x =and y = And substitutes t = 0 to find x = 1 and y = 2 M1 1.1b Makes an attempt to use Pythagoras’ Theorem to find the length of the line: M1 1.1b Correctly finds the length of the line segment,or states A1 1.1b (4) (8 marks) Notes Q Scheme Marks AOs Pearson Progression Step and Progress descriptor 2 Recognises that the identity can be used to find the cartesian equation. M1 2.2a 6th Convert between parametric equations and cartesian using trigonometry. Statesor Also states M1 1.1b Substitutesandinto M1 1.1b Solves to find, accept x < 1 or A1 1.1b (4 marks)
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