Junior Cert Maths Samples

_________________________________________ juniorcertsolutions.com ________________________________________ 2016 Question 4 – Higher Paper 2 Solution (a) A (–1,3), B (5,3), C (–1,8) ( 5 marks ) (b) Equation of line AB : y = 3 Equation of line AC : x = –1 Slope of 2 1 2 1 8 3 5 5 1 5 6 6 y y BC x x − − = = = = − − − − − Use ( ) 1 1 y y m x x − = − ( ) 5 3 5 6 y x − = − − Multiply everything by 6 and tidy it up 6 18 5 25 y x ⇒ − = − + 5 6 43 x y + = ( 10 marks ) (c) 1 5 opposite 5 5 tan tan 6 adjacent 6 6 θ θ − = = = ⇒ = 39.81 θ = D ( 10 marks ) (d) (i) Find BC using Pythagoras 2 2 2 5 6 61 61 BC BC + = = ⇒ = ( 5 marks ) (ii) Since 90 A BC ∠ = ⇒ D is the diameter of the circle The radius is 61 2 Area is 2 2 61 61 15.25 2 4 r π π π π   = = =     square units ( 5 marks ) (e) To find the equation of the line, we use ( ) 1 1 y y m x x − = − , 6 5 m = , since the new line is perpendicular to BC ( x , y ) is (–1,3) ( ) ( ) 6 6 3 1 3 1 5 5 y x y x − = − − ⇒ − = + Multiply everything by 5 and simplify 5 15 6 6 6 5 21 0 y x x y − = + ⇒ − + = ( 5 marks ) Comment: Very mixed up question. (–1,8) (–1,3) (5,3) C B A 5 6 θ