Leaving Cert Maths Samples

_________________________________________ leavingcertsolutions.com ________________________________________ 2018 Question 5 – Higher Paper 2 Solution (a) To verify that ( ) 2,1 A − is on m , substitute ( ) 2,1 − into m ( ) ( ) 2 2 3 1 1 0 4 3 1 0 − + + = ⇒ − + + = True. ∴ ( ) 2,1 − is on m ( 5 marks ) (b) AB is a line through A perpendicular to n Slope of n is 2 3 − ∴ Slope of AB is 3 2 ∴ Equation of AB is ( ) 3 1 2 2 2 3 6 3 2 8 0 2 y x y x x y − = − − ⇒ − = + ⇒ − + = Now solve: (1) 2 3 51 0 x y + − = (2) 3 2 8 0 x y − + = and eliminate y (1) ×2 and (2) ×3: 4 6 102 9 6 24 6 13 78 x y x y x x + = − = − ⇒ = = Now find y : ( ) 2 6 3 51 13 y y + = ⇒ = ∴ ( ) 6,13 B ( 10 marks ) (c) AB = JJJG up 8, up 12 1 4 AC AB = = JJJG JJJG up 2, up 3 ∴ C is ( ) 0, 4 Or use ( ) ( ) ( ) ( ) 1 2 1 2 3 2 1 6 3 1 1 13 , 0, 4 4 4 sx rx sy ry r s r s − + + + + ⇒ = = + + Midpoint is centre ( ) 2 0 1 4 , 1, 2.5 2 2 − + + = − Radius ( ) ( ) 2 2 13 0 1 4 2.5 4 = − − + − = The equation of the circle is ( ) ( ) 2 2 13 1 2.5 4 x y + + − = ( 10 marks ) Comment: A bit long. A (–2,1) 2 x + 3 y –51 = 0 C (6,13) (–2,1) 3 1 (–2,1) (0,4) (–1,2.5)