SECTION A(1) k k 1= = or (rejected) k 1. Suggested Solutions Marks Remarks. 1. x + 1 is the longest side of the triangle. 1M + 1A

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Hugh Warner
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1 SECTION A(). x + is the longest side of the triangle. ( x + ) = x + ( x 7) (Pyth. theroem) x x + x + = x 6x + 8 ( x )( x ) + x x + 9 x = (rejected) or x = +. AP and PB are in the golden ratio and AP > PB. AB AP = AP PB k + = k k k + = k k k = 0 Using the quadratic formula, ( ) ± ( ) ()( ) k = () + = or (rejected). Let (x, y) be the coordinates of P. Area of ABP = 8 sq. units (8 ) y = 8 y = 8 y = By substituting y = into y = x 7x +, we have x = x 7x + 0 ( x )( x ) 7x + x = or x = The coordinates of P are (, ) or (, ).

2 . The equation x + 7x + k has rational roots. We can let x + 7x + k = (x + m)( x + n), where m and n are integers. = x + ( m + n) x + mn By comparing the coefficients on both sides, we have 7 = m + n and k = mn The table below shows all the possible values of m and n, and the corresponding value of k. m n k The possible positive values of k are and. For k =, x + 7x + (x + )( x + ) x = or x =. The quadratic equation kx + 6x + k has equal roots. Δ 6 ( k )( k) 6 k k = 9 k = ± 6. The quadratic graph y = kx 6x does not intersect the x-axis. Δ < 0 ( 6) ( k )( ) < k < 0 9 k < The range of possible values of k is 9 k <. 7. (a) By substituting (a, 0) into y = x x + 9, we have 0 = a 0 = (a ) a = a + 9 y-intercept = 9 b = 9 (b) The axis of symmetry is x =.

3 (c) Let (c, 9) be the coordinates of C. B is reflected about the axis of symmetry to C. 0 + c = c = The coordinates of C are (, 9). 8. (a) g( = ( x )( x + ) = x + x = x + x f ( g( = x + x + ( x + x ) = x + x + x = x + x + By the remainder theorem, remainder = f ( ) = ( ) + ( ) + = 9. (a) p ( = 8x + x x + a By the remainder theorem, p = a = 7 a = 9 (b) 8x + x p( = (x ) q( + 7 x + 9 = (x ) q( + 7 q( = (8x = x + x + 8x + 6 x 8) (x ) 0. (a) g() = () + () + 6 = By the factor theorem, x is a factor of g (. (b) By long division, g( = x + x x + 6 = ( x )(x + x ) = ( x )( x + )(x )

4 . (a) x + is the factor of f (. ( ) + p( ) f ( ) + q( ) 8 x + is the factor of f (. ( ) + p( ) () (): we have p = p q = 9...() f ( ) + q( ) 8 p q = 8...() By substituting p = into (), we have q = 0. both correct (b) x + x + ( x + )( x + ) x + and x + are the factors of f (. f ( is divisible by x + x +.. (a) The equation of L is y ( ) ( 7) = x y + = x y + = x + x + y = 0 (b) Let (c, 0) be the coordinates of C. By substituting (c, 0) into the equation of L, we have c + 0 c = Coordinates of C =, 0. (a) L : x y () L :x y = 0...() () () : (9x y ) ( x y + 7) 8x 0 x = By substituting x = into (), we have () y = 0 y = Coordinates of P = (, )

5 (b) Slope of the straight line = = ( ) The equation of the straight line is y = ( x ) y = x x y + 7. Let (a, 0) be the coordinates of A. By substituting (a, 0) into the equation of L, we have a a = OA = PB = AP and PO AB OB = OA (prop. of isos. ) = Coordinates of B = (, 0) ( 6) y-intercept of L = = 6 Coordinates of P = (0, 6) 6 0 Slope of L = 0 ( ) = The equation of L is y = x + 6 x y + 6 SECTION A(). (a) f( x ax ( a + ) Δ = ( a) ()[ ( a + )] = a + a + = ( a + ) > 0 ( a ) The equation f( has two distinct real roots.

6 (b) Using the quadratic formula, ( a) ± ( a + ) x = () a ± ( a + ) = a + = or = a + or The roots of the equation f( are a + and. (c) The y-intercept of the quadratic graph y = f( is. By substituting x and y = into y = x ax ( a + ), we have = (0) a(0) ( a + ) a + = The x-intercepts of the graph are and. 6. (a) The axis of symmetry of the graph is x =. h = By substituting (0, 7) and h = into 7 = (0 ) 7 = 6 + k k = 9 (b) By substituting y, h = and k = 9 into ( x ) = 9 + k 0 = ( x ) + 9 y = ( x h) + k, we have y = ( x h) + k, we have x = ± x = or x = 7 The coordinates of A and B are (, 0) and (7, 0) respectively. + (c) y = ( x ) 9 Maximum value of y = 9 Maximum height of ABP = 9 Height of ABC = 7 Maximum height of ABP > height of ABC Also, base of ABP = base of ABC It is possible that the area of ABP is greater than that of ABC. 6

7 7. (a) y-intercept = c = By substituting (7, 9) and c = into 9 = 7 9 = 7b + 7 b = + b(7) + ( ) y = x + bx + c, we have (b) (i) When y, x x ( x + )( x 6) x = or x = 6 The x-intercepts are and 6. (ii) The axis of symmetry is + 6 x = x = (c) f () = () = 6 The coordinates of the vertex of the graph of y = f( are (, 6). k = 6 8. (a) g( = (x + )(6x x + 7) 6 = x 6x + x + g = = + + By the factor theorem, x is a factor of g (. (b) g( x 6x + x + (x )(6x x ) (x )(6x + )( x ) x = or 6 or all correct 7

8 9. (a) The remainder is when f ( is divided by x. f () = (a b) = a b = ± b = a + or a + (b) x is the factor of f (. f () ( a b) a b = ± a = b + or b (rejected) By substituting a = b + into the result of (a), we have When b = a +, b = ( b + ) + b = 8 b = Hence, a = + =. both correct When b = a, b = ( b + ) b = b = Hence, a = + =. both correct 0. (a) x is a factor of f (. f (0) mn + mn = m and n are integers and m < n < 0. m = and n = both correct 8

9 (b) f ( = ( x )( x )( x ) + = x 7x x 7x + x ( x + x f ( g( 9x + 0x + k) x + x k f ( g ( has real roots. Δ 0 () ()( k ) 0 k 8. (a) By long division, we have x + x x + x 7x + 6x x 8x x x + 6x x + x 8 Quotient = x + and remainder = x 8 + (b) (i) From the result of (a), we have f ( = (x + )( x x + ) + (x 8) g( = [(x + )( x = (x + )( x x + ) + (x 8)] ( rx + s) x + ) + (8 r) x (6 + s) g ( is divisible by x x +. 8 r and 6 + s, i.e. r = 8 and s = 6 + (ii) g( = (x + )( x x + ) = (x + )( x )( x ) 9

10 . (a) Slope of L = = ( ) Let m be the slope of OD. OD is the shortest distance from O to the line L. OD L m slope of L = m = m = The equation of OD is y = x x + y (b) (i) L : x y + () OD : y = x () By substituting () into (), we have x x + x + x = By substituting x = into (), we have y = ( ) = Coordinates of D = (, ) (ii) OD = = ( 0) units + ( 0) units = units The shortest distance from O to the line L is units. 0

11 . (a) Slope of 8 0 AB = 0 6 = The equation of AB is y 8 = ( x 0) y = x x + y CD AB Slope of CD slope of AB = Slope of CD = Slope of CD = The equation of CD is y 0 = [ x ( )] y = x + x y + (b) AB : x + y () CD : x y + () ( ) () :(x + 9y 7) (x 6y + 8) By substituting x + x = y = into (), we have y 0 y = Coordinates of D =,

12 Let (0, e) be the coordinates of E. e = y-intercept of CD = ( ) = Coordinates of E = (0, ) (c) Area of quadrilateral OBDE = area of AOB area of AED = (6)(8) (8 ) sq.units = ( 6) sq.units = 8sq.units. (a) E is the mid-point of AC Coordinates of E =, = (6, ) (b) (i) Slope of CD = = ( ) AB // DC Slope of AB = slope of CD = The equation of AB is y 7 = ( x ) x y + (ii) Let (b, ) be the coordinates of B. B(b, ) lies on AB. By substituting (b, ) into the equation of AB, we have ( b) + b = Coordinates of B = (, ) AB = ( ) + (7 ) units units = units Let G be a point on BC such that AG BC. AG = (7 ) units = units BC = (9 ) units = 8 units

13 By considering the area of parallelogram ABCD, AB CH = BC AG 8 CH = units 6 = units 6 = units. (a) Coordinates of E = (, ) Area of ABCD = 08 sq. units Area of BEA = 08 sq. units (9 )( b ) = 7 b = 9 b = E is the mid-point of BD. b + d = 0 = + d d = (b) Slope of AB = 9 = The equation of AB is y = ( x 9) y 0 = x + 7 x + y 7 CD // AB Slope of CD = slopeof AB = The equation of CD is y ( ) = ( x ) y + 8 = x + 9 x + y

14 6. (a) L : x + y...() L : x y...() () () : (x + y 0) (x y) y 0 y = By substituting y = into (), we have x + x = Coordinates of A = (, ) (b) (i) The equation of L is y = m( x ) y = mx m + Let (b, 0) and (0, c) be the coordinates of B and C respectively. By substituting (b, 0) into the equation of L, we have 0 = m( b) m + bm = m m b = m m Coordinates of B =, 0 m y-intercept of L = m + c = m + Coordinates of C = ( 0, m + ) (ii) P(, ) is the mid-point of BC. 0 + ( m + ) = = m + m = 7. (a) Slope of 0 BC = 0 = (b) P is the mid-point of BC Coordinates of P =, =,

15 Let m be the slope of PH. PH BC m slope of BC = m = m = The equation of PH is y = x 8y 6 = 6x 9 6x 8y + 7 Let (0, h) be the coordinates of H. 7 y-intercept of PH = ( 8) Coordinates of 7 h = 8 7 H = 0, 8 (c) O is the mid-point of AB and CO AB. CO is the perpendicular bisector of AB. Let Q be the mid-point of AC Coordinates of Q =, =, 0 Slope of AC = 0 ( ) Slope of = QH = = Slope of AC slope of QH = = AC QH QH is the perpendicular bisector of AC. The three perpendicular bisectors PH, QH and CO in ABC pass through the same point H.

COORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE. To find the length of a line segment joining two points A(x 1, y 1 ) and B(x 2, y 2 ), use

COORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE I. Length of a Line Segment: The distance between two points A ( x1, 1 ) B ( x, ) is given b A B = ( x x1) ( 1) To find the length of a line segment joining