Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 1 · Variant 2

dy 1 72 3 The equation of a curve is such that = 2x + . The curve passes through the point P 2, 8 . dx x4 (a) Find the equation of the normal to the curve at P. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the curve. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a)   M1 Tangent gradient must come from x = 2 substituted into the − 1  1 2  given expression. [Gradient of normal =]  −=−  11 11 11 Their   2  2  y − 8 2 2 x 92 A1 OE = − or 11 y + 2 x = 92 or y =− + x − 2 11 11 11 2 3(b)  1 2  72   x 2 24  B1, B1 One mark for each correct unsimplified { }. + c   y =   x  2  + 3 −3   + c   − 3  2  x   4 x  1 24 M1 Substitution of x = 2, y = 8 into their integrated expression, 8 = −4 + c defined by at least one correct power. Two terms and + c 4 8 needed.  1  2 24 A1 Both coefficients must be simplified but allow x− 3. Condone y = or 0.25 x − + 10   3  4  x c = 10 as line as long as either y or f(x) = is seen elsewhere. 4

More questions on Integration

4 A 2 cm C B The diagram shows the shape of a coin. The three arcs AB, BC and CA are parts of circles with centres C, A and B respectively. ABC is an equilateral triangle with sides of length 2cm. (a) Find the perimeter of the coin. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the area of the face ABC of the coin, giving the answer in terms of π and 3. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) π 60 B1 Finding one correct arc length – may be implied by correct  Arc length =  2  or  2π  2 final answer. 3 360  Perimeter =  2π or 6.28 B1 AWRT 2 4(b) 1 2 π 60 2  2π  B1 SOI AWRT [Area of one sector =]  2  or  π  2 = or 2.09   2 3 360  3  1 2  π  B1 AWRT [Area of triangle = ]  2  sin   or other valid method Allow use of 60 2  3   = 3 or1.73     2π  M1 OE − 3 + 3 [= 2.82]  Area of coin = 3 segments + triangle   3   2π    3  −2 3 Or 3 sectors – 2 triangles  3   or  3   2π  2π   Sector + 2 segments  + 2  − 3    3  3   2π − 2 3 or 2( π - 3 ) A1 Must be one of these simplified versions but equivalent decimal answers can score B1B1M1 4

More questions on Trigonometry

5 The first, second and third terms of a geometric progression are sin 1, cos 1 and 2 −sin 1 respectively, where 1 radians is an acute angle. (a) Find the value of 1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using this value of 1, find the sum of the first 10 terms of the progression. Give the answer in b the form c −1, where b and c are integers to be found. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) cos 2 − sin 2 *M1 OE. Forming a correct equation in  only using the terms of = leading to cos sin = sin( 2 − sin)sin the GP and an attempt to clear fractions. sin cos 2 2 1 DM1 Correct use of cos 2 + sin 2 = 1 and attempt to solve for cos + sin = 2sin leading to sin =  2 sin. A1 AWRT = π or 0.524 5π 6 A0 for = 30. Condone inclusion of and/or 6 3 5(b) 1 B1 OE SOI a = r = 3 Trigonometric values need to have been evaluated but allow 2 decimal equivalents (0.5 and 1.73 AWRT)  10  M1 Use of a correct formula for S10, with their value of . 1 − their 3  π  ( )  S10 = sin  their   Their 3 needs to come from their 3  6   1 − ( )  cos ( their) 2 − sin ( their)   or OE sin ( their) cos ( their) 121 A1 −121 [ S10 = ] or 165 AWRT scores B1M1A0. 3 − 1 1 − 3 3

More questions on Series

6 The equation of a curve is y = x2 −8x + 5. (a) Find the coordinates of the minimum point of the curve. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A 4 The curve is stretched by a factor of 2 parallel to the y-axis and then translated by . 1 (b) Find the coordinates of the minimum point of the transformed curve. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the equation of the transformed curve. Give the answer in the form y = ax2 + bx + c, where a, b and c are integers to be found. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: x − 8 x + 5 = 0  2 x − 8 = 0  Correct differentiation of x2 and equating their to 0.6(a) d ( 2 M1 d y ) dx d x Alternative method 1 for first mark of Question 6(a) 2 M1 2 y = ( x − 4 ) − 11 Attempt to complete the square as far as y = ( x − 4 )  k . Alternative method 2 for first mark of Question 6(a) −b 8 M1 x = = 2 a 2 x = 4, y = -11 A1 8  64 − 20 Answers from x = leading to x = 4  11 2 scores M0A0 2 6(b) x = ( their x value from a ) + 4 [=8] B1 FT Can be from finding the equation of the transformed curve, d y differentiating and putting = 0. d x y = ( their y value from a )  2 +1 [ −21] B1 FT Can be from putting x = 8 in the equation of the transformed curve. 2 If B0B0 scored, SC B1 for sight of ( 4, − 22 ) . 6(c) 2 2 B1 Can be implied if both transformations done together: 2 x − 8 x + 5 or 2{( x − 4 ) − 11} ( ) 2 ( x − 4 ) 2 − 8 ( x − 4 ) + 5 + 1 OE. ( ) M1 For the x translation, each x becomes ( x − 4 ) . + 1 or + 1 ( x − 4 ) 2 − 8 ( x − 4 ) + 5 ( x − 4 − 4 ) 2 − their11 ( )   M1 For the y translation of +1. y = 2 x 2 − 32 x + 107 or a = 2, b = −32, c = 107 A1 Evidence to support their answer may be in (b) but answer must be seen in (c). 4

More questions on Quadratics

7 (a) Verify the identity 2x −1 4x2 + 2x −1  8x3 −4x + 1. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ tan21 + 1 1 (b) Prove the identity  [3] tan21 −1 1 −2 cos21. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Using the results of (a) and (b), solve the equation tan21 + 1 = 4 cos 1, tan21 −1 for 0Å ≤1 ≤180Å. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4 x 2 + 2 x − 1 = 8 x 3 + 4 x 2 − 2 x − 4 x 2 − 2 x + 1 = 8 x 3 − 4 x + 1 B1 AG7(a) ( 2 x − 1)( ) Six correct terms leading to the correct answer. 1 7(b) sin 2  *M1 sin in the numerator and denominator. For use of tan= + 1  sin 2 + cos 2 cos cos 2  Starting with the LHS 2  = 2 2  sin   sin − cos  − 1 cos 2  1 DM1 For use of sin 2 + cos 2 = 1 twice, in a correct expression, = need to see clear evidence of this step 1 − cos 2 − cos 2  resulting in an expression in cos 2 . 1 A1 AG = 1 − 2cos 2  Alternative method 1 for Question 7(b) sin 2 + cos 2  sin 2 + cos 2 *M1 For use of sin 2 + cos 2 = 1 twice. Starting with the RHS 2 2 2  = 2 2  sin + cos − 2cos   sin − cos  sin 2  DM1 Dividing throughout by cos 2 . + 1 cos 2  = need to see clear evidence of this step sin 2  − 1 cos 2  tan 2 + 1 A1 AG = tan 2 − 1 7(b) Alternative method 2 for Question 7(b) sec 2  *M1 For use of 1+ tan 2 = sec 2  twice. Starting with the LHS sec 2 − 2 1 DM1 AG For multiplying throughout by cos 2  to give the RHS. Clear statement  1 − 2cos 2  A1 3 7(c) 1 2 B1 Replace LHS with RHS from (b) and clear fractions. 1 − 2cos  = 4cos leading to 1 = 4cos 2 ( ) 1 − 2cos  8cos 3 − 4cos+ 1 = 0    4cos + 2cos− 1 = 0  ( 2cos− 1)( 2 *B1 Use of the expression from (a) with x = cos. ) −2 4 + 16 DB1 OE  x or cos= 1 and OR 0.31, −0.81 AWRT For all three values. 2 8 = 60,72,144 B2,1,0 B2 for three correct answers only, B1 for two correct answers and no others (but allow 36 instead of 144) in the given range or 3 correct answers plus other values in the given range. Ignore answers outside of the given range. Accept AWRT 72.0, 144.0 . SC B1 for all 3 correct answers in radians and no others: π 2π 4π , and . 3 5 5 5

More questions on Trigonometry

9 y A 2 + 12 y = 3x−1 B 1 2 y = 2x 2 + 13x−1 x O 1 2 + 12. The curves intersect at The diagram shows curves with equations y = 2x 2 + 13x−1 2 and y = 3x−1 points A and B. (a) Find the coordinates of A and B. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the area of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(a) 1 − 1 − 1 1 1 *M1 OE 2 x 2 + 13 x 2 = 3 x 2 + 12 all  x 2  x − 6 x 2 + 5 = 0 Equating the two expressions in x and then multiplying each 1 1 term by x 2 or by their substitution for x 2 . Coefficients need to be retained but condone +/– sign errors. 1 Allow x 2 replaced by x.  1  1  6  36 − 4 1 5 DM1 OE 2 2  x − 1  x − 5  [= 0] or [x=] Solving their three-term quadratic.    2    Alternative method for first 2 marks of Question 9(a) 1 1 1 1 1 *M1 Equating the two expressions in x and isolating their term in − − 2 x 2 + 13 x 2 = 3 x 2 + 12 all  x 2 leading to 2 x + 10 = 12 x 2 1 x 2 . 2 2 DM1 OE (2 x + 10) = 144 x leading to x − 26 x + 25 = 0 [4]( ) Squaring both sides, rearranging and solving a three-term 26  676 −4 1 25 quadratic. leading to [4]( x − 25 )( x − 1) [= 0] or [x=] 2 3 A1, A1 A1 for both x-values and A1 for both y values. x = 1 and 25 , y = 15 and 12 If M1DM0 scored then SCB1B1 is available for final 5 answers. 4 Answers without working score 0/4 9(b)  − 1   1 − 1   1 − 1  M1 Attempt to integrate, defined by at least one correct fractional Area =  3 x 2 + 12  −  2 x 2 + 13 x 2   dx   = −2 x      2 + 12 − 10 x 2  power, and subtract – condone the wrong way round.        3   1  B1 B1 B1 for either { }.  2 x 2   10 x 2  B1 for completely correct integration of their expression =  −  + 12 x  −  following through +/– sign errors from the subtraction. 3 1      2   2   4 3 1  M1 OE − − ( their 25 ) 2 + 12 ( their 25 ) − 20 ( their 25 )  2  Substitution of their positive limits from part (a) in their  3  integrated expression, defined by at least one correct 3 1 4   fractional power, and subtraction. 2 − ( their 1) 2 + 12 ( their 1) − 20 ( their 1)    3  9(b) Alternative method for first 4 marks of Question 9(b)  − 1   1 − 1  M1 Attempt to integrate, defined by at least one correct fractional Area =  3 x 2 + 12   dx  −  2 x      2 + 13 x 2   dx  power, and subtract – condone the wrong way round.      1   3 1  B1 B1 OE  3 x 2   2 x 2 13 x 2  One mark for each correct expression. =  + 12 x  −  +  1 3 1      2   2 2    1   1   M1 OE their 25 ) 6 ( their 1) − Substitution of their positive limits from part (a) in both of   6 ( 2 + 12 ( their 25 )  −  2 + 12 ( their 1)          their integrated expressions, defined by at least one correct   4 3 1   4 3 1   fractional power, and subtraction. − their 25 ) 2 + 26 ( their 25 ) their 1) 2 + 26 ( their 1)   ( 2   ( 2     3   3   128 2 A1 AWRT [Area =] ,42 , 42.7 If M1B1B1M0 then SC B1 available for correct final answer. 3 3 Condone negative answer if corrected. 5 Condone the presence of π for the first 4 marks but use of y 2 scores 0/5

More questions on Integration

5 10 The equation of a curve is y = f x , where f x = 4x −3 3 −20 x. 3 (a) Find the x-coordinates of the stationary points of the curve and determine their nature. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) State the set of values for which the function f is increasing. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a) dy  5 2   20  B2,1,0 B2 Three correct unsimplified { } and no others. =  ( 4 x − 3 ) 3   4  −  B1 Two correct { }or three correct { } and an additional term dx  3   3  e.g. + c. B0 More than one error.  20 2 20  2 M1 d y 3 − = 0 4 x − 3 ) 4 x − 3) = k , k  0 leading to Equating their to 0 and using a valid method to arrive at  (  leading to (  3 3  d x 2 answers. 4 x −=3  m A1  4 x −=3 1  x = 1 ,1 2 d 2 y 40 − 1 B1 OE 3  4 = ( 4 x − 3 ) 2 dx 9  1  d 2 y  160  − 1 160 B1 d 2 y x = 3  0 or − or − 17.8 so max If evaluated the answers for both must be correct OR =   ( 4 x − 3 ) 2   2  2  d x  9  9 d x d 2 y  160  − 1 160 d y 3  0 or or 17.8 so min Clear use of change in sign of correctly for both B1’s. =   ( 4 x − 3 )  x = 1 2 dx  9  9 d x If B1M1A0B0B0 scored then SCB1 can be awarded for: dy  5 2   20  2 =  ( 4 x − 3 ) 3  −   leading to ( 4 x − 3) = 64 leading dx  3   3  5 11 to x = − , . 4 4 d 2 y 10 − 1 5 d 2 y = ( 4 x − 3 ) 2 3 , x = − , 2  0 so max, dx 9 4 dx 11 d 2 y x = ,  0 so min. 2 4 dx 6 10(b) x  12 , x  1 B1 Allow ⩽ and/or ⩾. FT only from special case x − 54 , x  114 Condone: 1  x  12 . 1

More questions on Differentiation

11 The coordinates of points A, B and C are 6, 4 , p, 7 and 14, 18 respectively, where p is a constant. The line AB is perpendicular to the line BC. (a) Given that p < 10, find the value of p. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A circle passes through the points A, B and C. (b) Find the equation of the circle. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the equation of the tangent to the circle at C, giving the answer in the form dx + ey + f = 0, where d, e and f are integers. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a)  7 − 4  18 − 7  *M1 Difference in the ys  their  their  = −1 Their gradients must both come from .  p − 6  14 − p  Difference in the xs OR Scalar product leading to (14 − p )( 6 − p ) − 33 = 0 p 2 − 20 p + 84 = 33 leading to p 2 − 20 p + 51 = 0 or p 2 − 20 p = −51 A1 Clearing of fractions and collecting terms to arrive at the three-term quadratic. Allow integer multiples. Alternative method for first 2 marks of Question 11(a) ( p − 6 )2 + ( 7 − 4 ) 2 + (14 − p )2 + (18 − 7 )2 = (14 − 6 )2 + (18 − 4 )2 *M1 For correct use of Pythagoras with A,B and C. OR OR For correct use of Pythagoras with the centre, B and one of 2 E.g. (10 − p ) + 42 = 42+72 the other two points. 2 p 2 − 40 p + 102  = 0 A1 OE Collecting terms to arrive at the three-term quadratic. 2 DM1 OE 20  20 − 4  51 [2]( p − 3 )( p − 17 ) or Solving their three-term quadratic. 2 p = 3 A1 If M1A1DM0 scored then SC B1 is available for final answer. 4 11(b) [Midpoint or Centre is] (10, 11) B1 SOI by final answer. 1 2 2 2 2 M1 Finding half of the length of AC or using their centre, which (14 − 6 ) + (18 − 4 ) or (18 − their11) + (14 − their10 ) or 2 2 cannot be A, B or C, to find r or r. Note: r = 65 is M0. ( their11 − 4 ) 2 + ( their10 − 6 ) 2  r 2 = 65 or r = 65    x − 6 )( x − 14 ) + ( y − 4 )( y − 18 ) = 0 scores 3/3. ( x − 10 ) 2 + ( y − 11) 2 = 65 or x 2 + y 2 − 20 x − 22 y + 156 = 0 A1 ( 3 11(c) 18 − their11 their11 − 4 18 − 4  7  *M1 Gradient of their centre, which cannot be A, B or C, from or or =   part (b), to A or C or the gradient of AC but working needed 14 − their10 their10 − 6 14 − 6  4  if incorrect centre. OR by clearly differentiating and substitution of (14,18). 1 DM1 OE y − 18 = − ( x − 14 ) 7 1 their Using (14,18) and − to form the equation of a 4 7 their 4 straight line. 4 x + 7 y − 182 = 0 A1 All terms on one side in any order. Allow multiples of this format by an integer only. 3

More questions on Coordinate geometry