Review exercise 2. 1 The equation of the line is: = 5 a The gradient of l1 is 3. y y x x. So the gradient of l2 is. The equation of line l2 is: y =

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Laurence Allen
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1 Review exercise The equation of the line is: y y x x y y x x y 8 x y 8 x+ 6 y x x + y 0 y ( ) ( x 9) y+ ( x 9) y+ x 9 x y 0 a, b, c Using points A and B: y y x x y y x x y x 0 k 0 y x k ky k x a The gradient of l is. So the gradient of l is. The equation of line l is: y ( x 6) y x+ y x+ b y x 6 y x+ x 6 x+ x+ x x 0 x y ( ) 6 x y 6 The point C is (, ). c Substituting point C into the equation: k(k) k (0) k k 0 0 (k + )(k ) 0 k or k a Using points (60, 7) and (80, 8): y y Gradient x x b l kh, where k is the gradient. So l 0.h c The model may not be valid for young people/children who are still growing. Where l meets the x-axis, y 0: 0 x 6 x 6 x The point A is (, 0). Where l meets the x-axis, y 0: 0 x + x x The point B is (, 0). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

2 c 8 Equation of circle with centre (, 8) and radius r: (x + ) + (y 8) r r distance from (, 8) to (0, 9) r (0 + ) + (9 8) The equation for C is: (x + ) + (y 8) 0 AB 0 The perpendicular height, using AB as the base is. Area of triangle ABC base height 0 6 Substituting y x into y + x 0: (x) + x 0 x 0 x y 6 The point P is (, 6). Distance from origin + 6 y y 7 Gradient of line x x Gradient of the perpendicular bisector is 6 Midpoint of line x + x, y + y + 7 8, (6, ) Equation of the perpendicular bisector is: y y m(x x) y (x 6) 6 y x + 6 This line crosses the x-axis at y 0: 6 x + 0 x 6 The point Q is ( 6, 0). 9 a Rearranging: x 6x + y + y 0 Completing the square: (x ) 9 + (y + ) 0 (x ) + (y + ) 0 a, b, r 0 b The circle has centre (, ) and radius 0. 0 a Rearranging x + y : y x Solving simultaneously using substitution: (x ) + ( x ) (x ) + ( x + ) x x + + 9x 66x + 0 0x 70x x 7x + 0 (x )(x ) 0 So x and x x : y x : y Point A is (, ) and point B is (, ). b Using Pythagoras theorem: Length AB ( ) + ( ) 0 The equation of the circle is x + y r. Solving simultaneously using substitution: x + (x ) r x + 9x x + r 0 0x x + r 0 Using the discriminant for no solutions: b ac < 0 ( ) (0)( r ) < r < 0 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

3 0r 6 < 0 When 0r 6 0 8(r ) 0 r r ± a AC (8 ) + ( ) Using Pythagoras theorem: AB + BC AC Therefore, ABC is 90. b As triangle ABC is a right-angled triangle, AC is a diameter of the circle. c AC is a diameter of the circle, so the midpoint of AC is the centre. < r < However, the radius cannot be negative. So 0 < r < a Equation of circle with centre (, ) and radius r: (x ) + (y ) r r distance from (, ) to (, ) r ( ) + ( ) The equation for C is: (x ) + (y ) 8 b Gradient of the radius of the circle at P y y 7 x x Gradient of the tangent 7 Equation of the tangent at P: y y m(x x) y + 7 (x ) x 7y 6 0 a AB (6 ) + ( ) + BC (8 6) + ( ) + 8 Midpoint x + x, y + y + 8+, (, ) Radius AC The equation of the circle is: (x ) + (y ) 0 x + 0x+ x + 0x+ x+ x x x+ x x ( x+ )( x+ 7) xx ( x 6) ( x+ )( x+ 7) xx ( + 7)( x 8) ( x + ) xx ( 8) a, b, c 8 a Using the factor theorem: f( ) ( ) 7( ) 7( ) So (x ) is a factor of x 7x 7x + 0. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

4 b x x 7x 7x+ 0 x x 0 6 c x + 0x 8 x x + x 8x+ x x 6 7 x 6 x + x x 0x + 0 0x x 9x 0x 8x 0x 0x 8x 8x + 0 x 7x 7x + 0 (x )(x x 0) (x )(x )(x + ) x + x 8x + (x )(x + 0x 8) (x )(x )(x + ) c (x )(x )(x + ) 0 So x, x or x So the curve crosses the x-axis at (, 0), (, 0) and (, 0). 7 a g(x) x x + g() () () So (x ) is a factor of g(x). When x 0, y 0 So the curve crosses the y-axis at (0, 0). x, y x, y b x + x x x 0x x+ x x x x x 9x x + x + 0 g(x) x x + (x )(x + x ) (x )(x + )(x ) 6 f(x) x + x 8x + c a f() 0 () + () 8() + c c 0 c b f(x) x + x 8x + f() 0, so (x ) is a factor of x + x 8x +. 8 a Example: When a 0 and b 0, (0 + 0). b (a + b) a + ab + b When a > 0 and b > 0, ab > 0 Therefore a + b < (a + b) When a < 0 and b < 0, ab > 0 Therefore a + b < (a + b) When a > 0 and b < 0, ab < 0 Therefore a + b > (a + b) When a < 0 and b > 0, ab < 0 Therefore a + b > (a + b) The conditions are a > 0 and b > 0 or a < 0 and b < 0. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

5 9 a p : + p 7: 7 9 () + p : () + p : 69 7() + p 7: 7 89 () + p 9: 9 6 () + b () + 7 and 7 is not a square number. 0 a Rearranging: x 0x + y 8y Completing the square: (x ) + (y ) 6 (x ) + (y ) 9 (x ) + (y ) a, b, r b Centre of circle C is (, ). Centre of circle D is (0, 0). Using Pythagoras theorem: Distance ( 0) + ( 0) c Radius of circle C Radius of circle D Distance between the centres + < Therefore, the circles C and D do not touch. a ( x) ( x) + ( x) 0 ( 7 ) + x (9) + 0( x) + ( x) + 0(9)(8) ( x) x + 80x 960x b (0.98) 0 ( (0.0)) 0 0(0.0) + 80(0.0) 960(0.0) ( d.p.) ( + x) + ( x) + ( x) +... () + ( x) + ( x) x + 0x +... ( x)( + x) ( x)( + 0x + 0x +...) + 0x + 80x +... x 0x x + 70x x + 70x a, b 9, c 70 ( x) q q x term ( x) q q q q x q qx q q q q 8 q q Using the sine rule: b c sin B sin C b sin sin 0 sin b sin 0 b b 0 AC 0 cm a Using the cosine rule: a + c b cos B ac (x ) + ( x+ ) cos 60 (x )() x x+ 9 + ( x + x+ ) 0(x ) (x ) x x + x x +8 0 x 8x Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

6 b x 8x (x ) 0 x 6 c Area ac sin B a c Area sin cm ( s.f.) 7 b Using the cosine rule: a + c b cos B ac ( ) (0)(0) B Angle ABC. ( d.p.) 8 Using the sine rule in triangle ABD: sin BDA sin 0...sin 0 sin BDA BDA.6 a Using the cosine rule x cos 70. x.9 km The distance of ship C from ship A is.9 km. b Using the sine rule: sin 70 sin A.9 sin A 0.90 A 70.9 The bearing of ship C from ship A is a If triangle ABC is isosceles, then two of the sides are equal. AB (6 + ) + (0 ) 00 0 BC AC (6 6) + (0 0) 00 0 (6 + ) + (0 ) AB BC Therefore ABC is an isosceles triangle. Using the angle sum of a triangle: ABD 80 ( ) 87.8 Using the sine rule in triangle ABD: AD. sin 87.8 sin 0 AD. cm AC AD + DC cm Area of triangle ABC..0 sin 0 9. cm 9 a (x ) + (y ) (x ) + (y ) b Substituting x 8 and y k into the equation of the circle: (8 ) + (k ) 9 + k k + 0 k k 0 (k + )(k 6) 0 k or k 6 k is positive, therefore k 6. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 9 c XY 0 a YZ XZ (0 ) + ( + ) 90 (8 0) + (6 ) 0 (8 ) + (6 + ) 98 Using the cosine rule: x + z y cos Y xz Rationalising the denominator: cos Y 0 So cos XYZ 0 b The curve y sin (x + ) crosses the y-axis when x 0. sin 0, Each of the four triangular faces is an equilateral triangle. Area of one triangle ac sin B s s sin 60 s s cm Total area area of triangles + area of square s + s s + s ( ) + s The total surface area of the pyramid is ( + ) s cm. b There are two solutions in the interval 0 x 60. a The curve y sin x crosses the x-axis at ( 60, 0), ( 80, 0), (0, 0), (80, 0) and (60, 0). y sin (x + ) is a translation of 0 so subtract from the x-coordinates. The curve crosses the x-axis at ( 0, 0), (, 0), (, 0), (, 0) and (, 0). ( 0, 0) is not in the range, so (, 0), (, 0), (, 0) and (, 0) a sin θ cos θ sinθ cosθ So tan θ b When tan θ θ or So sin θ cos θ when θ or tan x tan x ± For tan x x 0 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 6 a cos x sin x ( sin x) sin x sin x sin x sin x sin x + 0 (as required) So x 0 or x 0 For tan x x 0 (or 0 ) So x 0 or x 0 So x 0, 0, 0 or 0 sin (θ 0 ) sin (θ 0 ) θ 0 60 So θ 0 60 or θ 0 0 When θ 0 60 θ When θ 0 0 θ So θ 90 or 0 b Let sin x y y y + 0 (y )(y ) 0 So y or y When sin x, x 0 or x sin x is impossible. x 0 or 0 7 tan x tan x tan x tan x 0 Using the quadratic formula: ± ( ) ()( ) tan x () ± 7 When tan x + 7, x 7. or x When tan x 7, x. or x or x x 7., 7.,. or 7. 8 sin x 6( cos x) sin x + 6 cos x 6 0 ( cos x) + 6 cos x 6 0 cos x + 6 cos x 6 0 cos x 6 cos x + 0 ( cos x )(cos x ) 0 So cos x or cos x When cos x, x 78. or x When cos x, x 0 or 60 x 0, 78., 8. or 60 9 LHS cos x(tan x + ) sin x cos x + cos x sin x + cos x RHS Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 Challenge a Finding points B and C using y x : When y 0, x When x 0, y The point B is (, 0) and the point C is (0, ). Using Pythagoras theorem to find the length of the square: BC (0 ) + ( 0) 60 Area of square ( ) b The point A is ( 8, ) and the point D is (, 8). y y The gradient of line AD x x The equation of line AD is: y y m(x x) y (x + 8) y x When y 0, x 8 The point S is (, 0). Rearranging x + y + 8x 0y 9: x + 8x + y 0y 9 Completing the square: (x + ) 6 + (y ) 9 (x + ) + (y ) 00 Both circles have the same centre at (, ). The radius of one circle is 8 and the other is 0, so (x + ) + (y ) 8 lies completely inside x + y + 8x 0y 9. n n LHS + k k+ n! n! + k!( n k)! ( k+ )!( n k )! n!( k+ ) n!( n k) + ( k+ )!( n k)! ( k+ )!( n k)! n!(( k+ ) + ( n k)) ( k+ )!( n k)! n!( n+ ) ( k+ )!( n k)! ( n + )! ( k+ )!( n k)! n + k + RHS sin x sin x + cos x sin x sin x + sin x sin x + sin x sin x 0 sin x( sin x + sin x ) 0 sin x( sin x )(sin x + ) 0 So sin x 0, sin x or sin x When sin x 0, x 0, 80 or 60 When sin x, x 0 or x When sin x, x 70 So x 0, 0, 0, 80, 70 or 60 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9

Circles, Mixed Exercise 6

Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5