AQA Level 2 Further mathematics Coordinate geometry
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1 AQA Level Further mathematics Coordinate geometry Topic assessment. A line l has equation 5y4x. (i) Find the gradient of the line. [] (ii) Find the equation of the line l which is parallel to l and passes through the point (, -).. Describe fully the curve whose equation is x y 4. []. The coordinates of two points are A (-, -) and B (5, 7). (i) Calculate the equation of the perpendicular bisector of AB. (ii) Find the coordinates of the point C which divides AB in the ratio :. [] 4. Show that the line y = x 0 is a tangent to the circle x y The equation of a circle is x y 4x y 5 (i) Write down the coordinates of the centre C of the circle, and the radius of the circle. [] (ii) Describe the transformation that maps this circle on to the circle x y 0. [] (iii) Show that the point P (4, -5) lies on the circle. [] (iv) Find the equation of the tangent to the circle at the point P. 6. The coordinates of four points are P (-, -), Q (6, ), R (9, ) and S (, -). (i) Calculate the gradients of the lines PQ, QR, RS and SP. (ii) What name is given to the quadrilateral PQRS? [] (iii) Calculate the length SR. [] (iv) Show that the equation of SR is y = x 5 and find the equation of the line L through Q perpendicular to SR. (v) Calculate the coordinates of the point T where the line L meets SR. (vi) Calculate the area of the quadrilateral PQRS. 7. AB is the diameter of a circle. A is (, ) and B is (7, -). (i) Find the coordinates of the centre C of the circle. [] (ii) Find the radius of the circle. [] (iii) Find the equation of the circle. [] (iv) The line y + 5x = 8 cuts the circle at A and again at a second point D. Calculate the coordinates of D. (v) Prove that the line AB is perpendicular to the line CD. of 7 /07/4 MEI
2 8. The diagram shows triangle ABC. (i) Show that triangle ABC is right angled. (ii) Find the equation of the circle which goes through the points A, B and C. 9. The centre of a circle is (4, 0). The line x y is a tangent to the circle. Find the equation of the circle. Total 70 marks of 7 /07/4 MEI
3 Solutions to topic assessment. (i) 5 y 4x 5 y 4x y x Gradient of line = 5. 4 (ii) l is parallel to l, so it has gradient 5. 4 Equation of line is y ( ) 5 ( x ) 5( y ) 4( x ) 5 y 0 4x 4 5 y 4x 6 0. The curve is a circle, centre O and radius. y y (i) Gradient of AB x x 5 6 Gradient of line perpendicular to AB. 5 The line passes through the midpoint of AB 5 7, (,) Equation of line is y ( x ) 5 5( y ) ( x ) 5 y 0 x 6 5 y x 6 (ii) Distance from A to B in x-direction = 5 (-) = 6 x-coordinate of C Distance from A to B in y-direction = 7 (-) = 0 y-coordinate of C C = (.5, -0.5) [] [] [] 4. Substituting y x 0 into x y gives x (x 0) 0 x 9x 60x x 60x 90 0 x 6x 9 0 ( x ) 0 Since the equation has a repeated root, the line meets the circle just once, and so the 0 of 7 /07/4 MEI
4 line is a tangent to the circle. 5. (i) ( x ) ( y ) 0 The centre C of the circle is (, -) and the radius is 0. (ii) Translation through. (ii) Substituting x = 4 and y = -5: (4 ) ( 5 ) so the point (4, -5) lies on the circle. ( 5) 4 (iii)gradient of CP 4 Tangent at P is perpendicular to CP, so gradient of tangent. Equation of tangent is y ( 5 ) ( x 4) ( y 5 ) x 4 y 0 x 4 y x 4 y y 4 6. (i) Gradient of PQ x x 6 8 y y Gradient of QR x x 6 9 y y ( ) 4 Gradient of RS x x 9 8 y y ( ) Gradient of SP x x ( ) (ii) PQ is parallel to RS, and QR is parallel to SP, so the quadrilateral is a parallelogram. (iii) SR ( 9 ) ( ( )) (iv) From (i), gradient of SR Equation of SR is y ( ) ( x ) ( y ) x y 4 x y x 5 [] [] [] [] [] Line perpendicular to SR has gradient - 4 of 7 /07/4 MEI
5 Line L has gradient - and goes through (6, ) Equation of L is y ( x 6) y x y x 5 (v) Equation of L is y 5 x (vi) Substituting into equation of SR gives (5 x) x 5 When x = 7, y 5 7 Coordinates of T are (7, ) L 0 4x x x x 7 Q (6, ) T (7, ) R (9, ) P (-, -) S (, -) Length QT (7 6) ( ) 4 5 Area of parallelogram SR QT (i) C is the midpoint of AB. C 7 ( ), (4,) (ii) Radius of circle = CA (4 ) ( ) 9 4 (iii) Equation of circle is ( x 4) ( y ) (iv) Substituting y 5x 8into equation of circle: [] [] [] 5 of 7 /07/4 MEI
6 ( x 4) ( 5 x 8 ) ( x 4) ( 5 x 7) x 8 x 6 5 x 70x 49 6x 78 x 5 0 x x 0 ( x )( x ) 0 x or x x = is point A, so point D is x = When x =, y 5 8 The coordinates of D are (, -) ( ) 4 (v) Gradient of AB 7 6 Gradient of CD ( ) 4 Gradient of AB gradient of CD so AB is perpendicular to CD. 8. (i) Method Gradient of AB is 7 4 Gradient of BC is AB and BC are perpendicular to ABC is a right-angled triangle. Method AB ( 7 ) (4 ) 0 BC ( 7 6) (4 6) 5 AC (6 ) (6 ) 5 AB BC AC ABC is a right-angled triangle. (ii) Angle ABC is the angle in a semicircle so AC is a diameter. 6 6 The centre of the circle is at,, i.e, at (4, 4.5). The radius of the circle is half AC. AC (6 ) (6 ) of 7 /07/4 MEI
7 Radius =.5 The equation of the circle is ( x 4) ( y 4.5) The gradient of the tangent is -. The radius is perpendicular to the tangent and so it has gradient ½. The equation of the radius is y ( x 4) y x This line meets the tangent at a point on the circle. Substitute y x into x : x x.5 x 5 x y The radius of the circle is the length of the line joining (4, 0) to (, -). r 5 The equation of the circle is ( x 4) y 5. 7 of 7 /07/4 MEI
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