Additional Mathematics Lines and circles Topic assessment 1 The points A and B have coordinates ( ) and (4 respectively. Calculate (i) The gradient of the line AB [1] The length of the line AB [] (iii) The coordinates of the mid-point of the line AB. [] The points A B and C have coordinates (1 ) ( 5) and (6 respectively. Find (i) The equation of the line through A parallel to BC [] The equation of the line through C parallel to AB [] (iii) The point D where these two lines intersect. [] The points A B C and D have coordinates (0 6) (5 9) (6 4) and (1 respectively. (i) Find the coordinates of E the point where the lines AC and BD intersect. [5] Show that AD and BC are parallel and equal in length. [4] (iii) Show that E bisects AC and DB. [] 4 The coordinates of the points A B and C are (0 4) (9 ) and (8 0) respectively. The points L M N are on the sides BC CA and AB respectively of the triangle ABC such that AL is perpendicular to BC and BM is perpendicular to CA. The equation of the line AL is y + x = 8. (i) Find the gradient of the line AC and hence of the line BM. [] Find the equation of the line BM. [] (iii) Find the coordinates of the point P where AL and BM intersect. [] (iv) Show that CP is perpendicular to the line AB. [] 1 of 05/06/1 MEI
5 The vertices of a quadrilateral are A ( 1) B (5 1 C (10 4) and D (5). (i) Calculate the gradients of the diagonals AC and BD. Deduce a geometrical fact about these lines. [] Find the equation of the line AC and show that E the midpoint of BD lies on it. [4] 6 (i) Prove that the points A (- B ( ) and C (86) lie on a straight line. [] Find the equation of the line L perpendicular to ABC through C. [] (iii) Given that O is the origin find the equation of the line M parallel to OB through A. [] (iv) Find the coordinates of the point D at which the lines L and M intersect. [] A circle has equation x + y 8x 10y + 16 = 0. (i) Show that the circle touches the x-axis and find the coordinates of the points where it cuts the y-axis. [4] Write down the radius and the coordinates of the centre of the circle. [] Total: 60 of 05/06/1 MEI
Topic assessment Solutions 1. (i) Gradient AB yb ( 4 x x 4 A B Length of AB ( x x ) ( y y ) A B A B ( 4) ( ( ) ( ) 4 4 16 0 xa x y B B 4 1 6 (iii) Midpoint of AB ( yb yc 5 ( 6. (i) Gradient of BC xb xc 6 The equation of the line is y y1 m( x x where m = - and ( x1 y =(1 ) y ( x y x y x 4 yb 5 Gradient of AB xa xb 1 The equation of the line is y y1 m( x x where m and ( x1 y =(6 - y ( ( x 6) ( y ( x 6) y x 18 y x 0 (iii) Substitute y x 4 into y x 0 ( x 4) x 0 4x 8 x 0 x 8 x 4 A Substitute x = 4 into y x 4 gives y 4 4 8 4 4 The coordinates of D are (4-4). of 05/06/1 MEI
yc 6 4 1. (i) Gradient of AC xa xc 0 6 6 1 The equation of the line is y y1 m( x x where m and ( x1 y =(0 6) 1 y 6 x A1 ( y 6) x y x 18 yb yd 9 1 8 Gradient of BD xb xd 5 1 4 The equation of the line is y y1 m( x x where m = and ( x1 y =(1 y 1 ( x A1 y 1 x y x 1 Substituting y x 1 into y x 18 ( x x 18 6x x 18 x 1 x Substituting x = into y x 1 gives y 1 6 1 5 E is ( 5) yd 6 1 5 Gradient of AD 5 xa xd 0 1 1 yb yc 9 4 5 Gradient of BC 5 xb xc 5 6 1 so AD and BC are parallel. Length of AD (0 (6 1 5 6 Length of BC (5 6) ( 9 4) 1 5 6 so AD and BC are the same length. (iii) 0 6 6 4 6 10 Midpoint of AC (5) which is E 1 5 1 9 6 10 Midpoint of DB (5) which is E so E bisects AC and DB. 4 of 05/06/1 MEI
4 0 4 1 4. (i) Gradient of AC 0 8 8 Since BM is perpendicular to AC gradient of BM is. The equation of the line is y y1 m( x x where m = and ( x1 y =(9 ) y ( x 9) y x 18 y x 11 (iii) Substituting y x 11 into y x 8 ( x 1 x 8 14x x 8 15 x 105 x Substituting x = into y x 11 gives y 11 14 A 11 so P is ( ) 0 (iv) Gradient of CP 8 1 4 1 Gradient of AB 0 9 9 1 Gradient of CP gradient of AB 1 so CP and AB are perpendicular. 5. (i) 1 ( 4) 16 4 Gradient of AC 10 1 11 5 6 Gradient of BD 5 ( ) 8 4 The lines AC and BD are perpendicular since the product of their gradients is - 1. 4 The equation of the line is y y1 m( x x where m and ( x1 y =(- 1) 4 y 1 ( x ( )) E ( y 1) 4( x ) y 6 4x 8 y 4x 8 5 11 5 16 (18) 5 of 05/06/1 MEI
Substituting x = 1 and y = 8 into the equation of AC gives 8 4 1 4 4 8 so E lies on the line. 1 1 6. (i) Gradient of AB 4 6 1 Gradient of BC 8 6 Since AB and BC have the same gradient the point A B and C lie on a straight line. Gradient of L = - The equation of the line is y y1 m( x x where m and ( x1 y =(8 6) y 6 ( x 8) y 6 x 16 y x (iii) Gradient of OB The equation of the line is y y1 m( x x where m and ( x1 y =(- y 1 ( x ( )) ( y ( x ) y x 6 y x 8 (iv) Substituting y x into y x 8 ( x ) x 8 4x 44 x 8 x 6 x 6 6 6 8 Substituting x into y x gives y 6 8 D is A. (i) On x-axis y = 0 so x 8 x 16 0 ( x 4) 0 x 4 Since there is only one point at which y = 0 the circle touches the x-axis. 6 of 05/06/1 MEI
On y-axis x = 0 so y 10y 16 0 ( y )( y 8) 0 y or y 8 It cuts the y-axis at (0 ) and (0 8). Completing the square on x² and y²: x 8 x y 10y 16 0 ( x 4) 16 ( y 5) 5 16 0 ( x 4) ( y 5) 5 Radius = 5 and centre of circle is (4 5). of 05/06/1 MEI