Topical Revision Paper 9 Unit 9 Points Lines Shapes

Topical Revision Paper 9 Term 3 Unit 9: Points, Lines and Shapes Question 1: The vertices of a triangle ABC are A (5, 10), B (0, 0) and C (–6, 8). (i) Show that the triangle ABC is an isosceles triangle. [2] (ii) Calculate the area of triangle ABC. [2] (iii) Find the coordinates of the foot of perpendicular from A to BC. [1] (iv) Given that four points A, B, C and D form a rhombus, find the coordinates of the point D. [3] (v) Find the equation of AC. Hence or otherwise, find the coordinates of the foot of perpendicular from B to AC. [5] Solution: (i)             22 22 22 5 0 10 0 125 units 6 0 8 0 10 units 5 6 10 8 125 units AB BC AC                 AB = AC Hence, triangle ABC is isosceles. (ii) Area of triangle ABC 2 0 5 6 01 0 10 8 02 50 units   (iii) Foot of perpendicular from A to BC in the isosceles triangle ABC is the midpoint of BC. That is  6 0 8 0, 3, 422     

(iv) Let the coordinate of D be (x, y). ABDC is the rhombus. Since the diagonals of a rhombus bisect each other,  5 10, 3, 422 11, 2 : ( 11, 2) xy xy D        (v) Gradient of AC = 8 10 2 6 5 11   Equation of the line AC:  210 5 11 2 100 11 11 yx yx     Gradient of the line perpendicular to AC = 11 2 Equation of the line perpendicular to AC through B: 11 2yx Solving the simultaneous equations, we have 8 5 44 5 The coordinate of the foot of perpendicular 8 44are , . 55 x y    Question 2 : The diagram below shows triangle ANC. Point B has coordinates (8,0) , Point C has coordinates (4,7) and Point D has coordinates ( 2,5) . The equation of the straight line DB is 2x y k and this line intersects the y-axis at A. Point N lies on the line DB such that CN is perpendicular to line DB. Point F is on the x-axis such that DF is parallel to the y-axis.

(i) Find the value of k, in the equation of straight line DB. [2] (ii) Find (a) the gradient of CN, [2] (b) the equation of CN. [1] (iii) Calculate the coordinates of N. [2] (iv) Is CN a perpendicular bisector of BD? Explain your answer. [1] (v) Calculate the area of triangle DFB. [3] (vi) Show clearly that the distance between D and B is 11.18 units, correct to 2 decimal places. [2] (vii) Find the perpendicular distance of F to BD. [2] x y D ( 2, 5) x + 2y = kN A O C (4, 7) B (8, 0) F

Solution: (i) 2x y k Substitute ( 2,5) into the equation to find value of k, ( 2) 2(5) 8 k k     (ii) Rewriting the equation of DB, we have 1 42yx  . (a) Gradient of DB = 1 2 Gradient of CN = 2 (b) 7 2( 4)yx   Equation of CN is 21yx (iii) 12 1 4 2 2, 3 xx xy     : (2,3)N (iv) Midpoint of BD = 2 8 5 0 1, 3, 22 2 2              Therefore CN is not a perpendicular bisector of BD since N is not the midpoint of BD. (v) Coordinate of ( 2,0)F  Area of triangle DFB = 1 10 5 252   units2 (vi) Distance of DB = 22( 2 8) (5 0) 100 25 11.18       units (vii) 1 125 252 4.47 or 2 5 d d     or 1 11.18 252 4.47 d d     Question 3: Solutions to this question by accurate drawing will not be accepted. y

The diagram shows a rhombus ABCD with vertices A(10,k), B, C(–2,–2) and D in which the equation of line BC is given to be 8 14 0yx   and the equation of line BD is 2 3 16 0yx   . (i) Show that k = 6. [3] (ii) Find the coordinates of B and of D. [6] (iii) Find the area of the rhombus ABCD. [2] Solution: (i) Diagonals of rhombus bisect each other. Midpoint. is 2 3 16 0 2 10when 4, 2 2 3(4) 16 0 4 22 2 22 6 (shown) BD y x x y y k k            (ii) and intersect at 8 14 3 42 24 16 2 26 26 1 8 14 6 : (6, 1) BC BD B xy x y y y y x B            x A (10, )k C ( 2, 2) B D

Let D = ( , )xy 6 42 2 x x    1 22 5 : (2,5) y y D    (iii) 2 Area of 2 6 10 2 21 2 1 6 5 22 52 unit ABCD      Question 4: Solutions to this question by accurate drawing will not be accepted. The diagram shows a trapezium ABCD in which AB is parallel to DC and BC is a vertical line. The vertices of the trapezium are A(–2, 3), B(6, 6), C and D(–5, –3). The perpendicular from A to DC meets DC at the point P. (i) Find the coordinates of P and C. [6] The lines AB and DC cut the y-axis at E and F respectively. (-5, -3) (6, 6) (-2, 3) o x y F E P C A B D

(ii) Find the area of the parallelogram BCFE. [2] Solution: (i) 6 3 3 6 2 8 AB DCmm     Equation of AP: 83 ( 2)3yx   -------------------- (1) Equation of DC: 33 ( 5)8yx   ------------------------ (2) (2) – (1): 38( 5) ( 2) 683xx    3 8 15 16 68 3 8 3x     29 73x [or – 0.39726 ≈ – 0.397] 8 29 93 20( 2) 3 13 73 73 73y      [or – 1.2739 ≈ – 1.27]  29 20: , 1 73 73P  Sub x = 6 into (2): 3 7 1(6 5) 3 18 8 8y     [or 1.125]  the coordinates of C are 16,18  (ii) Area of BCFE = 1 117 16 6 1 29 8 4 4     units2