Questions
 Find the angle θ in degrees from the figure below
 In the diagram below, determine the equation of the line XY in the form y = mx + c
 Find the equation of a line which passes through the point (2, 3) and is perpendicular to y – 3x+ 1 = 0, giving your answer in the form y = mx + c
 T is the midpoint of line XY where X is point (1,4) and Y is the point (5, 10). Find the equation of a line, L_{2} which is perpendicular to line XY and goes through point T

 On the grid provided below, plot points A(2,1) B(4,3) and C(2,5)
 Given that the gradient of CD = 1 and CD =AD locate D and complete the quadrilateral ABCD
 What name is given to quadrilateral ABCD?
 In the figure below (not drawn to scale), PQRS is a rectangle and P and Q are the points (3, 2) and (1,4) respectively.
Given that the equation of the line PQ is y =3x 7, find: The equation of line QR
 The coordinates of point R
 The coordinates of point S
 OABC is a trapezium such that the coordinates of O, A , B and C are (0, 0), (2, 1), (4, 3) and (0, y)
 Find the value of y
 M is the midpoint of AB and N is the midpoint of OM. Find in column form
 the vector AN
 the vector
 Vector AC NC
 Hence show that A, N and C are collinear
 Use ruler and a pair of compasses only in this question.
 Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ∠ABC = 600.
 Measure
 side AC
 ∠ ACB
 Construct a circle passing through the three points A, B and C. Measure the radius of the circle.
 Construct ∆ PBC such that P is on the same side of BC as point A and ∠ PCB = ½ ∠ ACB, ∠ BPC = ∠ BAC measure ∠ PBC.
 ABCD is a parallelogram with vertices A (1,1) and C(8,10). AB has the equation 4x 5y = 1 and BC has the equation 5x – 2y = 20. Determine by calculation;
 the coordinates of the point M where the diagonals meet
 The coordinates of the vertices B and D
 the length of AB correct to 4 significant figures
 The table shows corresponding values of x and y for a certain curve;
x 1.0 1.2 1.4 1.6 1.8 2.0 2.3 y 6.5 6.2 5.2 4.3 4.0 2.6 2.4
Using 3 strips and midordinate rule estimate the area between the curve, xaxis
Answers
 2x  3y + 6 =0
3y = 2x – 6
y = ^{2x}/_{3} + 2
When y = 0 x = 3
x = 0 y = 2
∴ Coordinate of y – intercept is (0,2)
Coordinate of x – intercept is (3,0)
∴ ∠CAO = tan^{1} ^{2}/_{3}= 33.69^{o}∴∠θ = 180 – 33.69^{o}= 146.31^{o}  ^{}Point y(^{4 + 2}/_{2}, ^{7 + 1}/_{2}) = (1, 3)
grad AB = ^{7 + 1}/_{4 + 2} =^{ 8}/_{6}grad xy =  ¾
grad xy =  ¾
^{y – 3}/_{x – 2} =  ¾
y =  ¾ x + ^{15}/_{4}  _{} Y = 3x – 1
M =3
M_{1}M_{2 }= 1
M_{2 }= ^{1}/_{3}^{y – 3}/_{x 2}= ^{1}/_{3}3y – 9 =x +2
^{3y}/_{3} = ^{x}/_{3} + ^{11/}_{3}Y = ^{x}/_{3 }+ ^{11}/_{3}  _{} Pt T is ^{1 + 5}/_{2}, ^{4 + 10}/_{2} = (2, 7)
grad. of grid xy =^{ 10  4}/_{5 1} = ^{14}/_{6} = ^{7}/_{3}∴ grad of L_{2 }= ^{3}/_{7}Take a general pt P(x,y) on L2
→^{y  7}/_{x  2} = ^{3}/_{7}→7y – 49 = 3x + 6
7y = 3x + 55
Or y = 3x + 55 (equation of L_{2}  a, b
(c) Name : a kite  Grad of line QP = ^{42}/_{13}= ^{2}/_{2}= 1
Grad of line QR = 1
Take a pt Q(1,4) and T(x,y) on line QR
^{y 4}/_{x – 1}= 1
y – 4 = x 1
y = x + 3 .....equ. of QR  y= x+3 …(i) Equ of QR
y = 3x 7 ...(ii) Equ. of Pr
Solving simultaneously ;:
x +3 = 3x 7
2x = 10
x = 5
Substituting ; y = 8
∴R is the pt (5,8) 
 Grad of line QP = ^{42}/_{13}= ^{2}/_{2}= 1
 Gradient OA = Gradient of CB
^{1 – 0}/_{2 – 0} =  ½
Gradient of CB
^{y – 3}/_{0 – 4}=  ½
2y – 6 = 4
2y = 10
y = 5 

4AN = AC And A is a common point hence A, N, C lie on a straight line.
 Gradient OA = Gradient of CB

 ∆ ABC line AB = 7 cm and BC = 8 cm ∡ ABC = 60^{o}
Construction of ∡60^{o}  ^{}AC = 7.6 + 0.1 and
∡ ACB = 53 + 1^{o}  2 sides bisector 1
Circle drawn radius 4.4. ± 0.1  Bisect ∡ ACB
Bisection line to cut the circle to identify P
∡BPC = ∡BAC = 67^{o}
∡ PBC = 88 ± 0.1^{o}
 ∆ ABC line AB = 7 cm and BC = 8 cm ∡ ABC = 60^{o}

M (^{1+8}/_{2}, ^{1+10}/_{2}) = M (4.5, 5.5)  AB: 4x – 5y = 1 x 2
BC: 5x – 2y = 20 x 5
8x – 10y = 2
25x – 10y = 100
17x = 102
x=^{102}/_{17 }= 6.0
24 – 5y = 1
5y = 25
Y = 5
∴B(6,5)
^{x+ 6.0}/_{2} = 4.5 x = 3
^{y + 5}/_{2} = 5.5 y = 6
∴D (3,6)  AB = √(16 – 1)^{2 }+ (51)^{2}√25 + 16
√41 = 6.40 (units)

 Mid ordinate
Area = 1.2 (6.2 + 4.3 + 2.6)
= 15.72
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