Questions
 A solid right pyramid has a rectangular base 10cm by 8cm and slanting edge 16cm.
calculate: The vertical height
 The total surface area
 The volume of the pyramid
 The line passing through the points A (1, 3K) and B (K, 3) is parallel to the line whose
equation is 2y + 3x = 9. Write down the coordinates of A and B  Find the value of a if the gradient of the graphs of the function y = x^{2}  x^{3} and y = x  ax are equal at x = ^{1}/_{3}
 Two perpendicular lines meet at the point (4,5). If one of the lines passes through the point (2,1), determine the equation of the second line in the form ax + by + c =0.
 Find the equation of the line passing through (5, 2) and with Xintercept as 3. Leave your answer in the form of Y = mX + C.

 copy and complete the table below:
x 0 1 2 3 4 5 6 y=2x4 y = 12 – 2x 
 On the grid provided and using the same axes, draw the lines y = 2x + 4 and y = 12 – 2x
 Hence use your graphs to solve the simultaneous equations
½ x – ¼ y = 1
x + ½ y = 6
 By use of substitution method, solve the simultaneous equations;
6x + 4y = 36
x + 3y = 13
 copy and complete the table below:
 Find the equation of a line through point 2, 4 which is parallel to 3y = 2x + 8. Express your answer in the form y = mx + c.
 Determine the equation of a line passing through (1, 3) and parallel to the line whose equation is 3x 5y = 10
 A straight line passing through point (3,4) is perpendicular to the line whose equation is 2y5x=11 and intersects the xaxis and yaxis at the points P and Q respectively. Find the coordinates of P and Q.
 A triangle ABC is formed by the points A(3, 4), B(7, 2) and C(1, 2)
 Find the coordinates of the midpoints K of AB and P of AC
 Find the equation of the perpendicular bisector of the KP
 The equation of line L_{1} is ^{3}/_{5}x + 3y = 6. Find the equation of a line L_{2} passing through point T (1, 2) and perpendicular to line L_{1}.
 Determine the equation of a line passing through (1, 3) and parallel to the line whose equation is 3x 5y = 10.
 A straight line through the points A (2, 1) and B (4, m) is perpendicular to the line, whose equation is 3y = 52x. Determine the value of m.
 Determine the equation of a line which is perpendicular to the line 2x + 3y + 4 = 0 and passes through P(1,1)
 Koech bought 144 pineapples at shs.100 for every six pineapples. She sold some of them at shs.72 for every three and the rest at shs.60 for every two. If she made a profit of 40%; Calculate the number of pineapples sold at 72 for every three.
 Solve the equation
x + 2  x – 1 = 5
3 2
Answers

 Length of diagonal = √(10^{2} + 8^{2})
= √164
Vertical height = √16^{2} – (√^{164}/_{2})^{2}
= 14.66cm  Height of the slant surfaces
√16^{2} – 4^{2} = √240
√16^{2} – 5^{2} = √231
Area of slant surfaces
( ½ x 8 x √240 x 2) = 124.0 cm2
(½ x 10x √231 x 2) = 152.0cm2
Area of the rectangular base= 8 x 10 = 80cm^{2}
Total surface area = 356cm^{2}  Volume
= ( ^{1}/_{3} x 80 x 14.66) = 391.0cm^{3}
 Length of diagonal = √(10^{2} + 8^{2})
 Gradient of line AB =
3  3k
K +1
Equation of other line can be written as
Y = ^{3x}/_{2} + ^{9}/_{2}
its gradient = ^{3}/_{2}
Hence
3  3k = ^{3}/_{2}
K +1
66K = 3k 3
3K = 9
K= 3
A(1, 9), B (3,3)  M_{1} = 2x – 3x^{2}
M_{2} = 1 – 2ax
M_{1} = M_{2} at x = ^{1}/_{3}
2x – 3x^{2} = 11 – 2ax
^{2}/_{3} – 3 (^{1}/_{3})^{2} = 1 – 2ax^{1}/_{3}
^{2}/_{3} – ^{1}/_{3} = 1 – ^{2}/_{3}a
^{3}/_{2} = ^{2}/_{3}a
^{9}/_{4} = a  M_{1} =
5 – 1 = ^{4}/_{6} = ^{2}/_{3}
4 2
M_{2} = ^{3}/_{2}
i.e. ^{3}/_{2} = ^{(y – 5)}/_{(x4)}
2 (y – 5) = 3 (x – 4)
2y – 10  3x + 12
3x + 2y – 22  Points (3, 0) and (5, 2)
M =  ¼
y – 0 =  ¼
x – 3
y =  ¼ x + ¾  Grad = ^{2}/_{3}
y – 4 = ^{2}/_{3}
x + 2
y = ^{2}/_{3}x + ^{16}/_{3}  3y – 5x = 4 or equivalence
5y = 3x – 10
y = ^{3}/_{5}x – 2
Gradient = ^{5}/_{3}
5 = y – 3
x + 1
3y – 9 = 5x – 5  2y – 5x = 11
Y= ^{5}/_{2} x + ^{11}/_{2}
g = ^{5}/_{2}^{5}/_{2}m = 1
M= ^{2}/_{5}Y – 4 = ^{2}/_{5}
X + 4
5y + 2x = 14
P(x,o)
5 X o + 2x = 14
X = 7
Q(o, y)
5y + 2 X o = 14
Y = 2.8
P (7,0)
Q (0, 2.8) 
 K ( ^{(37)}/_{2}, ^{(4+2)}/_{2}) (2, 3)
P (^{(3+1)}/_{2}, ^{(4–2)}/_{2}) = (2,1)  K_{1} =
3 – 1 =  ½
2 – 2
= 2
 K ( ^{(37)}/_{2}, ^{(4+2)}/_{2}) (2, 3)
 Gradient of L_{1} = ^{1}/_{5}
Gradient of L_{2} = 5
Y= mx + c
2 = 5 (1) t c
2 = 5tc
C = 7
Epuding L2
Y= 5x + 7  3y – 5x = 4 or equivalence
5y = 3x – 10
y = ^{3}/_{5} x – 2
Gradient = ^{5}/_{3}5 = y – 3
x + 1
3y – 9 = 5x – 5  Gradient = g = ^{(m–1)}/_{(42)}= ^{(m – 1)}/_{2}
3y = 5 – 2x
y = ^{5}/_{3} – ^{2x}/_{3} g_{1} = ^{2}/_{3}
g x g1= m – 1 – 2 = 1
2 3
2(m – 1) = 6
2m + 2 = 6
2m = 8
M = 4  L_{1}y = ^{2}/_{3}x – ^{4}/_{3}
M_{1} = ^{2}/_{3}
M_{2} = ^{3}/_{2}
L_{2}y = ^{3}/_{2}x + c x = 1, y = 1
1 = ^{3}/_{2} + c
c =  ½
L_{2}y = ^{3}/_{2}x  ^{1}/_{2}  BP = shs. ^{144}/_{6} x 100
SP = shs. ^{140}/_{100} x ^{144}/_{6} x 100
Let pineapples sold at shs. 72 for every shs. 3 be x
∴ At shs. 60 for every 2 will be 144 – x
^{x}/_{3} x 72 + ^{(144 – x)}/_{3} = 3360
24x + 30 (144 – x) = 3360
6x = 960
x = 60  ^{(x + 2)}/_{3} – ^{(x – 1)}/_{2} = ^{5}/_{1}
2(x + 2) – 3(x – 1) = 30
22x + 4 – 3x + 3 = 30
x + 7 = 30
x = 23
x = 23
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