Mathematics

This document discusses various concepts in mathematics, including equations of lines, intersection points, and Pappus's Hexagon Theorem. It also includes solved problems on topics like distance, speed, and fractions. The document provides step-by-step explanations and calculations for each problem.

Running head: MATHEMATICS 1
Mathematics
Name
Institution

MATHEMATICS 2
Question 1
Part a
A ( 3,14 ) , B ( 7,26 ) ,∧C(9,32)
AB= ( 7−3 , 26−14 ) = ( 4 , 12 ) =4(1,3)
BC= ( 9−7 , 32−26 ) = ( 2 , 6 ) =2(1,3)
Which implies that AB=2 BC. Since AB is a multiple of BC, the points lie on a straight.
Part b
Equation of AE
A ( 3,14 ) ∧E(5,0)
Gradient= 0−14
5−3 =−14
2 =−7
Let the line pass through point A and an arbitrary point with coordinates (x , y ).
Gradient= y−14
x −3 =−7
y−14=−7 ( x−3)
y=−7 x +21+14=−7 x +35
y=−7 x +35
Equation of AF
A ( 3,14 )∧F(14,0)

MATHEMATICS 3
Gradient= 0−14
14−3 =−14
11
Let the line pass through point A and an arbitrary point with coordinates (x , y ).
Gradient= y−14
x −3 =−14
11
y−14=−14
11 ( x−3 ) =−14
11 x + 42
11
y=−14
11 x+ 42
11 +14
y=−14
11 x+ 196
11
Equation of BD
B (7,26 )∧D(0,0)
Gradient= 0−26
0−7 =−26
−7 = 26
7
Let the line pass through point D and an arbitrary point with coordinates (x , y ).
Gradient= y−0
x−0 = 26
7
y= 26
7 x
Equation of BF
B ( 7,26 ) ∧F (14,0)
Gradient= 0−26
14−7 =−26
7

MATHEMATICS 4
Let the line pass through point B and an arbitrary point with coordinates (x , y ).
Gradient= y−26
x−7 =−26
7
y−26=−26
7 ( x−7)
y=−26
7 x+ 26+26=−26
7 x+ 52
y=−26
7 x+52
Equation of CD
C ( 9,32 ) ∧D(0,0)
Gradient= 0−32
0−9 =−32
−9 = 32
9
Let the line pass through point D and an arbitrary point with coordinates (x , y ).
Gradient= y−0
x−0 = 32
9
y= 32
9 x
Equation of CE
C (9,32)∧E(5,0)
Gradient= 0−32
5−9 =−32
−4 =8
Let the line pass through point E and an arbitrary point with coordinates (x , y ).

MATHEMATICS 5
Gradient= y−0
x −5 =8
y=8( x−5)
y=8 x −40
Part c
Coordinates of x
X is the point of intersection between lines AE and BD
AE: y =−7 x+ 35
y +7 x=35 … equation 1
BD : y= 26
7 x … equation 2
Substitute the value of y in equation 2 into equation 1 to get:
26
7 x +7 x=35
26
7 x +7 x=35
75
7 x=35
x=35 × 7
75 = 49
15
y= 26
7 x=26
7 × 49
15 = 182
15

MATHEMATICS 6
x= ( 49
15 , 182
15 )
Coordinates of y
y is the point of intersection between lines CD and AF
CD : y =8 x−40 … equation 1
AF : y=−14
11 x+ 196
11 … equation2
Substitute the value of y in equation 1 into equation 2 to get:
8 x−40=−14
11 x+ 196
11
8 x +14
11 x= 196
11 + 40
102
11 x= 636
11
x= 636
11 × 11
102 = 106
17
y=8 x −40=8 × 106
17 −40=168
17
y= ( 106
17 , 168
17 )
Coordinates of z
z is the point of intersection between lines BF and CE
BF : y=−26
7 x +52 … equation 1

MATHEMATICS 7
CE : y=8 x−40 … equation2
Substitute the value of y in equation 1 into equation 2 to get:
−26
7 x +52=8 x−40
8 x + 26
7 x=52+40
8 x + 26
7 x=52+40
82 x
7 =92
x=92 × 7
82 = 322
41
y=8 x −40=8 ( 322
41 )−40= 936
41
z= (322
41 , 936
41 )
Part d
x= ( 49
15 , 182
15 )
y= ( 106
17 , 168
17 )
z= (322
41 , 936
41 )

MATHEMATICS 8
xy= ( 106
17 − 49
15 , 168
17 −182
15 )=( 757
255 ,− 574
255 )
yz= ( 322
41 −106
17 , 936
41 −168
17 )=( 1128
697 , 9024
697 )
757
255 ÷ 1128
697 =1.8343
−574
255 ÷ 9024
697 =−0.1739
Since 1.8343 ≠−0.1739, the points x,y, and z don’t lie on a straight line.
Part e
A ( 3,14 ) , C ( 9,32 ) , D ( 0,0 ) , F (14,0)
Gradient AC =32−14
9−3 =3
y−14
x−3 =3
y−14=3 (x−3)
AC : y=3 x+ 5
Gradient DF = 0−0
14−0 =0
y−0
x−0 =0
DF : y=0

MATHEMATICS 9
Gradient xz =
936
41 − 182
15
322
41 − 49
15
= 506
217
y− 182
15
x− 49
15
= 506
217
y− 182
15 = 506
217 ( x− 49
15 )
xz : y= 506
217 x +140
31
Lines AC and DF meet when y=3 x +5=0
3 x+5=0
x=−5
3
Lines AC and DF meet at (−5
3 ,0 ). Substituting the point into the equation of line xz we get:
0 ≠ 506
217 (−5
3 )+ 140
31 ≠− 410
651
Since (−5
3 ,0 ) is not a solution to the line xz, the lines do not meet at the same point.
Part f

MATHEMATICS 10
“Pappus’s Hexagon Theorem states that given two sets of collinear points A, B, C and D, E, F
the intersection points X, Y, Z of the line pairs AE and BD, AF and CD, and CE and BF lie on
the Pappus line hence collinear (Bogomolny, 2018). To prove the theorem, we need to test
whether the line pairs A, B, C and D, E, F are collinear. However, since they the points D, E, F
are not collinear we can adjust their coordinates to make them lie on a straight line before
proving the theorem (Wolfram, 2007).
Question 2
Day Pages Left Pages Read
0 P(to solve) 0
1 p− ( 1
5 p+12 )= 4
5 p−12 1
5 p+12
2
( 4
5 p−12 )− ( 1
5 p+12 )= 3
5 p−24 1
4 ( 4
5 p−12 ) +15= 1
5 p+12
3
( 3
5 p−24 )− ( 1
5 p+10 )= 2
5 p−34 1
3 ( 3
5 p−24 )+18= 1
5 p+10
2
5 p−34=14
2
5 p=34 +14=48
p=48 × 5
2 =120 pages
Therefore, the book had 120 pages.
Question 3

MATHEMATICS 11
Let the total number of cakes be x. Each of them eats x
3 cakes. Similarly, let the total amount of
money be y. Each of them needs to take a third of the total amount. That is y
3 dollars. Therefore,
Sylvia should give them y
3 dollars each where y is the total amount of money she has.
Question 4
Let the distance between point A and B be x kilometers.
Time taken ¿ A ¿ B= Distance
Speed = x
60
Time taken ¿ B ¿ A= Distance
Speed = x
90
Total distance covered=x + x=2 x kilometers
Total time travelled= x
60 + x
90 = 90 x+ 60 x
5400 = 150 x
5400 = x
36 hours
Average speed= Total distance covered
Total timetravelled =2 x kilometers
x
36 hours
=2 x × 36
x =72km h−1
Question 5
xy= 1
3
y=3 x
2(x− y )2
2(x+ y)2 = 2(x−3 x)2
2( x+3 x)2 =2(−2 x)2
2(4 x)2 = 24 x2
216 x2

MATHEMATICS 12
¿ 24 ×2x2
216 ×2x2 = 24
216 =24 −16 =2−12= 1
4096
Question 6
To test whether it is a “selfie fraction”, we perform the following operations
2019
5384 = 201+9
538+4 =210
542 =0.38745
2019
5384 = 201−9
538−4 = 192
534 =0.35955
since 0.38745≠ 0.35955 , 2019
5384 is not a “selfie fraction”

MATHEMATICS 13
References
Bogomolny, A. (2018). Pappus' Theorem. Retrieved from
https://www.cut-the-knot.org/pythagoras/Pappus.shtml
Wolfram. (2007, December 1). Pappus's Hexagon Theorem -- from Wolfram MathWorld.
Retrieved from http://mathworld.wolfram.com/PappussHexagonTheorem.html