WMAT123 Term 2 2019: Foundations of Mathematics Assignment 1

This document presents a complete solution set for the WMAT123 Assignment 1, a mathematics assignment from Macquarie University International College, Term 2, 2019. The assignment covers a range of mathematical concepts, including coordinate geometry, algebraic equations, and problem-solving. The solutions demonstrate step-by-step approaches to questions involving geometric proofs, linear equations, and quadratic equations. The document provides detailed explanations for each problem, including calculations for areas of shapes, solving for unknown variables, and applying mathematical formulas. The assignment contains 13 questions, each with a detailed solution. The document also includes the application of algebraic formulas to solve various mathematical problems. The solutions are presented in a clear and organized manner to aid in understanding and learning. The solutions are relevant for students taking the WMAT123 course at Macquarie University and are a valuable resource for exam preparation.

1) a)
Considering the rectangular coordinates,
The coordinates of B can be found out from A,
A (a, b), now if we move q units right and p units down we reach to B. So, the coordinates of B is
(a+q, b-p). Again we move q units down from B and p units left of B we reach C (c, d).
We can write c and d as, d = b – p – q and c = a + q – p
p + q = b – d
p – q = a – c
Solving this two equations we get,
p= 1
2 ( b +a ) −1
2 (d +c )
q= 1
2 ( b−a ) −1
2 (d −c)
Coordinates of B is,
( a+ q , b− p )=( ( 1
2 ( b+ a )− 1
2 ( d−c ) ), ( 1
2 ( b−a ) +1
2 ( d +c ) ) )
Coordinates of D is,
( a− p , b−q )= (( 1
2 ( a−b ) + 1
2 ( d +c ) ), ( 1
2 ( b+a )+ 1
2 ( d −c ) ) )
b)
The sides of the square ABCD is,
√ p2+ q2= √ ( 1
2 ( b+ a )− 1
2 ( d +c ) )2
+ ( 1
2 ( b−a )−1
2 ( d−c ) )2
¿ √ 1
4 (b+a)2+ 1
4 (d +c )2−1
2 (b+ a)(d +c )+ 1
4 ( b−a )2+ 1
4 ( d−c )2− 1
2 (b−a)(d−c)
¿ √ 1
2 ( b2 +a2 ) + 1
2 ( d2+ c2 ) − ( bd +ac )
√ 1
2 ( ( b2+a2 ) + ( d2 +c2 )−2 ( bd +ac ) )

√ 1
2 ((b−d)¿¿ 2+ ( a−c ) 2 )¿
Therefore area of the square is,
√ p2+ q22
=¿ ¿
¿ 1
2 ((b−d )¿ ¿2+ ( a−c )2 )¿
1
2 ( b−d)2 + 1
2 ( a−c ) 2
is the area.
2) a)
Equation of the line AB,
y−a2= b2−a2
b−a (x−a)
y−a2=(b+ a)(x −a)
y−a2= ( b+a ) x−ab−a2= ( b+ a ) x−ab
Equation of the line OC,
y= c2
c ( x )=¿ y =cx
Since OC and AB are parallel theirs slope will be equal,
( b+ a )=c
b)
Let the coordinates of DE is
D ( d , d2 ) ∧E(e , e2)
y−d2= e2−d2
e−d (x−d)
y−d2=(e+d )(x−d )
y−d2= ( e+ d ) x−ed−d2= ( e+ d ) x−ed
Midpoint of AB,
( a+b
2 , a2 +b2
2 )

Putting the value of the y coordinate in the equation line of AB
a2 +b2
2 −a2= ( b+ a ) ( x−a )
x= b+a
2
Similarly for the line OC,
Putting the value of the y coordinate in the equation line of OC,
x= c
2
As from the 1st part we have derived,
( b+ a )=c
So the mid points of the line AB and OC lie on the same straight line.
As DE is parallel to AB, the slopes will be equal,
So,
( b+ a )=c= ( e+ d )
Now midpoint of DE is,
( e+ d
2 , e2 +d2
2 )
Putting the value of the y coordinate in the equation line of DE
d2+e2
2 −d2 = ( d +e ) ( x−d )
x= d +e
2
As we have derived,
( b+ a )= ( d+e )=¿ b +a
2 = d +e
2
So the mid points of the line AB and DE lie on the same straight line.
Therefore, it is proved that midpoints of AB, OC and DE lie on the same straight line parallel to y axis.
c)
For any 3 parallel lines cutting through the parabola y=x2we can say that the midpoints of the
parallel lines lie on the same straight line parallel to the y axis and passing through a specific point in
the x axis.
It is true for any other parabola also, as it will be shifted by some value or it will be inclined at some
angle to the axis.

3)
Coordinate of D (0, 0)
Coordinate of C (0, -5)
Coordinate of B (-5, -5)
Coordinate of A (-5, 0)
Coordinate of E (7, 0)
Coordinate of F (7, 7)
Coordinate of G (0, 7)
Coordinate of T is the midpoint of CF, (7/2, 1)
Area of the triangle ATG,
Coordinate of A (-5, 0), T (7/2, 1), G (0, 7)
Area=1
2 |
−5 0 1
7
2 1 1
0 7 1
|=1
2 (−5 ( 1−7 ) + ( 7
2 × 7 )) =1
2 ( 30+ 49
2 )= 109
4
Area = 27.25 sq.cm
4)
Let the number be x
So she was supposed to this calculation, 3 x+ 6
But she did 3(x +6)=2019
( x +6)=673
x=667
Her answer would have been,
3 x+6=3 ( 667 )+ 6=20 01+6=2007
5)
x=25
1
t −2 =5
2
t −2 =5
t
t−2
2
t = y
2
t
6)
a
b = 2
3

a2
b2 = 4
9
3 a2 +2 b2
3 a2−2 b2 = ( 3 a¿¿ 2+2 b2)+(3 a2−2b2)
(3 a ¿¿ 2+ 2b2)−(3 a2 −2b2)= 6 a2
4 b2 = 6
4 × 4
9 = 2
3 ¿
¿
7)
Using the formula,
a2−b2=(a−b)(a+b)
Here,
a=( 20192019+ 2019−2019
2 )
b=( 20192019−2019−2019
2 )
a+ b=20192019
a−b=2019−2019
( a−b ) ( a+ b ) =20192019× 2019−2019=20192019−2019=20190 =1
8)
x≠ y ≡ 2 xy
x− y
a≠1≡ 2 a
a−1
a≠(a≠1)≡
2a × 2 a
a−1
a− 2 a
a−1
4 a2
a2−a−2 a =3
4 a2
a2−3 a =3

4 a2=3 ( a2−3 a )=3 a2−9 a
a2=−9 a=¿ a=0 ,−9
9)
Total student = 24
Class average = 90 % = 0.9
Therefore,
0.9= classtotal
24
class total=21.6
Boys average = 85% = 0.85
Therefore,
0. 85= boys total
1 4
boys total =11.9
girls total=class total−boys total=21.6−11.9=9.7
girl average= girls total × 100
1 0 = 9.7 ×100
10 =97 %
10)
Let number of days to exam is x
So total problem he would solve is 15x
In (x-3) days he solved 18 problems
So total problems he solved till that time is 18*(x-3)
Now in 3 days he will solve only 4 problems, so 18*(x-3) + 4
18 × ( x−3 )+4=15 × x
18 x−5 4 +4=15 x=¿ 3 x=50
x= 50
3
Total problem he would solve is ¿ 50
3 ×15=250 problems

11)
x2− y2=7 ( x− y )
x + y=7
x2+ y2=9 ( x+ y )
x2+ y2=63
12)
Let the sides are a, b and c the hypotenuse
Let a be the shorter side
a+ c=49=¿ c=49−a
a2+ b2=c2
a2+ b2= ( 49−a ) 2
b2= ( 49−a ) 2−a2
b2= ( 49−2 a ) ( 49 )
b2=2401−98 a
Now,
b2>0 as well as b >a
Satisfying both these conditions,
When a = 12, b = 35 satisfies the condition
When a = 20, b = 21 satisfies the condition
Therefore b can take two values 35 and 21, so the sum of the possible values is 56.
13)
M friends, Total cost is Y.
Therefore cost per friend is Y/M
If P friend does not attend then the number of friends become = M – P
Cost per friend becomes = Y/ (M – P)
Therefore the additional cost which they have to bear is
Y
M −P − Y
M
C= Y (M −M + P)
(M −P) M = YP
( M −P ) M

Thus the additional cost is,
YP
( M −P ) M