MATH 132 Assignment 5: Introduction to Mathematical Thinking Problems

This document presents a complete solution for a mathematics assignment from Victoria University, focusing on linear equations. The assignment covers fundamental concepts such as finding equations of lines given various conditions (perpendicular or parallel to other lines, passing through specific points), determining x and y-intercepts, and manipulating equations into the slope-intercept form (y = ax + b). The solutions demonstrate the application of key principles including the relationship between gradients of perpendicular lines (M1M2 = -1), the concept of parallel lines having the same gradient, and the use of point-slope form to derive the equation of a line. The problems involve calculations of gradients from given points, and the ability to rearrange equations to determine intercepts. The assignment also includes tutorial exercises that reinforce these concepts through additional problem-solving opportunities.

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
1st May 2019
1

Assignment 5 Questions
Q1.a) For two perpendicular lines the product of their gradients equals to negative one.
i.e. M1M2=-1 where M1 and M2 are the gradients
of line 1and line 2 respectively.
- 1
3M2= -1 , M2= 3
y−0
x+ 2
3
=3
y= 3(x + 2
3 ¿
y= 3x+2
The perpendicular line to y=−1
3 x +1 and passing
(−2
3 ,0) has the equation y= 3x+2.
y-3x=2
Dividing through by 2 we have y
2 - 3
2x=1
y
2 +
x
−2
3
=1
X-intercept =−2
3 , y-intercept=2
b) Gradient of straight line passing through
(1,-1) and (-1,-2) is
change∈ y
change∈ x = Δ y
Δ x = −2+1
−1−1 =−1
−2= 1
2
For two perpendicular lines the product of their
2

gradients equals to negative one.
i.e. M1M2=-1 where M1 and M2 are the gradients
of line 1 and line 2 respectively.
1
2M2= -1 , M2= -2
y−1
x−0 =−2 ,
y-1=-2(x-0)
y-1=-2x
y=-2x +1
y
1 +
x
−1
2
=1
X-intercept =−1
2 , y-intercept=1
c) Two or more lines are parallel if they have the same gradient.
y=2x+2
The gradient of line 2= The gradient of line 1 = 2
y−1
x+ 1
2
=2
y-1=2(x + 1
2 )
y-1=2x+1
y=2x+2
y-2x=2
y
2 + x
−1=1
X-intercept =−1 , y-intercept=2
d) The gradient through (-1,0) and (1,-2)
3

change∈ y
change∈ x = Δ y
Δ x = 0+2
−1−1 = 2
−2= −1
The gradient of line 1= The gradient of line 2 = -1
y−1
x−0 =−1
y-1= -x
y= -x +1
y+x =1
y
1 + x
1 =1
X-intercept = 1, y-intercept=1
Q 2a) y=3x-1
3M2= -1 , M2= - 1
3
y−1
x−3 =¿- 1
3
3(y-1)= -1(x-3)
3y-3=-x+3
3y= -x+6
y
2 + x
6 =1
x-intercept =6, y-intercept=2
b)
y +1
x−1
2
=¿-2
y+1= -2¿ )
y+1=-2x+1
y=- 2x
x-intercept =0, y-intercept=0
4

c) The gradient through (-1,-3) and (2,0)
change∈ y
change∈ x = Δ y
Δ x = 0+3
2+1 = 3
3= 1
1 M2 =−1, M 2=−1
y−4
x−1 =¿-1
y-4= - 1(x-1)
y-4= -x +1
y= -x+5
y+x =5
y
5 + x
5 =1
x-intercept =5, y-intercept=5
Q3 a)
x- y
2 =1
2x-y=2
y=2x-2
b) 2x+3y=6
3y=-2x+6
y= - 2
3x +2
c) y +1
x−2 =¿3
5

y+1=3(x-2)
y+1=3x-6
y=3x-7
Q4 a) y=2x-1 and y=-x+2
2x-1=-x+2
3x=3, x=1
y=-x +2= - 1 +2=1
point of meeting (1,1)
b) y=x+2 and 2x+3y=1
2x+3(x+2)=1 , 2x+3x+6=1
5x=-5 , x= -1
Y=x+2=-1+2=1
point of meeting (1,1)
c) - 1
2m=-1, m=2
y+ 1
x +1 =¿2
y+1=2x+2
y=2x+1
Solving for y=2x+1 and y=-x+1 we have
2x+1=-x+1
3x=0 ,x=0 and y=-x+1=o+1=1
point of meeting (0,1)
6

Tutorial exercises
Q1.a) For two perpendicular lines the product of their gradients
Equals to negative one.
i.e. M1M2=-1 where M1 and M2 are the gradients
of line 1and line 2 respectively.
2
3M2= -1 , M2= - 3
2
y−0
x−2
3
=−3
2
2( y−0)= -3(x− 2
3 ¿
2y= -3x+2
The perpendicular line to y= 2
3x +1 and passing ( 2
3 ,0) has
the equation 2y= -3x+2
2y+3x=2
7

Dividing through by 2 we have y+ 3
2x=1
y
1 +
x
2
3
=1
X-intercept = 2
3 , y-intercept=1
b) Gradient of straight line passing through (1,-1) and (4,-2) is
change∈ y
change∈ x = Δ y
Δ x = −2+1
4−1 = −1
3
For two perpendicular lines the product of their gradients
equals to negative one.
i.e. M1M2=-1 where M1 and M2 are the gradients of line 1
and line 2 respectively.
- 1
3M2= -1 , M2= 3
y−1
x−0 =3
y-1=3x
y-1=3x
y=3x +1
y-3x=1
y
1 +
x
−1
3
=1
x-intercept = −1
3 , y-intercept=1
8

c) Two or more lines are parallel if they have the same gradient.
y= - 1
2x+2
The gradient of line 2= The gradient of line 1 = - 1
2
y−1
x−2 =−1
2
2(y-1)= -1(x-2)
2y-2=-x+2
2y+x=4
y
2 + x
4 =1
x-intercept =4 , y-intercept=2
d) The gradient through (1,0) and (-1,-2)
change∈ y
change∈ x = Δ y
Δ x = 0+2
1+1 = 2
2= 1
The gradient of line 1= The gradient of line 2 = 1
y−1
x−0 =1
y-1= x
y= x +1
y-x =1
y
1 + x
−1=1
x-intercept = −1, y-intercept=1
Q 2a) y= -3x-1
-3M2= -1 , M2= 1
3
9

y−2
x +3 = 1
3
3(y-2)= 1(x+3)
3y-6=x+3
3y= x+9
3y-x=9
1
3y+ x
−9 =1
y
3 + x
−9 =1
x-intercept = −9, y-intercept=3
b) y +1
x−1 =¿-2
y+1= -2¿ )
y+1= -2x+2
y = - 2x +1
y + 2x =1
y
1 +
x
1
2
=1
X-intercept = 1
2, y-intercept=1
c) The gradient through (1,3) and (-2,0)
change∈ y
change∈ x = Δ y
Δ x = 3−0
1+2 = 3
3= 1
1 M2 =−1, M 2=−1
y−4
x−1 =¿-1
y-4=- x+1
10

y= -x +5
y+x=5
y
5 + x
5 =1
X-intercept = 5, y-intercept=5
Q3 a) x
2 + y
2 =1
x+y=2
y= -x +2
b) 3x-2y=6
2y=3x-6
y= 3
2x – 3
c) y +1
x−2 =1
2
2(y+1)=1(x-2)
2y+2=x-2
2y=x-4
y = 1
2x -2
Q4 a) y=2x+2 and y= -2x+2
2x+2= -2x+2
4x=0, x= 0
y= -x +2= 0 +2= 2
point of meeting (0, 2)
11

b) y=3x-1 and x - y=-1
x -(3x-1)=-1 , x-3x+1=-1
-2x=-2 , x= 1
y=3x-1=3-1=2
point of meeting (1,2)
c) x+y=-1 , and
1
2x- 1
3 y=2, multiplying this equation by 6
we have 3x-2y=12
2y=3x-12
y = 3
2x-6
Solving x+y = -1 and y = 3
2x-6 we have
X+( 3
2x-6)=-1, X+ 3
2x-6)=-1
5
2x=5, x=2
When x=2, y= 3
2x-6= 3
2(2)-6= -3
point of meeting (2, −3)
d) 1
3m=-1, m=-3
y−2
x−1 =−¿3 , y-2=-3(x-1)
y-2=-3x+1), y=-3x+3
solving for y=-3x+3 and y=-x+1
-3x+3=-x+1
12

-2x=-2 ,x=1
When x=1, y=-x +1=-1+1=0
point of meeting (1,0)
13