Victoria Uni MATH 132 Assignment 5: Linear Equations and Graphs

This assignment solution addresses problems related to linear equations, focusing on finding equations of lines, determining x and y-intercepts, and understanding the concepts of parallel and perpendicular lines. The questions involve finding the equation of a line given its relationship (parallel or perpendicular) to another line and a point it passes through. The solution also covers calculating the gradient of a line and using it to find the equation of a perpendicular line, as well as determining x and y intercepts for various linear equations. The assignment covers different scenarios involving linear equations, providing a comprehensive understanding of the concepts. The solution provides step-by-step explanations and calculations to arrive at the correct answers for each question, making it a useful resource for students studying linear equations. The assignment covers topics such as finding equations of lines, calculating intercepts, and understanding parallel and perpendicular lines, and it's designed to help students improve their understanding and problem-solving skills in linear algebra.

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

2
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
3x +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 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= −2+1
−1−1 =−1
−2= 1
2
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.

3
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)
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= 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

4
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 (𝑥 −1
2 )
y+1=-2x+1
y=- 2x
x-intercept =0, y-intercept=0
c) The gradient through (-1,-3) and (2,0)
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= 0+3
2+1 =3
3= 1
1M2 = −1 , M2 = −1
𝑦 − 4
𝑥−1 = -1
y-4= - 1(x-1)
y-4= -x +1
y= -x+5

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
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

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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)

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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
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 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= −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.

8
- 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
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 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= 0+2
1+1 =2
2= 1
The gradient of line 1= The gradient of line 2 = 1

9
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 − 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 (𝑥 − 1)
y+1= -2x+2
y = - 2x +1
y + 2x =1
𝑦
1 + 𝑥
1
2
=1
X-intercept =𝟏
𝟐, y-intercept=1

10
c) The gradient through (1,3) and (-2,0)
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
= Δ𝑦
Δ𝑥= 3−0
1+2 =3
3= 1
1M2 = −1 , M2 = −1
𝑦 − 4
𝑥−1 =- 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
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

11
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)
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

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solving for y=-3x+3 and y=-x+1
-3x+3=-x+1
-2x=-2 ,x=1
When x=1, y=-x +1=-1+1=0
point of meeting (1,𝟎)