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 (paralle...Read More

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

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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

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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

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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)