Algebra Homework: HW 2 (Sec 1.4) - Equations, Lines, and Slopes

This document presents the solutions to an Algebra homework assignment (HW 2, Section 1.4). It covers several key concepts including finding equations of lines, determining parallel and perpendicular lines, calculating slopes, and writing equations in slope-intercept form. The solutions demonstrate step-by-step calculations and explanations for problems involving points, gradients, and linear equations. Furthermore, the assignment includes problems related to real-world applications, such as calculating the cost of a computer and determining fixed and variable charges. The document provides a comprehensive guide to solving various types of linear equations and understanding the properties of lines, making it a valuable resource for students studying algebra.

HW 2 (Sec 1.4) 1
HW 2 (Sec 1.4)
Find the equation of the lines
1. Parallel to 6 x +5 y=−5 ; x intercept =−3
The line 6 x +5 y=−5 can be written as
5 y=−6 x−5
y=−6
5 x−1
Since the lines are parallel the gradient is equal, the x intercept is the point at which y=0
Hence we need the equation of the line passing through (−3,0 ) ∧have gradient− 6
5
The equation is given by
y=mx+c
0=−6
5 ∗−3+ c
c=−18
5
Hence the equation is y=−6
5 x−18
5
Also written as 5 y +6 x=−18
Choice B
2. Parallel to the line x=−3 , passin point ( 6,8 )
There are many lines that are parallel to the line x=−3 , but only the line x=6 will
contain the point (6,8)
The solution is
Choice A
3. Perpendicular to line −3 x+ 8 y=51 , passing point ( 7 ,−9 )
The line −3 x+8 y=51 can be writen as 8 y =3 x+51
Also, as y= 3
8 x+ 51
8
The gradient is 3
8 , the gradient of the perpendicular line is obtained as

HW 2 (Sec 1.4) 2
3
8∗m=−1
m=−1∗8
3
m=−8
3
Nowwe need the line passing through (7, -9) with a gradient of -8/3
Equation of a line is y=mx+c
−9=−8
3 ∗7+ c
c= 2 9
3
Hence y=−8
3 x+ 2 9
3 is the equation of the line
Also written as
−3 y−8 x=−29
Choice D
4. Parallel to line y=−3 x , passing through poin t (5,5)
Now the gradient is -3
The equation is
y=mx+c
5=−3∗5+ c
c=20
is
y=−3 x +20
Choice C
5. Perpendicular to line 5 x−6 y=5 , y intercept=−6
5 x−6 y=5 can be written as
−6 y=−5 x +5
y= 5
6 x−5
6
Gradient is obtained by m1∗m 2=−1

HW 2 (Sec 1.4) 3
5
6∗m 2=−1
m 2=−6
5
Passing through point (0 ,−6)
The equation is given by y=mx+c
−6=−6
5 ∗0+ c
c=−6
y=−6
5 x−6
5 y +6 x=−30
Or
−5 y−6 x=30
Choice C
6. Perpendicular to line y= 1
4 x +7 passingthroiugh ( 3 ,−4 )
m1∗m 2=−1
m 2=−4
y=mx+c
−4=−4∗3+ c
c=8
y=−4 x +8
Choice A
Find the slope and y- intercept of the lines
7. Line x=−4
Slope undefined, does not cut the y axis hence no y- intercept
Choice B
8. Line 2 x−7 y=14
Can be written as
−7 y=−2 x +14

HW 2 (Sec 1.4) 4
¿ y= 2
7 x −¿2
The slopeis 2
7 , y intercept is−¿2
Choice B
9. Line 3 x−4 y=12
Can be written as
−4 y=−3 x +12
y= 3
4 x−3
Slope 3
4 , y intercept =−3
Choice A
10. Line y=−12
Slope is 0, y intercept is -12
Choice C
11. Line y=−2
Slope is o, y intercept is -2
Choice C
12. Line x=−1
Slope undefined, y intercept does not exist
Choice D
Decide whether the lines are parallel, perpendicular or neither
13. Lines 12 x+ 4 y=16
Written as 4 y=−12 x+16
y=−3 x + 4
And line
9 x +3 y=13
Written as 3 y=−9 x +13
y=−3 x + 13
3
The lines have similar gradient hence are parallel
Choice A

HW 2 (Sec 1.4) 5
14. Line 3 x−6 y=9
Written as −6 y=−3 x +9
y= 1
2 x− 3
2
¿ line
18 x+ 9 y=7
Written as
9 y=−18 x +7
y=−2 x + 7
9
Now m1∗m2=−2∗1
2 =−1
The towlines are thus perpendicular to each other
Choice B
Find the general form of the equation of the lines
15. Slope −8
9 , passing through(2 ,2)
The line is given by y=mx+c
2=−8
9 ∗2+ c
c= 34
9
Hence the line is y=−8
9 x+ 34
9
9 y +8 x=34
Choice C
16. Slope 5
7 , passingthrough (0,2)
The line is given by y=mx+c
2= 5
7∗0+c
c=2

HW 2 (Sec 1.4) 6
y= 5
7 x+2
7 y−5 x=14
Choice B
Find an equation for the line in the indicated form
17. Passing through points (−6 , 10 ) ∧(−4 ,−4)
Gradient 10−−4
−6−−1 =−14
5
Now the equation is given by
y=mx+c
10=−14
5 ∗−6+ c
c=−34
5
Equation y=−14
5 x− 34
5
5 y +14 x=−34
Choice C
18. Passing through points ( −3 ,−7 ) ∧(3 , 0)
Gradient is −7−0
−3−3 =7
6
The equation is
y=mx+c
−7=−7
2 + c
c=−¿7/2
y= 7
6 x−7
2
6 y−7 x=−21
−6 y +7 x=21
Choice C

HW 2 (Sec 1.4) 7
Find the equation in slope intercept form
19. 8 x−7 y=3
−7 y=−8 x+3
y= 8
7 x−3
7
Choice D
20. 5 x+ 7 y=13
7 y=−5 x +13
y=−5
7 x+ 13
7
Choice D
Find the slope intercept form of the equation of the lines
21. Passing through point ( 2 , 0 ) ∧(0 , 7)
Gradient is 0−7
2−0 =−7
2
Equation of the line is given by y=mx+c
7=c
Now the line is given by
y=−7
2 x+7
Choice B
22. Cost of the computer $ 16,000
Linear equation
Passing through points ( 0 , 16000 ) ∧(5,0)
Gradient 160000−0
0−5 =−3200
Equation y=mx+c
0=−3200∗5+c
c=16000
Equation is given by y=−3200 x +16000
After 4 years

HW 2 (Sec 1.4) 8
y= (−3200∗4 ) +16000
y=$ 3200
Choice D
23. Points ( 10 , 50 )∧(30,86)
Gradient= 50−86
10−30 = 9
5
Equation y=mx+c
86= 9
5∗30+c
c=32
y= 9
5 x+32
28=9
5 x+ 32
28−32= 9
5 x
9
5 x=−4
x=−4∗5
9
x=−20
9
Choice B
24. Fixed charge $ 29
variable charge $ 0.11
Linear equation
C=29+0.11 x
C=0.11 x+29
When x=170
C= ( 0.11∗170 ) +29
C=$ 47.70
Choice B
25. Linear equation C=0.09 x+27

HW 2 (Sec 1.4) 9
When x=190
C= ( 0.09∗190 ) +27
C=$ 44.10
Choice C
Find the slope of the line and sketch it
26. 2 x−3 y=−8
Written as
−3 y=−2 x−8
y= 2
3 x + 8
3
Slopeis 2
3
Choice B