Equation of the Lines Assignment: Questions and Answers

Pre-cal homework, multiple choice.

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