State whether the lines are parallel, perpendicular,or neither.
1 answer:
Answer:
Please check the explanation.
Step-by-step explanation:
- Two lines are parallel if their slopes are equal.
- Two lines are perpendicular if the product of their slope is -1
We also know that the slope-intercept form of the line equation is
where m is the slope and b is the y-intercept
Given the lines
1)
- y = 6х - 3
Comparing with y=mx+b, the slope of y = 6х - 3:
m₁=6
- y = - 1/6x + 7
Comparing with y=mx+b, the slope of y = - 1/6x + 7:
m₂=-1/6
As
m₁ × m₂ = -1
6 × - 1/6 = -1
-1 = -1
Thus, the lines y = 6х - 3 and y = - 1/6x + 7 are perpendicular.
2)
- y = 3x + 2
Comparing with y=mx+b, the slope of y = 3x + 2:
m₁=3
- 2y = 6x - 6
simplifying to write in slope-intercept form
y=3x-3
Comparing with y=mx+b, the slope of y=3x-3:
m₂=3
As the slopes of y = 3x + 2 and 2y = 6x - 6 are equal.
i.e. m₁ = m₂ → 3 = 3
Thus, the lines y = 3x + 2 and 2y = 6x - 6 are paralle.
3)
- 8x - 2y = 3
simplifying to write in slope-intercept form
y = 4x - 3/2
Comparing with y=mx+b, the slope of y = 4x - 3/2:
m₁=4
- x + 4y = - 1
simplifying to write in slope-intercept form
y=-1/4x-1/4
Comparing with y=mx+b, the slope of y=-1/4x-1/4:
m₂=-1/4
As
m₁ × m₂ = -1
4 × - 1/4 = -1
-1 = -1
Thus, the lines 8x - 2y = 3 and x + 4y = - 1 are perpendicular.
4)
- 3x+2y = 5
simplifying to write in slope-intercept form
y = -3/2x + 5/2
Comparing with y=mx+b, the slope of y = -3/2x + 5/2:
m₁=-3/2
- 3y + 2x = - 3
simplifying to write in slope-intercept form
y = -2/3x - 1
Comparing with y=mx+b, the slope of y = -2/3x - 1:
m₂=-2/3
As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.
5)
- y - 5 = 6x
simplifying to write in slope-intercept form
y=6x+5
Comparing with y=mx+b, the slope of y=6x+5:
m₁=6
- y - 6x = - 1
simplifying to write in slope-intercept form
y=6x-1
Comparing with y=mx+b, the slope of y=6x-1:
m₂=6
As the slopes of y - 5 = 6x and y - 6x = -1 are equal.
i.e. m₁ = m₂ → 6 = 6
Thus, the lines y - 5 = 6x and y - 6x = -1 are paralle.
6)
- y = 3х + 9
Comparing with y=mx+b, the slope of y = 3х + 9:
m₁=3
- y = -1/3x - 4
Comparing with y=mx+b, the slope of y = 1/3x - 4:
m₂=1/3
As m₁ and m₂ are neither equal nor their product is -1, hence the lines neither perpendicular nor parallel.
You might be interested in
The pair of equations that has (0, 8) as its solution is: A. equation A and equation C.
<h3>What is the Solution to a System of equations?</h3>The solution to a system of linear equations that are graphed is the pair of x and y coordinates of the points where two lines intersect.
In the graph given, the lines representing equations A and C intersect at point (0, 8). Therefore, the pair of equation that has a solution of (0, 8) is: A. equation A and equation C.
Learn more about solution of a system of equations on:
#SPJ1
Yay, easy peasy, nice refresher
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞