Question

find the distance between the parallel lines with the given equations. 2x-3y+4=0 and y=2/3x+5

Answer 1

you might have some formula for this right?

Answer 2

first equation is 2x - 3y = -4
and second one is 3y = 2x + 15 => 2x - 3y = -15
{ we just took those equation in a*x + b*y = constant.

Answer 3

now we will ensure that the coefficients of x and y in both equations are same { this is important }, if they are not equal, you would've to multiply either of equations with some number.
here they are already there, it's ok

Answer 4

then in this form the difference between the two constants -4 and -15 is the distance between them.
here -4 - ( -15) = -4 + 15 = 11 units.

Answer 5

y= (2/3)(2) + (4/3) =8/3
y= (2/3)(2) +5 = 19/3

19/3 - 8/3 =8/3

Answer 6

Because the two lines are parallel you need only find the y values of each of the two equations and then subtract the two.

Answer 7

We hope this helps.


Regards,

Academic GurusInc.

Twitter (@Academic_Gurus)
Facebook (AcademicGurusInc)
Youtube (Academic Gurus Inc)

Answer 8

thank you so much guys :)

Answer 9

now we got 11 and the coefficient of y is 3,so we need to divide by 3. thus the
answer 11/3.

Answer 10

the answer is 11sqrt13/13 because were working with the "normal"

Answer 11

@AcademicGurusInc your explanation is smooth +1, but you oversaw a little error, 19/3 - 8/3 is 11/3 not 8/3 :P

Answer 12

as for the second question i have two more lines 4x-y+1=0 and 4x-y-8=0

Answer 13

@SandeepReddy thank you for the correction.
@Sophie_dofi sorry for the oversight. I hope you were able to understand the concept behind the solution.

Answer 14

AS for the second question: The idea is very similar. First decide on the relationship between the two lines. Are they also parallel? (Hint: take a look at the slopes)

Answer 15

yea i understood it :) i should have mentioned I'm working with normal it was my bad

Answer 16

yea they are parellel

Answer 17

Ok no worries. So are you okay to solve the second question or do you need my help?

Answer 18

well im used to finding the normal with just one equation im a bit lost with two

Answer 19

Ok, so let's take it step by step. You found that the two lines are parallel because they both have the same slope (4). That means regardless which x value you choose, you can find the distance by solving for the respective y values. Try using the x value 3, for example. What y values do you get for the each of the two equations?

Answer 20

-y=-1-4x?
-y=8-4x sorry i dont have paper so thats probably wrong actually im pretty sure i didnt do that correctly

Answer 21

You're on the right track. You now need to multiply the equation through by -1 to remove the negative sign. So they the two equations should look like this:

4x-y+1=0 -----> y =4x + 1

4x-y-8=0 ------> y = 4x-8

Answer 22

now solve for y using x=3

Answer 23

y= 13
y=4

Answer 24

perfect. now subtract the two. What do you find

Answer 25

you mean subtract them from each other? 9

Answer 26

That is correct. The distance is 9

Answer 27

so 9 squt17/17?

Answer 28

so 9 squt17/17? Sorry, are you trying to find the distance of the two lines you mentioned? I'm not sure of where you got your result from

Answer 29

If you are trying to find the distance between the two lines, 9 is correct.

Answer 30

i used normal form

Answer 31

Unfortunately I will have to leave. If there the question was asking for something other than the distance please feel free to send us a message. We hope you found the solution helpful.

Regards,

Academic Gurus Inc.

Answer 32

thanks :)

Answer 33

Here is the idea:
Say we have the two parallel lines:
y=mx+b
y=mx+a
We need to find a perpendicular line to both. The slope of a perpendicular line will be -1/m.
So the easier perpendicular line we can choose is
y=-1/m x .
Now we just need to find the solutions to the two systems:
y=mx+b
y=-1/m x

and also

y=mx+a
y=-1/m x


So let's solve the first system by substitution:
y=mx+b
y=-1/m x
so we have -1/m x=mx+b
Solving for x....
Subtract mx on both sides
\[- \frac{1}{m}x- mx =b \\
-\frac{1}{m}x-\frac{m^2}{m}x=b\\
\frac{-1-m^2}{m}x=b \\
x=\frac{mb }{-1-m^2} \\
x=\frac{-mb}{1+m^2} \]
\[y=\frac{-1}{m} \cdot \frac{-mb}{1+m^2}=\frac{b}{1+m^2}\]

We can solve the other system in a similar way:
y=mx+a
y=-1/m x
\[x=\frac{-ma}{1+m^2}\]
\[y=\frac{a}{1+m^2}\]

Now find the distance between these two points:
\[d=\sqrt{(\frac{-mb}{1+m^2}-\frac{-ma}{1+m^2})^2+(\frac{b}{1+m^2}-\frac{a}{1+m^2})^2} \\ d=\sqrt{(\frac{-mb+ma}{1+m^2})^2+(\frac{b-a}{1+m^2})^2} \\ d=\sqrt{(\frac{m(-b+a)}{1+m^2})^2+(\frac{b-a}{1+m^2})^2} \\ d=\sqrt{ m^2 (\frac{(-1)(b-a)}{1+m^2})^2+(\frac{b-a}{1+m^2})^2} \\ d=\sqrt{m^2(-1)^2 (\frac{b-a}{1+m^2})^2+(\frac{b-a}{1+m^2})^2}\]
\[d=\sqrt{m^2(\frac{b-a}{1+m^2})^2+(\frac{b-a}{1+m^2})^2} \\ d=\sqrt{(\frac{b-a}{1+m^2})^2(m^2+1)} \\ d=\sqrt{(\frac{b-a}{1+m^2})^2 } \sqrt{m^2+1} \\ d=|\frac{b-a}{1+m^2}| \sqrt{m^2+1}\]
d is the shortest distance between the two parallel lines y=mx+a and y=mx+b