Question

Distance between parallel lines 3x-4y=8 and 6x-8y+5=0​

Answer 1

Answer:

Let the two parallel lines be:

ax + by + C1 = 0

ax + by + C2 = 0

So the distance between two lines is |c1-c2|

According to the question, parallel lines are:

3x-4y+5=0

6x-8y-15=0

Now as mentioned above the two lines in question to be parallel then a and b of both have to be same.

So the lines will be now,

3x-4y+5=0

3x-4y+(-15/2)=0

So the distance is |c1-c2|

Hence |c1-c2| = |5-(-15/2)| = |5+(15/2)|

= |5+7.5| = | 12.5|

Hence the distance between parallel lines is 12.5 units.

I hope that you have got your answer and understood it

Answer 2

Answer:

The distance between the two given parallel lines is 11/10.

Explanation:

Distance between two parallel lines:

Let the two parallel lines be

   ax+by+c = 0 and ax+by+c₁ = 0

then the distance between two parallel lines is given by

            d = [|c-c₁|]/[√a²+b²]

Given two parallel lines are

          3x-4y = 8  --------(i)

and 6x-8y+5 = 0 --------(ii)

 The above two lines can be written as

        3x-4y-8 = 0 -------(iii)

     3x-4y-5/2 = 0 -------(iv)

   now apply distance formula for lines (iii) and (iv)

    where a = 3, b = -4, c = -8, c₁ = -5/2

       distance d = [|-8-(-5/2)|]/[√(3)²+(-4)²]

                          = [|-8+(5/2)|]/√(9+16)]

                          = [|(-16+5)/2]/√25

                          = [|-11|]/2/5

                          = 11/2×5

                       d = 11/10

 Hence, the distance between the two given parallel lines is 11/10.

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