Question

Find the distance between the parallel lines $5x-3y-4=0$ and $10x-6y-9=0$.

Answer 1

multiply 1 st eq by 2,both side

10x- 6y - 8 = 0

SO DISTANCE = 1/ √100+64
= 1√164

Answer 2

Answer:

$\frac{1}{2\sqrt{34}}$

Step-by-step explanation:

Concept:

The distance between the parallel lines

$ax+by+c_1=0 \:and\: ax+by+c_2=0\: is\\\\d =\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Given lines are

5x -3y - 4 = 0

10x - 6y - 9 = 0

Divide by 2

5x -3y -( 9/2) = 0

Distance between the parallel lines

$d =\frac{|-4-(\frac{-9}{2})|}{\sqrt{5^2+(-3)^2}}\\\\d =\frac{|-4+(\frac{9}{2})|}{\sqrt{25+9}}\\\\d =\frac{|(\frac{1}{2})|}{\sqrt{34}}\\\\d =\frac{1}{2\sqrt{34}}$