find the perpendicular distance between the parallel lines 3x + 5y - 7 = 0 and 3x + 5y +10 = 0. - find...

Given the parallel lines
:

L1: 3x + 5y - 7 = 0 ,
and

L2: 3x+ 5y + 10 = 0

We
need to find the perpendicular distance between both
lines.

First we will find any point on
L1

==> We will substitute with x = 0 in
L1.

==> 0 + 5y - 7 =
0

==>  5y =7 ==> y=
7/5

Then one point of line L1 is ( 0,
7/5)

Now we need to find the distance between the point (
0, 7/5) and the line L2 : 3x + 5y + 10 = 0

We know that the
distance between a point and line is given by :

D = l ax +
by + c l / sqr(a^2 + b^2)

==> D = l 0*3 + 5*7/5 + 10
l / sqrt(3^2 + 5^2)

==> D = l 7+10l /
sqr(34)

==> D =
17/sqr34

Then, the perpendicular distance
between L1 and L2 is 17/sqrt34 units.