find the perpendicular distance between the parallel lines 3x + 5y - 7 = 0 and 3x + 5y +10 = 0.- choose any point on the first line - find perpendicular distance between that point and the other line

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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.

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