Question

find the distance between the following parallel lines
Question
3x-4y=12,3x-4y=7​

Answer 1

$\huge \mathtt{ \fbox{Solution :)}}$

Given ,

The equation of parallel lines are 3x - 4y = 12 and 3x - 4y = 7

Here A = 3 , B = - 4 , C1 = - 12 and C2 = - 7

We know that , the distance between parallel lines is given by

$\large \mathtt{ \fbox{d = \frac{ | c_{1} - c_{2} | }{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}$

Substitute the known values , we get

$\sf \mapsto d = \frac{ | - 12 - ( - 7)| }{ \sqrt{ {(3)}^{2} + { ( - 4)}^{2} } } \\ \\ \sf \mapsto d = \frac{ | - 5| }{ \sqrt{25} } \\ \\ \sf \mapsto d = \frac{ | - 5| }{5} \\ \\ \sf \mapsto d = 1$

Hence , the distance between parallel lines is 1 unit

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Answer 2

hope this helps you...

thanks