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# Find the distance between the lines $3x-4y+2=0$ & $3x-4y+10=0$?

$\begin{array}{1 1}(a)\;\large\frac{8}{3}&(b)\;\large\frac{8}{5}\\(c)\;\large\frac{6}{5}&(d)\;\large\frac{4}{5}\end{array}$

Perpendicular distance between two parallel lines is equal to $\large\frac{\big| c_1-c\big|}{\sqrt{a^2+b^2}}$
Hence for these two lines $3x-4y+2=0$ & $3x-4y+10=0$
Perpendicular distance between these two lines is $\large\frac{\big|10-2\big|}{\sqrt{3^2+4^2}}$
$\Rightarrow \large\frac{8}{5}$