The distance between the parallel lines 8x+6y+5=0 and 4x+3y-25=0 is

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The distance between the parallel lines 8x+6y+5=0 and 4x+3y-25=0 is

Options:

A .

\frac{7}{3}

B .

\frac{9}{2}

C .

\frac{11}{2}

D .

\frac{5}{4}

Answer: Option C
:
C

Distance between parallel lines ax + by + $c_{1}$ = 0 and ax + by + $c_{2}$ = 0 is $\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

Given lines are 8x+6y+5 = 0 and 4x + 3y - 25 = 0

$\Rightarrow$ 8x + 6y - 50 = 0

$∴$ Required distance = $\frac{5 + 50}{\sqrt{8^{2} + 6^{2}}} = \frac{55}{10} = \frac{11}{2}$

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