The distance between the parallel lines 3 x-4 y+7=0 and 3 x-4 y+5=0 is a/b . Value of a+b is

Q42 / 125 · BITSAT 2021 · Mathematics

The distance between the parallel lines $3 x-4 y+7=0$ and $3 x-4 y+5=0$ is $\frac{a}{b}$. Value of $a+b$ is

A
2
B
5
Correct answer
C
7
D
3

Solution

Given parallel lines are $3 x-4 y+7=0$ and $3 x-4 y+5=0$ Required distance $=\frac{|7-5|}{\sqrt{(3)^{2}+(-4)^{2}}}=\frac{2}{5}$ $\Rightarrow a=2, b=5$

Hence, the correct answer is option: 'C'.

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