The distance between the lines 3x + 4y = 9 an | vedclass

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(B) To find the distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$,we first rewrite them in the form $ax + by + c = 0$.
Dividing the second e

Vedclass · Accepted Answer

$\frac{3}{10}$

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(B) To find the distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$,we first rewrite them in the form $ax + by + c = 0$.
Dividing the second equation by $2$,we get $3x + 4y = 7.5$ or $3x + 4y - 7.5 = 0$.
The first equation is $3x + 4y - 9 = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} ight|$.
Here,$a = 3$,$b = 4$,$c_1 = -9$,and $c_2 = -7.5$.
$d = \left| \frac{-9 - (-7.5)}{\sqrt{3^2 + 4^2}} ight| = \left| \frac{-1.5}{\sqrt{9 + 16}} ight| = \left| \frac{-1.5}{5} ight| = \frac{1.5}{5} = \frac{3}{10} 	ext{ units}$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is

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