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

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(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$.First,rewrite the equations in the same form by multiplying the first equation by $2$:$6x + 8

Vedclass · Accepted Answer

$3/10$

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(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$.First,rewrite the equations in the same form by multiplying the first equation by $2$:$6x + 8y = 18$ and $6x + 8y = 15$.These are parallel lines of the form $ax + by = c_1$ and $ax + by = c_2$.The distance $d$ between two parallel lines $ax + by = c_1$ and $ax + by = c_2$ is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here,$a = 6$,$b = 8$,$c_1 = 18$,and $c_2 = 15$.Substituting these values into the formula:$d = \frac{|18 - 15|}{\sqrt{6^2 + 8^2}} = \frac{3}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10}$.

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

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