The distance between the lines 4x + 3y = 11 a | vedclass

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(C) The given lines are $4x + 3y = 11$ and $8x + 6y = 15$. First,we rewrite the second equation to have the same coefficients for $x$ and $y$ as the f

Vedclass · Accepted Answer

$\frac{7}{10}$

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(C) The given lines are $4x + 3y = 11$ and $8x + 6y = 15$. First,we rewrite the second equation to have the same coefficients for $x$ and $y$ as the first equation by dividing by $2$: $4x + 3y = \frac{15}{2}$. 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,$a = 4$,$b = 3$,$c_1 = 11$,and $c_2 = \frac{15}{2}$. $D = \frac{|11 - \frac{15}{2}|}{\sqrt{4^2 + 3^2}} = \frac{|\frac{22 - 15}{2}|}{\sqrt{16 + 9}} = \frac{\frac{7}{2}}{\sqrt{25}} = \frac{\frac{7}{2}}{5} = \frac{7}{10}$.

The distance between the lines $4x + 3y = 11$ and $8x + 6y = 15$ is:

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