The distance between the lines 5x + 3y - 7 = | vedclass

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(C) The given lines are $5x + 3y - 7 = 0$ $(i)$ and $15x + 9y + 14 = 0$ $(ii)$.To find the distance between parallel lines $ax + by + c_1 = 0$ and $ax

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$\frac{35}{3\sqrt{34}}$

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(C) The given lines are $5x + 3y - 7 = 0$ $(i)$ and $15x + 9y + 14 = 0$ $(ii)$.To find the distance between parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$,we first make the coefficients of $x$ and $y$ identical.Divide equation $(ii)$ by $3$: $5x + 3y + \frac{14}{3} = 0$.Now,the distance $d$ between the parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here,$a = 5$,$b = 3$,$c_1 = -7$,and $c_2 = \frac{14}{3}$.$d = \frac{|-7 - \frac{14}{3}|}{\sqrt{5^2 + 3^2}} = \frac{|-\frac{21}{3} - \frac{14}{3}|}{\sqrt{25 + 9}} = \frac{|-\frac{35}{3}|}{\sqrt{34}} = \frac{35}{3\sqrt{34}}$.

The distance between the lines $5x + 3y - 7 = 0$ and $15x + 9y + 14 = 0$ is

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