What is the distance between the parallel lin | vedclass

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(D) The given equations of the lines are $y = 2x + 4$ and $6x = 3y + 5$.Rewriting them in the standard form $ax + by + c = 0$:$2x - y + 4 = 0$$6x - 3y

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$17\sqrt{5} / 15$

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(D) The given equations of the lines are $y = 2x + 4$ and $6x = 3y + 5$.Rewriting them in the standard form $ax + by + c = 0$:$2x - y + 4 = 0$$6x - 3y - 5 = 0 \Rightarrow 2x - y - \frac{5}{3} = 0$Since the lines are parallel,the distance $d$ between them is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Here,$a = 2$,$b = -1$,$c_1 = 4$,and $c_2 = -\frac{5}{3}$.$d = \frac{|4 - (-\frac{5}{3})|}{\sqrt{2^2 + (-1)^2}}$$d = \frac{|4 + \frac{5}{3}|}{\sqrt{4 + 1}}$$d = \frac{|\frac{12 + 5}{3}|}{\sqrt{5}}$$d = \frac{17}{3\sqrt{5}}$Rationalizing the denominator:$d = \frac{17\sqrt{5}}{3 	imes 5} = \frac{17\sqrt{5}}{15}$

What is the distance between the parallel lines $y = 2x + 4$ and $6x = 3y + 5$?

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