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

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(C) To find the distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$, we first rewrite the equations with the same coeffici

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$0.3$

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(C) To find the distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$, we first rewrite the equations with the same coefficients for $x$ and $y$.Multiply the first equation $3x + 4y = 9$ by $2$ to get $6x + 8y = 18$.Now the lines are $6x + 8y - 18 = 0$ and $6x + 8y - 15 = 0$.The distance $d$ is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Substituting the values: $d = \frac{|-18 - (-15)|}{\sqrt{6^2 + 8^2}} = \frac{|-3|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3 	ext{ units}$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is (in $\text{ units}$)

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