The distance between the lines given by 3x + | vedclass

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(D) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$. To find the distance between parallel lines,we first make the coefficients of $x$ and $y$ id

Vedclass · Accepted Answer

$0.3$

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(D) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$. To find the distance between parallel lines,we first make the coefficients of $x$ and $y$ identical. Multiply the first equation by $2$: $2(3x + 4y) = 2(9) \Rightarrow 6x + 8y = 18$. The distance $d$ between two 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,$a = 6$,$b = 8$,$c_1 = -18$,and $c_2 = -15$. $d = \frac{|-18 - (-15)|}{\sqrt{6^2 + 8^2}} = \frac{|-3|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3$ units.

The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

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