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

\frac{7}{2}

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4

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\frac{7}{10}

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None of these

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Solution

The correct option is C $\frac{7}{10}$
Given equation of lines can be written as
$4 x + 3 y - 11 = 0, 4 x + 3 y - \frac{15}{2} = 0$
Since, these lines are parallel .
So, $d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$
$\Rightarrow d = \frac{7 / 2}{5} = \frac{7}{10}$

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4 x + 3 y = 11

8 x + 6 y = 15

The distance between the parallel lines $8 x + 6 y + 5 = 0$ and $4 x + 3 y - 25 = 0$ is

Determine the distance between the following pair of parallel lines:

(i) $4 x - 3 y - 9 = 0$ and $4 x - 3 y - 24 = 0$

(ii) $8 x + 15 y - 34 = 0$ and $8 x + 15 y + 31 = 0$

(iii) $y = m x + c$ and $y = m x + d$

(iv) $4 x + 3 y - 11 = 0$ and $8 x + 6 y = 15$

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