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

\frac{7}{2}

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\frac{9}{2}

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\frac{11}{2}

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\frac{5}{4}

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Solution

The correct option is C $\frac{11}{2}$
Given,

$8 x + 6 y + 5 = 0$

$\Rightarrow 4 x + 3 y = - \frac{5}{2}$

$4 x + 3 y - 25 = 0$

$\Rightarrow 4 x + 3 y = 25$

Distance between 2 parllel lines

$= \frac{c_{1} - c_{2}}{\sqrt{a^{2} + b^{2}}}$

$= \frac{25 - (\frac{- 5}{2})}{\sqrt{4^{2} + 3^{2}}}$

$= \frac{25 + \frac{5}{2}}{5}$

$= \frac{55}{2 \times 5} = \frac{11}{2}$

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Similar questions

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

The distance between the parallel lines 4x-3y+5=0 and 4x-3y+15 =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 $8 x + 6 y = 15$

4 x + 3 y = 11

8 x + 6 y = 15

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