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

\frac{7}{2}

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\frac{7}{3}

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\frac{7}{5}

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\frac{7}{10}

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Solution

The correct option is C $\frac{7}{10}$
$G i v e n - t h e e q u a t i o n s o f t w o p a r a l l e l l i n e s a r e 4 x + 3 y = 11 ⟹ 4 x + 3 y - 11 = 0 \equiv a_{1} x + b_{1} y {+ c}_{1} = 0 . . . . . . . (i) a n d 8 x + 6 y = 15 ⟹ 8 x + 6 y - 15 = 0 \equiv a_{2} x + b_{2} y {+ c}_{2} = 0 . . . . . . . . (i i) . T o f i n d o u t - t h e p e r p e n d i c u l a r d i s t a n c e b e t w e e n t w o p a r a l l e l l i n e s = ? S o l u t i o n - T h e p e r p e n d i c u l a r d i s t a n c e d b e t w e e n t w o p a r a l l e l l i n e s = d i f f e r e n c e d b e t w e e n t h e i r d i s t a n c e s f r o m t h e o r i g i n . N o w t h e p e r p e n d i c u l a r d i s t a n c e = p o f a l i n e a x + b y + c = 0 f r o m t h e o r i g i n i s p = \frac{c}{\sqrt{a^{2} + b^{2}}} . F r o m (i) a_{1} = 4, b_{1} = 3 & c_{1} = - 11. ∴ p_{1} = \frac{c_{1}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2}}} = ∣ ∣ ∣ ∣ \frac{- 11}{\sqrt{4^{2} + 3^{2}}} ∣ ∣ ∣ ∣ u n i t s = \frac{11}{5} u n i t s . A g a i n, f r o m (i i), a_{2} = 8, b_{2} = 6 & c_{1} = - 15. ∴ p_{2} = \frac{c_{1}}{\sqrt{{a_{2}}^{2} + {b_{2}}^{2}}} = ∣ ∣ ∣ ∣ \frac{- 15}{\sqrt{8^{2} + 6^{2}}} ∣ ∣ ∣ ∣ u n i t s = \frac{15}{10} u n i t s . S o d = | p_{1} - p_{2} | = ∣ ∣ ∣ \frac{11}{5} - \frac{15}{10} ∣ ∣ ∣ u n i t s = \frac{7}{10} u n i t s . ∴ T h e d i s t a n c e b e t w e e n t h e g i v e n l i n e s = \frac{7}{10} u n i t s . A n s - O p t i o n D .$

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