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

\frac{7}{10}

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\frac{7}{2}

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4

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{7}{10}$
We know that one of the points on the line $4 x + 3 y = 11 i s (0, \frac{11}{3})$
Therefore all we need to do is find the length of perpendicular from this point to the other line. Required distance = $\frac{∣ ∣ 0 \times 8 + 6 \times \frac{11}{3} - 15 ∣ ∣}{\sqrt{64 + 36}} = \frac{7}{10}$

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Similar questions

4 x + 3 y = 11

8 x + 6 y = 15

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 $8 x + 6 y = 15$

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

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