The ratio in which the line $3 x + 4 y + 2 = 0$ divides the distance between $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 5 = 0$ .

7 : 3

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3 : 7

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2 : 3

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None of these

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Solution

The correct option is B $3 : 7$
Lines $3 x + 4 y + 2 = 0$ & $3 x + 4 y + 5 = 0$ are on the same side of the origin. The distance between these Iines is

$d_{1} = ∣ ∣ ∣ ∣ \frac{2 - 5}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = \frac{3}{5}$

Lines 3x + 4y + 2 = 0 and 3x + 4y - 5 = 0 are on the opposite sides of the origin. The distance between these lines is

$d_{2} = ∣ ∣ ∣ ∣ \frac{2 + 5}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = \frac{7}{5}$

Thus, $3 x + 4 y + 2 = 0$ divides the distance between $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 5 = 0$ in the ratio $d_{1} : d_{2}$ , i.e. $3 : 7$ .

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The ratio in which the line $3 x + 4 y + 2 = 0$ divides the distance between the lines $3 x + 4 y = 0$ and $3 x + 4 y - 5 = 0$ is

Find the ratio in which the line $3 x + 4 y + 2 = 0$ divides the distance between the lines $3 x + 4 y + 5 = 0$ and $3 x + 4 y - 5 = 0$ .

3 x + 4 y = 2

3 x + 4 y = 7

6 x + 8 y + 19 = 0

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