First slide

Pair of straight lines

Question

The distance between the parallel lines given by $(x + 7 y)^{2} + 4 \sqrt{2} (x + 7 y) - 42 = 0$ is

Moderate

A

4/5
B

$4 \sqrt{2}$
C

2
D

$10 \sqrt{2}$

Solution

The lines given by the equation are $(x + 7 y - 3 \sqrt{2}) (x + 7 y + 7 \sqrt{2}) = 0$
$\Rightarrow x + 7 y - 3 \sqrt{2} = 0$ and $x + 7 y + 7 \sqrt{2} = 0$
distance between these lines $= |\frac{7 \sqrt{2} - (- 3 \sqrt{2})}{\sqrt{1^{2} + 7^{2}}}|$
$= \frac{10 \sqrt{2}}{5 \sqrt{2}} = 2$ .

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