Find the distances between the following pair of parallel lines:
$3 x + 4 y = 13, 3 x + 4 y = 3$

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Solution

Distance between parallel lines $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ is

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

Here, $c_{1} = 13, c_{2} = 3, a = 3, b = 4$

Distance between parallel lines is

$∣ ∣ ∣ ∣ \frac{13 - 3}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = 2.$

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