Find the distance between 2x + y + 4 = 0 and 2x + y + 8 = 0: - Vidyadip

MCQ

Find the distance between 2x + y + 4 = 0 and 2x + y + 8 = 0:

A
$\frac{4}{\sqrt5}$
B
$\frac{3}{\sqrt5}$
C
$\frac{9}{\sqrt5}$
D
$\frac{3}{\sqrt5}$

✓

Answer

$\frac{4}{\sqrt5}$

Solution:

Distance between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is $\mid\frac{\text{c}_{1}-\text{c}_{2}}{\sqrt{\text{a}^2+\text{b}^2}}\mid$

So, distance 2x + y + 4 = 0 and 2x + y + 8 = 0 is $\mid\frac{8-4}{\sqrt2^2+1^2}\mid=\frac{4}{\sqrt5}$

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