MCQ
Find the distance between $2x + y + 4 = 0$ and $2x + y + 8 = 0:$
  • $\frac{4}{\sqrt5}$
  • B
    $\frac{3}{\sqrt5}$
  • C
    $\frac{9}{\sqrt5}$
  • D
    $\frac{3}{\sqrt5}$

Answer

Correct option: A.
$\frac{4}{\sqrt5}$
Distance between parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ 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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