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$\frac{3}{\sqrt5}$
Find the distance between 2x + y + 4 = 0 and 2x + y + 8 = 0:
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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