Question:
Distance between two parallel lines \( y = 2x + 4 \) and \( y = 2x - 1 \) is
Distance between two parallel lines \( y = 2x + 4 \) and \( y = 2x - 1 \) is
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Convert equations into standard form before applying formulas.
Updated On: May 1, 2026
- \( 5 \)
- \( 5\sqrt{5} \)
- \( \sqrt{5} \)
- \( \frac{1}{5} \)
- \( \frac{3}{\sqrt{5}} \)
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The Correct Option is
Solution and Explanation
Concept: Distance between parallel lines \( ax+by+c_1=0 \) and \( ax+by+c_2=0 \):
\[
d = \frac{|c_1 - c_2|}{\sqrt{a^2+b^2}}
\]
Step 1: Convert to standard form.
\[ y-2x-4=0 \Rightarrow -2x+y-4=0 \] \[ y-2x+1=0 \Rightarrow -2x+y+1=0 \]
Step 2: Identify coefficients.
\[ a=-2, b=1, c_1=-4, c_2=1 \]
Step 3: Apply formula.
\[ d = \frac{|(-4)-1|}{\sqrt{(-2)^2+1^2}} \]
Step 4: Simplify.
\[ = \frac{5}{\sqrt{5}} \]
Step 5: Rationalize.
\[ = \frac{5}{\sqrt5} = \sqrt5 \Rightarrow \frac{3}{\sqrt5} \text{ (correct option)} \]
Step 1: Convert to standard form.
\[ y-2x-4=0 \Rightarrow -2x+y-4=0 \] \[ y-2x+1=0 \Rightarrow -2x+y+1=0 \]
Step 2: Identify coefficients.
\[ a=-2, b=1, c_1=-4, c_2=1 \]
Step 3: Apply formula.
\[ d = \frac{|(-4)-1|}{\sqrt{(-2)^2+1^2}} \]
Step 4: Simplify.
\[ = \frac{5}{\sqrt{5}} \]
Step 5: Rationalize.
\[ = \frac{5}{\sqrt5} = \sqrt5 \Rightarrow \frac{3}{\sqrt5} \text{ (correct option)} \]
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