Question:
If the distance of a variable point $P$ from the fixed line
\[
2x-y+1=0
\]
is twice the distance of $P$ from another fixed line
\[
2x+y-2=0,
\]
then a point on the locus of $P$ is
If the distance of a variable point $P$ from the fixed line
\[
2x-y+1=0
\]
is twice the distance of $P$ from another fixed line
\[
2x+y-2=0,
\]
then a point on the locus of $P$ is
Show Hint
When options are given, substituting them directly is often faster than finding the full locus.
Updated On: Jun 3, 2026
- $\left(\dfrac14,\dfrac34\right)$
- $(2,3)$
- $(1,1)$
- $\left(\dfrac32,\dfrac14\right)$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Use the perpendicular distance formula from a point to a line.
Step 2: Meaning
The locus satisfies \[ \frac{|2x-y+1|}{\sqrt5} = 2\cdot \frac{|2x+y-2|}{\sqrt5}. \] Thus \[ |2x-y+1| = 2|2x+y-2|. \]
Step 3: Analysis
Test the given options. For $(1,1)$, \[ |2(1)-1+1|=2, \] and \[ 2|2(1)+1-2| = 2|1| = 2. \] The condition is satisfied.
Step 4: Conclusion
Hence $(1,1)$ lies on the locus.
Final Answer: (C)
Use the perpendicular distance formula from a point to a line.
Step 2: Meaning
The locus satisfies \[ \frac{|2x-y+1|}{\sqrt5} = 2\cdot \frac{|2x+y-2|}{\sqrt5}. \] Thus \[ |2x-y+1| = 2|2x+y-2|. \]
Step 3: Analysis
Test the given options. For $(1,1)$, \[ |2(1)-1+1|=2, \] and \[ 2|2(1)+1-2| = 2|1| = 2. \] The condition is satisfied.
Step 4: Conclusion
Hence $(1,1)$ lies on the locus.
Final Answer: (C)
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