Question:
If a particle moving along
\[
x-2y-3=0
\]
gets reflected in a perpendicular direction upon hitting the line
\[
3x-2y-5=0,
\]
then the line of movement of the particle after reflection is
If a particle moving along
\[
x-2y-3=0
\]
gets reflected in a perpendicular direction upon hitting the line
\[
3x-2y-5=0,
\]
then the line of movement of the particle after reflection is
Show Hint
If two lines are perpendicular, then the product of their slopes is \(-1\). Also, the reflected path must pass through the point where the particle hits the reflecting line.
Updated On: Jun 25, 2026
- \(2x+y+1=0\)
- \(2x+y-1=0\)
- \(2x+y-3=0\)
- \(2x+y+3=0\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Find the point where the particle hits the reflecting line.
The particle moves along \[ x-2y-3=0 \] It hits the line \[ 3x-2y-5=0 \] Solving both equations, \[ x-2y-3=0 \] and \[ 3x-2y-5=0 \] Subtracting the first equation from the second, \[ 2x-2=0 \] So, \[ x=1 \] Substituting \(x=1\) in \[ x-2y-3=0, \] we get \[ 1-2y-3=0 \] \[ -2y-2=0 \] Hence, \[ y=-1 \] Therefore, the point of reflection is \[ (1,-1) \]
Step 2: Find the slope of the original path.
The original path is \[ x-2y-3=0 \] Writing in slope form, \[ 2y=x-3 \] \[ y=\frac{x}{2}-\frac{3}{2} \] So, its slope is \[ m_1=\frac{1}{2} \]
Step 3: Find the slope of the reflected path.
The reflected direction is perpendicular to the original direction.
Hence, \[ m_1m_2=-1 \] So, \[ \frac{1}{2}m_2=-1 \] Therefore, \[ m_2=-2 \]
Step 4: Form the equation of the reflected line.
The reflected line passes through \[ (1,-1) \] and has slope \[ -2 \] Using point-slope form, \[ y-y_1=m(x-x_1) \] \[ y+1=-2(x-1) \] \[ y+1=-2x+2 \] \[ 2x+y-1=0 \]
Step 5: Final conclusion.
Hence, the line of movement after reflection is \[ \boxed{2x+y-1=0} \]
The particle moves along \[ x-2y-3=0 \] It hits the line \[ 3x-2y-5=0 \] Solving both equations, \[ x-2y-3=0 \] and \[ 3x-2y-5=0 \] Subtracting the first equation from the second, \[ 2x-2=0 \] So, \[ x=1 \] Substituting \(x=1\) in \[ x-2y-3=0, \] we get \[ 1-2y-3=0 \] \[ -2y-2=0 \] Hence, \[ y=-1 \] Therefore, the point of reflection is \[ (1,-1) \]
Step 2: Find the slope of the original path.
The original path is \[ x-2y-3=0 \] Writing in slope form, \[ 2y=x-3 \] \[ y=\frac{x}{2}-\frac{3}{2} \] So, its slope is \[ m_1=\frac{1}{2} \]
Step 3: Find the slope of the reflected path.
The reflected direction is perpendicular to the original direction.
Hence, \[ m_1m_2=-1 \] So, \[ \frac{1}{2}m_2=-1 \] Therefore, \[ m_2=-2 \]
Step 4: Form the equation of the reflected line.
The reflected line passes through \[ (1,-1) \] and has slope \[ -2 \] Using point-slope form, \[ y-y_1=m(x-x_1) \] \[ y+1=-2(x-1) \] \[ y+1=-2x+2 \] \[ 2x+y-1=0 \]
Step 5: Final conclusion.
Hence, the line of movement after reflection is \[ \boxed{2x+y-1=0} \]
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