Question:
The lines
\[
L_1:2x+y+1=0
\]
and
\[
L_2:x-2y+4=0
\]
intersect at \(A\). Let \(P\) be a point at a distance \(5\) units from \(L_1=0\) and \(\alpha\) units from \(L_2=0\). If \(M\) and \(N\) are the feet of the perpendiculars from \(P\) on the lines \(L_1=0\) and \(L_2=0\) respectively, and the area of the quadrilateral \(AMPN\) is \(25\) sq. units, then the point \(P\) lies on the line
The lines
\[
L_1:2x+y+1=0
\]
and
\[
L_2:x-2y+4=0
\]
intersect at \(A\). Let \(P\) be a point at a distance \(5\) units from \(L_1=0\) and \(\alpha\) units from \(L_2=0\). If \(M\) and \(N\) are the feet of the perpendiculars from \(P\) on the lines \(L_1=0\) and \(L_2=0\) respectively, and the area of the quadrilateral \(AMPN\) is \(25\) sq. units, then the point \(P\) lies on the line
Show Hint
If a point is equidistant from two intersecting lines, it lies on one of their angle bisectors. Use
\[
\frac{L_1}{\sqrt{a_1^2+b_1^2}}
=
\pm
\frac{L_2}{\sqrt{a_2^2+b_2^2}}
\]
to obtain the angle bisectors directly.
Updated On: Jul 29, 2026
- \[ x+3y-3=0 \]
- \[ 3x-y+12=0 \]
- \[ x+3y=0 \]
- \[ 3x-y=0 \]
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
If \(M\) and \(N\) are the feet of perpendiculars from a point \(P\) to two intersecting lines, then the quadrilateral \(AMPN\) can be divided into two right triangles:
\[
\triangle AMP
\quad \text{and} \quad
\triangle ANP.
\]
Hence,
\[
\text{Area}(AMPN)
=
\frac12(AM)(PM)
+
\frac12(AN)(PN).
\]
Also, the perpendicular distances of \(P\) from the lines are
\[
PM=5,
\qquad
PN=\alpha.
\]
Step 1: Find the angle between the two lines. For \[ L_1:2x+y+1=0, \] slope \[ m_1=-2. \] For \[ L_2:x-2y+4=0, \] slope \[ m_2=\frac12. \] Since \[ m_1m_2=-1, \] the lines are perpendicular. Therefore, \[ \angle MAN=90^\circ. \]
Step 2: Express the area of quadrilateral \(AMPN\). Because the lines are perpendicular, \[ AM=PN=\alpha, \] and \[ AN=PM=5. \] Hence, \[ \text{Area}(AMPN) = \frac12(\alpha)(5) + \frac12(5)(\alpha). \] \[ = 5\alpha. \] Given area \(=25\), \[ 5\alpha=25. \] \[ \alpha=5. \]
Step 3: Interpret the result geometrically. Thus \(P\) is at equal distances from the two lines: \[ d(P,L_1)=5, \qquad d(P,L_2)=5. \] Hence \(P\) lies on an angle bisector of the lines \[ 2x+y+1=0 \] and \[ x-2y+4=0. \] Therefore, \[ \frac{2x+y+1}{\sqrt5} = \pm \frac{x-2y+4}{\sqrt5}. \] \[ 2x+y+1 = \pm(x-2y+4). \]
Step 4: Find the angle bisectors. Taking the positive sign, \[ 2x+y+1=x-2y+4. \] \[ x+3y-3=0. \] Taking the negative sign, \[ 2x+y+1=-x+2y-4. \] \[ 3x-y+5=0. \] Among the given options, only \[ x+3y-3=0 \] appears.
Step 5: Write the final answer. \[ \boxed{x+3y-3=0} \]
Step 1: Find the angle between the two lines. For \[ L_1:2x+y+1=0, \] slope \[ m_1=-2. \] For \[ L_2:x-2y+4=0, \] slope \[ m_2=\frac12. \] Since \[ m_1m_2=-1, \] the lines are perpendicular. Therefore, \[ \angle MAN=90^\circ. \]
Step 2: Express the area of quadrilateral \(AMPN\). Because the lines are perpendicular, \[ AM=PN=\alpha, \] and \[ AN=PM=5. \] Hence, \[ \text{Area}(AMPN) = \frac12(\alpha)(5) + \frac12(5)(\alpha). \] \[ = 5\alpha. \] Given area \(=25\), \[ 5\alpha=25. \] \[ \alpha=5. \]
Step 3: Interpret the result geometrically. Thus \(P\) is at equal distances from the two lines: \[ d(P,L_1)=5, \qquad d(P,L_2)=5. \] Hence \(P\) lies on an angle bisector of the lines \[ 2x+y+1=0 \] and \[ x-2y+4=0. \] Therefore, \[ \frac{2x+y+1}{\sqrt5} = \pm \frac{x-2y+4}{\sqrt5}. \] \[ 2x+y+1 = \pm(x-2y+4). \]
Step 4: Find the angle bisectors. Taking the positive sign, \[ 2x+y+1=x-2y+4. \] \[ x+3y-3=0. \] Taking the negative sign, \[ 2x+y+1=-x+2y-4. \] \[ 3x-y+5=0. \] Among the given options, only \[ x+3y-3=0 \] appears.
Step 5: Write the final answer. \[ \boxed{x+3y-3=0} \]
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