Question:
If \(A,B\) are the feet of the perpendiculars drawn from \(P(3,1)\) on the pair of lines
\[
2x^2+3xy-2y^2+3x+y+1=0
\]
and \(Q\) is the point of intersection of the pair of lines, then the area of the quadrilateral \(PAQB\) is
If \(A,B\) are the feet of the perpendiculars drawn from \(P(3,1)\) on the pair of lines
\[
2x^2+3xy-2y^2+3x+y+1=0
\]
and \(Q\) is the point of intersection of the pair of lines, then the area of the quadrilateral \(PAQB\) is
Show Hint
Whenever a quadratic equation represents a pair of straight lines, factorize it first. If the two lines turn out to be perpendicular, the geometry becomes much simpler and area calculations often reduce to products of perpendicular distances.
Updated On: Jul 9, 2026
- \(25\)
- \(\dfrac{36}{5}\)
- \(\dfrac{25}{6}\)
- \(\dfrac{49}{8}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
Factor the given second-degree equation into two straight lines. The quadrilateral \(PAQB\) is the union of two right triangles \(PAQ\) and \(PBQ\). Its area is the sum of their areas.
Step 1: Factorize the given pair of lines. Given \[ 2x^2+3xy-2y^2+3x+y+1=0. \] Factorizing, \[ (2x-y+1)(x+2y+1)=0. \] Hence the two lines are \[ L_1: 2x-y+1=0, \] \[ L_2: x+2y+1=0. \]
Step 2: Find the point of intersection \(Q\). Solving \[ 2x-y+1=0 \] and \[ x+2y+1=0, \] from the first, \[ y=2x+1. \] Substituting into the second, \[ x+2(2x+1)+1=0. \] \[ 5x+3=0. \] \[ x=-\frac35. \] \[ y=-\frac15. \] Therefore, \[ Q\left(-\frac35,-\frac15\right). \]
Step 3: Find the perpendicular distance of \(P(3,1)\) from each line. Distance from \(L_1\): \[ d_1 = \frac{|2(3)-1+1|} {\sqrt{2^2+(-1)^2}} = \frac{6}{\sqrt5}. \] Distance from \(L_2\): \[ d_2 = \frac{|3+2(1)+1|} {\sqrt{1^2+2^2}} = \frac{6}{\sqrt5}. \] Thus, \[ PA=PB=\frac{6}{\sqrt5}. \]
Step 4: Find \(QA\) and \(QB\). Distance of \(Q\) from \(L_1\) measured along \(L_2\): Take a point on \(L_1\), \[ (-\tfrac12,0). \] Then \[ QA = \text{distance from }Q\text{ to }L_1\text{ along }L_2 = \frac{\text{distance}(Q,L_1)} {\sin\theta}, \] where \(\theta\) is the angle between the lines. For \[ m_1=2,\qquad m_2=-\frac12, \] \[ \tan\theta = \left| \frac{2+\frac12}{1-1} \right| =\infty. \] Hence \[ \theta=90^\circ. \] Therefore the lines are perpendicular. So \[ QA=\text{distance}(Q,L_1)=0, \] and the two triangles \(PAQ\) and \(PBQ\) are right triangles with legs along the two perpendicular lines.
Step 5: Use coordinates of \(P\) and \(Q\). \[ PQ = \sqrt{ \left(3+\frac35\right)^2 + \left(1+\frac15\right)^2 } = \sqrt{\left(\frac{18}{5}\right)^2+\left(\frac65\right)^2} = \frac{6\sqrt{10}}{5}. \] Since the lines are perpendicular, \[ \text{Area}(PAQB) = PA\cdot PB. \] \[ = \left(\frac6{\sqrt5}\right) \left(\frac6{\sqrt5}\right). \] \[ = \frac{36}{5}. \]
Step 6: Write the final answer. \[ \boxed{\frac{36}{5}} \]
Step 1: Factorize the given pair of lines. Given \[ 2x^2+3xy-2y^2+3x+y+1=0. \] Factorizing, \[ (2x-y+1)(x+2y+1)=0. \] Hence the two lines are \[ L_1: 2x-y+1=0, \] \[ L_2: x+2y+1=0. \]
Step 2: Find the point of intersection \(Q\). Solving \[ 2x-y+1=0 \] and \[ x+2y+1=0, \] from the first, \[ y=2x+1. \] Substituting into the second, \[ x+2(2x+1)+1=0. \] \[ 5x+3=0. \] \[ x=-\frac35. \] \[ y=-\frac15. \] Therefore, \[ Q\left(-\frac35,-\frac15\right). \]
Step 3: Find the perpendicular distance of \(P(3,1)\) from each line. Distance from \(L_1\): \[ d_1 = \frac{|2(3)-1+1|} {\sqrt{2^2+(-1)^2}} = \frac{6}{\sqrt5}. \] Distance from \(L_2\): \[ d_2 = \frac{|3+2(1)+1|} {\sqrt{1^2+2^2}} = \frac{6}{\sqrt5}. \] Thus, \[ PA=PB=\frac{6}{\sqrt5}. \]
Step 4: Find \(QA\) and \(QB\). Distance of \(Q\) from \(L_1\) measured along \(L_2\): Take a point on \(L_1\), \[ (-\tfrac12,0). \] Then \[ QA = \text{distance from }Q\text{ to }L_1\text{ along }L_2 = \frac{\text{distance}(Q,L_1)} {\sin\theta}, \] where \(\theta\) is the angle between the lines. For \[ m_1=2,\qquad m_2=-\frac12, \] \[ \tan\theta = \left| \frac{2+\frac12}{1-1} \right| =\infty. \] Hence \[ \theta=90^\circ. \] Therefore the lines are perpendicular. So \[ QA=\text{distance}(Q,L_1)=0, \] and the two triangles \(PAQ\) and \(PBQ\) are right triangles with legs along the two perpendicular lines.
Step 5: Use coordinates of \(P\) and \(Q\). \[ PQ = \sqrt{ \left(3+\frac35\right)^2 + \left(1+\frac15\right)^2 } = \sqrt{\left(\frac{18}{5}\right)^2+\left(\frac65\right)^2} = \frac{6\sqrt{10}}{5}. \] Since the lines are perpendicular, \[ \text{Area}(PAQB) = PA\cdot PB. \] \[ = \left(\frac6{\sqrt5}\right) \left(\frac6{\sqrt5}\right). \] \[ = \frac{36}{5}. \]
Step 6: Write the final answer. \[ \boxed{\frac{36}{5}} \]
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