Question:
If the equation
\[
\lambda x^2-5xy+6y^2+x-3y=0
\]
represents a pair of straight lines, then the point of intersection of these straight lines is
If the equation
\[
\lambda x^2-5xy+6y^2+x-3y=0
\]
represents a pair of straight lines, then the point of intersection of these straight lines is
Show Hint
For
\[
ax^2+2hxy+by^2+2gx+2fy+c=0,
\]
the intersection point of the pair of lines is obtained from
\[
ax+hy+g=0,
\qquad
hx+by+f=0.
\]
First find any unknown parameter using the determinant condition, then solve these two linear equations.
Updated On: Jul 29, 2026
- \((1,3)\)
- \((1,-3)\)
- \((3,-1)\)
- \((-3,-1)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
A second-degree equation representing a pair of straight lines can be written as
\[
(ax+by+c)(dx+ey+f)=0.
\]
The point of intersection of the two lines satisfies both linear factors simultaneously.
For
\[
ax^2+2hxy+by^2+2gx+2fy+c=0,
\]
the point of intersection \((x,y)\) is obtained from
\[
ax+hy+g=0,
\]
\[
hx+by+f=0.
\]
Step 1: Compare with the general equation. Given \[ \lambda x^2-5xy+6y^2+x-3y=0. \] Comparing with \[ ax^2+2hxy+by^2+2gx+2fy+c=0, \] we get \[ a=\lambda, \qquad 2h=-5 \Rightarrow h=-\frac52, \] \[ b=6, \qquad 2g=1 \Rightarrow g=\frac12, \] \[ 2f=-3 \Rightarrow f=-\frac32. \]
Step 2: Form the equations of the intersection point. Using \[ ax+hy+g=0, \] \[ \lambda x-\frac52y+\frac12=0. \] Also, \[ hx+by+f=0, \] \[ -\frac52x+6y-\frac32=0. \] Multiplying the second equation by \(2\), \[ -5x+12y-3=0. \] \[ 5x-12y+3=0. \]
Step 3: Determine \(\lambda\) using the condition for a pair of straight lines. For the equation to represent a pair of straight lines, \[ \begin{vmatrix} \lambda & -\frac52 & \frac12\\ -\frac52 & 6 & -\frac32\\ \frac12 & -\frac32 & 0 \end{vmatrix} =0. \] Evaluating, \[ \lambda\left(0-\frac94\right) +\frac52\left(0+\frac34\right) +\frac12\left(\frac{15}{4}-3\right)=0. \] \[ -\frac94\lambda+\frac{15}{8}+\frac38=0. \] \[ -\frac94\lambda+\frac94=0. \] \[ \lambda=1. \]
Step 4: Find the intersection point. Substituting \(\lambda=1\), \[ x-\frac52y+\frac12=0. \] Multiplying by \(2\), \[ 2x-5y+1=0. \] Together with \[ 5x-12y+3=0, \] solve the system. Multiplying the first equation by \(5\), \[ 10x-25y+5=0. \] Multiplying the second equation by \(2\), \[ 10x-24y+6=0. \] Subtracting, \[ y+1=0. \] \[ y=-1. \] Substituting into \[ 2x-5y+1=0, \] \[ 2x+5+1=0. \] \[ 2x=-6. \] \[ x=-3. \]
Step 5: Write the final answer. \[ \boxed{(-3,-1)} \]
Step 1: Compare with the general equation. Given \[ \lambda x^2-5xy+6y^2+x-3y=0. \] Comparing with \[ ax^2+2hxy+by^2+2gx+2fy+c=0, \] we get \[ a=\lambda, \qquad 2h=-5 \Rightarrow h=-\frac52, \] \[ b=6, \qquad 2g=1 \Rightarrow g=\frac12, \] \[ 2f=-3 \Rightarrow f=-\frac32. \]
Step 2: Form the equations of the intersection point. Using \[ ax+hy+g=0, \] \[ \lambda x-\frac52y+\frac12=0. \] Also, \[ hx+by+f=0, \] \[ -\frac52x+6y-\frac32=0. \] Multiplying the second equation by \(2\), \[ -5x+12y-3=0. \] \[ 5x-12y+3=0. \]
Step 3: Determine \(\lambda\) using the condition for a pair of straight lines. For the equation to represent a pair of straight lines, \[ \begin{vmatrix} \lambda & -\frac52 & \frac12\\ -\frac52 & 6 & -\frac32\\ \frac12 & -\frac32 & 0 \end{vmatrix} =0. \] Evaluating, \[ \lambda\left(0-\frac94\right) +\frac52\left(0+\frac34\right) +\frac12\left(\frac{15}{4}-3\right)=0. \] \[ -\frac94\lambda+\frac{15}{8}+\frac38=0. \] \[ -\frac94\lambda+\frac94=0. \] \[ \lambda=1. \]
Step 4: Find the intersection point. Substituting \(\lambda=1\), \[ x-\frac52y+\frac12=0. \] Multiplying by \(2\), \[ 2x-5y+1=0. \] Together with \[ 5x-12y+3=0, \] solve the system. Multiplying the first equation by \(5\), \[ 10x-25y+5=0. \] Multiplying the second equation by \(2\), \[ 10x-24y+6=0. \] Subtracting, \[ y+1=0. \] \[ y=-1. \] Substituting into \[ 2x-5y+1=0, \] \[ 2x+5+1=0. \] \[ 2x=-6. \] \[ x=-3. \]
Step 5: Write the final answer. \[ \boxed{(-3,-1)} \]
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