Question:
If
\[
ax^2-34xy-5y^2+2x+26y-5=0
\]
represents a pair of straight lines, then the value of \(a\) is:
If
\[
ax^2-34xy-5y^2+2x+26y-5=0
\]
represents a pair of straight lines, then the value of \(a\) is:
Show Hint
A second-degree equation represents a pair of straight lines if
\[
\begin{vmatrix}
A & H & G\\
H & B & F\\
G & F & C
\end{vmatrix}
=0.
\]
Always compare the equation with
\[
Ax^2+2Hxy+By^2+2Gx+2Fy+C=0.
\]
Updated On: Jun 26, 2026
- \(7\)
- \(5\)
- \(2\)
- \(13\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Compare with the general second-degree equation.
The given equation is \[ ax^2-34xy-5y^2+2x+26y-5=0 \] Comparing with \[ Ax^2+2Hxy+By^2+2Gx+2Fy+C=0, \] we obtain \[ A=a,\quad H=-17,\quad B=-5, \] \[ G=1,\quad F=13,\quad C=-5. \]
Step 2: Use the condition for a pair of straight lines.
For the equation to represent a pair of straight lines, \[ \begin{vmatrix} A & H & G\\ H & B & F\\ G & F & C \end{vmatrix} =0 \] Substituting the values, \[ \begin{vmatrix} a & -17 & 1\\ -17 & -5 & 13\\ 1 & 13 & -5 \end{vmatrix} =0 \]
Step 3: Expand the determinant.
\[ a \begin{vmatrix} -5 & 13\\ 13 & -5 \end{vmatrix} +17 \begin{vmatrix} -17 & 13\\ 1 & -5 \end{vmatrix} +1 \begin{vmatrix} -17 & -5\\ 1 & 13 \end{vmatrix} =0 \] \[ a(25-169) +17(85-13) +(-221+5) =0 \] \[ -144a+17(72)-216=0 \] \[ -144a+1224-216=0 \] \[ -144a+1008=0 \] \[ 144a=1008 \] \[ a=7 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{a=7} \] Hence, the correct option is \[ \boxed{(1)\ 7} \]
The given equation is \[ ax^2-34xy-5y^2+2x+26y-5=0 \] Comparing with \[ Ax^2+2Hxy+By^2+2Gx+2Fy+C=0, \] we obtain \[ A=a,\quad H=-17,\quad B=-5, \] \[ G=1,\quad F=13,\quad C=-5. \]
Step 2: Use the condition for a pair of straight lines.
For the equation to represent a pair of straight lines, \[ \begin{vmatrix} A & H & G\\ H & B & F\\ G & F & C \end{vmatrix} =0 \] Substituting the values, \[ \begin{vmatrix} a & -17 & 1\\ -17 & -5 & 13\\ 1 & 13 & -5 \end{vmatrix} =0 \]
Step 3: Expand the determinant.
\[ a \begin{vmatrix} -5 & 13\\ 13 & -5 \end{vmatrix} +17 \begin{vmatrix} -17 & 13\\ 1 & -5 \end{vmatrix} +1 \begin{vmatrix} -17 & -5\\ 1 & 13 \end{vmatrix} =0 \] \[ a(25-169) +17(85-13) +(-221+5) =0 \] \[ -144a+17(72)-216=0 \] \[ -144a+1224-216=0 \] \[ -144a+1008=0 \] \[ 144a=1008 \] \[ a=7 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{a=7} \] Hence, the correct option is \[ \boxed{(1)\ 7} \]
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