
If \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+k=0\] represents a pair of straight lines , then k is equal to-
A. -15
B. 6
C. -10
D. 4
Answer
164.1k+ views
Hint: The given equation has similarity with the equation of curve. Comparing the given equation with the general equation of curve we can easily determine the value of h. Here we have to use the condition for a straight line.
Formula used: $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
Complete step by step solution: The given equation is \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+k=0\]
This equation can be written in terms of the general equation of the curve. The general equation of the curve is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation of curve we can get the following values of the coefficients.
\[\text{a}=6,\text{b}=-10,\text{g}=\dfrac{1}{2},\text{f}=\dfrac{31}{2},h=\dfrac{11}{2}\]
The value of $c$ is given as $k$ .
The condition for straight line is given as follows-
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Putting the values of $a,b,c,g,f$ in the condition for straight line we get-
$
abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
6\times \left( -10 \right)\times k+2\times \dfrac{1}{2}\times \dfrac{31}{2}\times \dfrac{11}{2}-6\times {{\dfrac{31}{2}}^{2}}-\left( -10 \right)\times {{\left( \dfrac{1}{2} \right)}^{2}}-\left( k \right){{(\dfrac{11}{2})}^{2}}=0 \\
-60k+\dfrac{341}{4}-\dfrac{5766}{4}+\dfrac{10}{4}-\dfrac{121k}{4}=0 \\
\dfrac{-240k+341-5766+10-121k}{4}=0 \\
-361k-5415=0 \\
-361k=5415 \\
k=\dfrac{-5415}{361} \\
k=-15 \\
$
Thus we get the value of k by solving the above equation as $-15$ .
Thus we can write that the value of k for which the equation \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+k=0\] represents a pair of straight lines is $-15$ .
Thus, Option (A) is correct.
Note: The other method to solve the above problem is by forming a matrix with the coefficients of the equation. The matrix is given as $\left( \begin{matrix}
a & h & g \\
h & b & f \\
g & f & c \\
\end{matrix} \right)$ . The condition for the straight line is that the determinant of the matrix is 0.
Formula used: $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
Complete step by step solution: The given equation is \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+k=0\]
This equation can be written in terms of the general equation of the curve. The general equation of the curve is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation of curve we can get the following values of the coefficients.
\[\text{a}=6,\text{b}=-10,\text{g}=\dfrac{1}{2},\text{f}=\dfrac{31}{2},h=\dfrac{11}{2}\]
The value of $c$ is given as $k$ .
The condition for straight line is given as follows-
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Putting the values of $a,b,c,g,f$ in the condition for straight line we get-
$
abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
6\times \left( -10 \right)\times k+2\times \dfrac{1}{2}\times \dfrac{31}{2}\times \dfrac{11}{2}-6\times {{\dfrac{31}{2}}^{2}}-\left( -10 \right)\times {{\left( \dfrac{1}{2} \right)}^{2}}-\left( k \right){{(\dfrac{11}{2})}^{2}}=0 \\
-60k+\dfrac{341}{4}-\dfrac{5766}{4}+\dfrac{10}{4}-\dfrac{121k}{4}=0 \\
\dfrac{-240k+341-5766+10-121k}{4}=0 \\
-361k-5415=0 \\
-361k=5415 \\
k=\dfrac{-5415}{361} \\
k=-15 \\
$
Thus we get the value of k by solving the above equation as $-15$ .
Thus we can write that the value of k for which the equation \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+k=0\] represents a pair of straight lines is $-15$ .
Thus, Option (A) is correct.
Note: The other method to solve the above problem is by forming a matrix with the coefficients of the equation. The matrix is given as $\left( \begin{matrix}
a & h & g \\
h & b & f \\
g & f & c \\
\end{matrix} \right)$ . The condition for the straight line is that the determinant of the matrix is 0.
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