Question

Find the value of k so that the following equations may represent pairs of straight lines:
$kxy-8x+9y-12=0$

Answer 1

Hint: General equation of conic i.e. $a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$ will represent pairs of straight lines if
$\begin{align}
  & \left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|=0 \\
 & or \\
 & abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
\end{align}$

We have equation given;
$kxy-8x+9y-12=0$……………….(1)
Now, as we know that general conic equation is given as;
$a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$…………(2)
Above equation of conic will represent a pair of straight lines, circle, ellipse, parabola etc. with some condition among the coefficients of the given equation (2).
We know that conic will represent pair of straight lines if;
$\left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|=0$
On expanding the above determinant, we get;
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$…………………..(3)
Now, comparing equation given or equation (1) with equation (2), we get;
$\begin{align}
  & a=0 \\
 & b=0 \\
 & 2h=k\text{ or }h=\dfrac{k}{2} \\
\end{align}$
$\begin{align}
  & 2g=-8\text{ or }g=-4 \\
 & 2f=9\text{ or }f=\dfrac{9}{2} \\
\end{align}$
$And\text{ }c=-12$
Now, we have values of all the coefficients required for equation (3). Putting values in it, we get;
$\begin{align}
  & abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
 & \left( 0 \right)\left( 0 \right)\left( -12 \right)+2\left( \dfrac{9}{2} \right)\left( -4 \right)\left( \dfrac{k}{2} \right)-\left( 0 \right){{\left( \dfrac{9}{2} \right)}^{2}}-\left( 0 \right){{\left( -4 \right)}^{2}}-\left( -12 \right){{\left( \dfrac{k}{2} \right)}^{2}}=0 \\
\end{align}$
On simplifying the above relation, we get;
$\begin{align}
  & 0-18k-0-0+3{{k}^{2}}=0 \\
 & 3{{k}^{2}}-18k=0 \\
\end{align}$
Taking out ‘k’ common from the above equation, we get;
$\left( k \right)\left( 3k-18 \right)=0$
Therefore,
k = 0 and 3k = 18 or k = 6
Now, we have two values of k, i.e.
k = 0 and k = 6.
Putting k = 0, in equation one, we get equation $-8x+9y-12=0$, which represents one straight in coordinate plane.
Hence, k = 0 is not possible for the representation of a pair of straight lines.
Hence, k = 6 is the answer which gives pairs of straight line as;
$6xy-8x+9y-12=0$

Note: One can go wrong while comparing the given equation with general conic $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ to get the coefficients. So need to be very careful while substituting the values in $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$.
One need to verify k = 0 by putting in the given equation for checking the given equation is of the pairs of straight lines.