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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.

$\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.

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