Question

For what value of k does the equation \[{\text{12}}{{\text{x}}^2} + 2{\text{kxy + 2}}{{\text{y}}^2} + 11x - 5{\text{y + 2 = 0}}\] represent a pair of lines?

Answer 1

Hint: In order to solve to this problem, we must remember the condition when general equation of 2nd degree ${\text{a}}{{\text{x}}^2} + 2{\text{hxy + b}}{{\text{y}}^2} + 2gx + 2{\text{fy + c = 0}}$ represent a pair of lines.

Complete step-by-step answer:
We know that,
A general equation of 2nd degree ${\text{a}}{{\text{x}}^2} + 2{\text{hxy + b}}{{\text{y}}^2} + 2gx + 2{\text{fy + c = 0}}$ represent a pair of lines if ${\text{D = }}\left| {\begin{array}{*{20}{c}}
  {\text{a}}&{\text{h}}&{\text{g}} \\
  {\text{h}}&{\text{b}}&{\text{f}} \\
  {\text{g}}&{\text{f}}&{\text{c}}
\end{array}} \right| = 0$
We have,
\[{\text{12}}{{\text{x}}^2} + 2{\text{kxy + 2}}{{\text{y}}^2} + 11x - 5{\text{y + 2 = 0}}\]
So on comparing this with general equation of 2nd degree ${\text{a}}{{\text{x}}^2} + 2{\text{hxy + b}}{{\text{y}}^2} + 2gx + 2{\text{fy + c = 0}}$ we found

∴a=12, h=k, b=2, g=$\dfrac{{11}}{2}$, f = $\dfrac{{ - 5}}{2}$ and c=2
∴ ${\text{D = }}\left| {\begin{array}{*{20}{c}}
  {12}&{\text{k}}&{\dfrac{{11}}{2}} \\
  {\text{k}}&2&{\dfrac{{ - 5}}{2}} \\
  {\dfrac{{11}}{2}}&{\dfrac{{ - 5}}{2}}&2
\end{array}} \right| = 0$
Expanding we get

$ \Rightarrow 12\left( {4 - \dfrac{{25}}{4}} \right) - {\text{k}}\left( {2{\text{k + }}\dfrac{{55}}{4}} \right) + \dfrac{{11}}{2}\left( {\dfrac{{ - 5{\text{k}}}}{2} - 11} \right) = 0$
On further solving
$ \Rightarrow 12\left( { - \dfrac{9}{4}} \right) - 2{{\text{k}}^2} - \dfrac{{55{\text{k}}}}{4} - \dfrac{{55{\text{k}}}}{4} - \dfrac{{121}}{2} = 0$
$ \Rightarrow - 27 - 2{{\text{k}}^2} - {\text{k}}\left( {\dfrac{{55}}{4} + \dfrac{{55}}{4}} \right) - \dfrac{{121}}{2} = 0$
On simplifying
$ \Rightarrow - 27 - 2{{\text{k}}^2} - {\text{k}}\left( {\dfrac{{110}}{4}} \right) - \dfrac{{121}}{2} = 0$
multiply both sides by Minus 1
$ \Rightarrow 2{{\text{k}}^2}{\text{ + k}}\left( {\dfrac{{110}}{4}} \right) + \dfrac{{175}}{2} = 0$
\[ \Rightarrow {\text{8}}{{\text{k}}^2}{\text{ + 110k}} + 350 = 0\]
On taking 2 common
\[ \Rightarrow 4{{\text{k}}^2}{\text{ + 55k}} + 175 = 0\]

Here we have, a quadratic equation \[ \Rightarrow 4{{\text{k}}^2}{\text{ + 55k}} + 175 = 0\] in terms of k
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] , where a not equal to 0 , & a, b, c are real coefficients of the equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]
Being quadratic it has 2 roots.
X = $\dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}$ (1)

On comparing the given equation \[ \Rightarrow 4{{\text{k}}^2}{\text{ + 55k}} + 175 = 0\] with the general quadratic equation \[{\text{a}}{{\text{k}}^2}{\text{ + bk + c = 0}}\] we got values of coefficients a = 4, b = 55, c = 175
On putting the value of coefficients a, b, c in equation (1)
${\text{k = }}\dfrac{{\left( { - (55){\text{ + }}\sqrt {{{(55)}^2} - 4 \times (4) \times (175)} } \right)}}{{2 \times 4}}{\text{ & }}\dfrac{{\left( { - (55){\text{ - }}\sqrt {{{(55)}^2} - 4 \times (4) \times (175)} } \right)}}{{2 \times 4}}$
${\text{k = }}\dfrac{{\left( {{\text{ - 55 + }}\sqrt {225} } \right)}}{8}{\text{ & }}\dfrac{{\left( {{\text{ - 55 - }}\sqrt {225} } \right)}}{8}$
${\text{k = }}\dfrac{{\left( {{\text{ - 55 + 15}}} \right)}}{8}{\text{ & }}\dfrac{{\left( {{\text{ - 55 - }}15} \right)}}{8}$
${\text{k = - 5 & }}\dfrac{{\left( {{\text{ - 35}}} \right)}}{4}$
We know that if discriminant D $ \geqslant {\text{0}}$ then it will give real and distinct roots.
Here D= ${\text{ = }}\sqrt {{{(55)}^2} - 4 \times 8 \times (175)} {\text{ = }}\sqrt {225} {\text{ }} \geqslant {\text{0}}$ Therefore we got two distinct real roots ${{\text{k}}_1}{\text{ = }} - 5{\text{ & }}{{\text{k}}_2}{\text{ = }}\dfrac{{ - 35}}{4}$
${{\text{k}}_1}{\text{ = - 5 & }}{{\text{k}}_2}{\text{ = - 8}}{\text{.75}}$
Hence we got two values of k for a pair of straight lines.

Note: Whenever we came up with this type of problem where we are given general equation of 2nd degree then to find the unknown values just like k in above question, we can also use the condition $\vartriangle = {\text{ abc + 2fgh - a}}{{\text{f}}^2} - {\text{b}}{{\text{g}}^2} - {\text{c}}{{\text{h}}^2} = 0$. Hence we can get the required value of k.