Question

A general equation of second degree $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of straight lines if
1. ${{h}^{2}}=2ab$
2. ${{h}^{2}}>2ab$
3. $\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=0$
4. None of these

Answer 1

At first we take the general equation that is $S=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ which represents a straight line for that we have to consider two line equation ${{l}_{1}}x\text{ }+\text{ }{{m}_{1}}y\text{ }+~{{n}_{1~}}=~0\text{ }$and${{l}_{2}}x\text{ }+~{{m}_{2}}y\text{ }+~{{n}_{2~}}=~0$. Therefore, $({{l}_{1}}x\text{ }+\text{ }{{m}_{1}}y\text{ }+~{{n}_{1~}})({{l}_{2}}x\text{ }+~{{m}_{2}}y\text{ }+~n)=0$ and simplify further in this problem.

Complete step by step answer:
Suppose $S=0$ represents pair of straight line but there are two line of equation are:
${{l}_{1}}x\text{ }+\text{ }{{m}_{1}}y\text{ }+~{{n}_{1~}}=0---(1)$
${{l}_{2}}x\text{ }+~{{m}_{2}}y\text{ }+~{{n}_{2~}}=~0---(2)$
It represents a pair of straight line that means
$S=({{l}_{1}}x\text{ }+\text{ }{{m}_{1}}y\text{ }+~{{n}_{1~}})({{l}_{2}}x\text{ }+~{{m}_{2}}y\text{ }+~n_2)$
By simplifying this above equation we get:
$S={{l}_{1}}{{l}_{2}}{{x}^{2}}\text{+}{{l}_{1}}~{{m}_{2}}xy+{{l}_{1}}{{n}_{2}}x\text{+}{{l}_{2}}{{m}_{1}}xy+{{m}_{1}}{{m}_{2}}{{y}^{2}}\text{+}{{m}_{1}}{{n}_{2}}y\text{+}{{l}_{2}}{{n}_{1~}}x+{{m}_{2}}{{n}_{1~}}y\text{ }+~{{n}_{2}}{{n}_{1~}}$
By rearranging the term we get:
$S={{l}_{1}}{{l}_{2}}{{x}^{2}}+{{m}_{1}}{{m}_{2}}{{y}^{2}}+{{l}_{1}}~{{m}_{2}}xy+{{l}_{2}}{{m}_{1}}xy+{{l}_{1}}{{n}_{2}}x+{{l}_{2}}{{n}_{1~}}x+{{m}_{1}}{{n}_{2}}y+{{m}_{2}}{{n}_{1~}}y+~{{n}_{2}}{{n}_{1~}}$
By solving further we get:
$S={{l}_{1}}{{l}_{2}}{{x}^{2}}+{{m}_{1}}{{m}_{2}}{{y}^{2}}+({{l}_{1}}~{{m}_{2}}+{{l}_{2}}{{m}_{1}})xy+({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1~}})x+({{m}_{1}}{{n}_{2}}+{{m}_{2}}{{n}_{1~}})y+~{{n}_{2}}{{n}_{1~}}--(3)$
By comparing the general equation that is $S=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c$ with equation $(3)$
We get,
$a={{l}_{1}}{{l}_{2}},\,\,b={{l}_{1}}{{l}_{2}},\,\,2h={{l}_{1}}~{{m}_{2}}+{{l}_{2}}{{m}_{1}},\,\,c={{n}_{2}}{{n}_{1~}},\,\,2g={{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1~}},\,\,2f={{m}_{1}}{{n}_{2}}+{{m}_{2}}{{n}_{1~}}$
Now, $8fgh$ can be written as $(2f)(2g)(2h)$
$\therefore 8fgh=(2f)(2g)(2h)$
Substitute the value of h, g and f we get;

& 8fgh=({{m}_{1}}{{n}_{2}}+{{m}_{2}}{{n}_{1~}})({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1~}})({{l}_{1}}~{{m}_{2}}+{{l}_{2}}{{m}_{1}}) \\\ & \\\ \end{aligned}$$ After simplifying the bracket we get: $$8fgh={{l}_{1}}{{l}_{2}}({{m}_{1}}^{2}{{n}_{2}}^{2}+{{m}_{2}}^{2}{{n}_{1~}}^{2})+{{m}_{1}}{{m}_{2}}({{l}_{1}}^{2}{{n}_{2}}^{2}+{{l}_{2}}^{2}{{n}_{1~}}^{2})+{{n}_{1~}}{{n}_{2~}}({{l}_{1}}^{2}~{{m}_{2}}^{2}+{{l}_{2}}^{2}{{m}_{1}}^{2})+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}}$$ By arranging the term we get: $$8fgh={{l}_{1}}{{l}_{2}}\left[ ({{m}_{1}}^{2}{{n}_{2}}^{2}+{{m}_{2}}^{2}{{n}_{1~}}^{2}+2{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}})-2{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}} \right]+{{m}_{1}}{{m}_{2}}\left[ ({{l}_{1}}^{2}{{n}_{2}}^{2}+{{l}_{2}}^{2}{{n}_{1~}}^{2}+2{{l}_{1}}{{l}_{2}}{{n}_{1~}}{{n}_{2~}})-2{{l}_{1}}{{l}_{2}}{{n}_{1~}}{{n}_{2~}} \right]+{{n}_{1~}}{{n}_{2~}}\left[ ({{l}_{1}}^{2}~{{m}_{2}}^{2}+{{l}_{2}}^{2}{{m}_{1}}^{2}+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}})-2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}} \right]+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}}$$ As we know the value of $$a,\,\,b,\,\,c\,\,$$substitute in these equation we get: And apply the formula for $${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$ $$8fgh=a\left[ {{({{m}_{1}}{{n}_{2}}+{{m}_{2}}{{n}_{1~}})}^{2}}-2{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}} \right]+b\left[ {{({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1~}})}^{2}}-2{{l}_{1}}{{l}_{2}}{{n}_{1~}}{{n}_{2~}} \right]+c\left[ {{({{l}_{1}}~{{m}_{2}}+{{l}_{2}}{{m}_{1}})}^{2}}-2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}} \right]+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1~}}{{n}_{2~}}$$ By substituting the value of $$h,\,\,f,\,\,g\,\,$$in this above equation: $$8fgh=a(4{{f}^{2}}-2bc)+b(4{{g}^{2}}-2ca)+c(4{{h}^{2}}-2ab)+2abc$$ By simplifying this we get: $$8fgh=4(a{{f}^{2}}+b{{g}^{2}}+c{{h}^{2}}-abc)$$ Divide by 4 on both sides we get: $$2fgh=a{{f}^{2}}+b{{g}^{2}}+c{{h}^{2}}-abc$$ By rearranging the term we get: $$2fgh=a{{f}^{2}}+b{{g}^{2}}+c{{h}^{2}}-abc$$ or $$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$$ If we notice this above equation then it is a determinant of the matrix which is generally it is represented as $$\Delta $$ that is the above equation becomes $$\Delta =0$$ This is the required condition for $$S=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$$ represents a straight line $$\Delta =0$$ can also be written as: $$\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=0$$ Because $$\Delta =\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|=abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$$ **So, the correct answer is “Option 3”.** **Note:** In this particular problem we have to remember that $$S=0$$ only for the equation which represents a pair of straight lines. And its determinant will be 0. So, while simplifying the brackets, don't make silly mistakes. So the above solution is referred for such types of problems.