If the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represented two lines $y={{m}_{1}}x \\$
$y={{m}_{2}}x \\$ the
A. ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .
B. ${{m}_{1}}+{{m}_{2}}=\dfrac{2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .
C. ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{-a}{b}$ .
D. ${{m}_{1}}+{{m}_{2}}=\dfrac{2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{-a}{b}$ .

Hint: The equation of a pair of straight lines can be determined from their slopes also. Here determining the equation of the pair of straight lines in terms of slopes and comparing it with the given equation that is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ we can determine the sum and product of the two slopes.



Formula Used:$y=mx+c$
$a{{x}^{2}}+2hxy+b{{y}^{2}}=0$



Complete step by step solution:The given equation of the straight line is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
Let the lines of the given pair of straight lines are given as-
$ y={{m}_{1}}x \\ $
 $ y={{m}_{2}}x \\ $
Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now the lines can be written as follows-

 $ y-{{m}_{1}}x=0 \\ $
  $ y-{{m}_{2}}x=0 \\ $

Thus the equation of the straight line in terms of slopes can be given as-

   $ (y-{{m}_{1}}x)(y-{{m}_{2}}x)=0 \\ $
$ {{y}^{2}}-{{m}_{1}}xy-{{m}_{2}}xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0 \\$
$ {{y}^{2}}+(-{{m}_{1}}-{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0 \\ $

Comparing this new equation of straight line with the given equation that is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ we get the result as follows-
$\dfrac{a}{{{m}_{1}}{{m}_{2}}}=\dfrac{2h}{-({{m}_{1}}+{{m}_{2}})}=\dfrac{b}{1}$
Now taking the product of the slopes equal to $b$ we get-

  $ \dfrac{a}{{{m}_{1}}{{m}_{2}}}=\dfrac{b}{1} \\$


${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$
Thus the product of the slopes is ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$.
Now taking the sum of the slopes is equal to b we get-
 $\dfrac{2h}{-({{m}_{1}}+{{m}_{2}})}=\dfrac{b}{1}$
${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$
Thus the value of the sum of the slopes is ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$.
Thus we can write that if the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represented two lines
 $ y={{m}_{1}}x \\ $
 $ y={{m}_{2}}x \\ $
the ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .



Option ‘A’ is correct



Note: The condition for two parallel line is that the slopes of the two lines must be equal. If two lines of equation $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$are parallel to each other then we can say that ${{m}_{1}}={{m}_{2}}$ .