Question

If the equation $6x^2+2hxy+12y^2+22x+31y+20=0$ represents two straight lines then the value of h is

• A. $\frac{171}{20}$ or $\frac{17}{2}$
• B. $\frac{171}{20}$ or $\frac{17}{20}$
• C. ${}_{-}^{+}\frac{17}{20}$
• D. None of these

Answer 1

The correct option is A $\frac{171}{20}$ or $\frac{17}{2}$

The given equation is
$6x^2+2hxy+12y^2+22x+31y+20=0$
The general equation of the second degree is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
Equating the coefficients of (1) and (2) , we get a=6, h=h, b=12, g=11, f$\frac{31}{2}$, c=20
The equation (1) will represent two straight lines if
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
$\Rightarrow$ 6.12.20 + 2. $\frac{31}{2}.11h-6.\frac{{31}^2}{4}-12.{11}^2-20.h^2=0$
$\Rightarrow 1440+341h-\frac{2883}{2}-1452-20.h^2=0$
$\Rightarrow 40h^2-682h+2907=0$

$\Rightarrow h=\frac{{682}_{-}^{+}\sqrt{465124-465120}}{80}=\frac{{682}_{-}^{+}2}{80}=\frac{684}{80}$ or $\frac{680}{80}=\frac{171}{20}$ or $\frac{17}{2}$