If an equation hxy + gx + fy + c = 0 represen | vedclass

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(A) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.Comparing this with the given equation $hxy + gx + fy +

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$fg = ch$

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(A) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.Comparing this with the given equation $hxy + gx + fy + c = 0$,we get:$A = 0, B = 0, C = c, H = \frac{h}{2}, G = \frac{g}{2}, F = \frac{f}{2}$.For the equation to represent a pair of lines,the condition is $\Delta = \begin{vmatrix} A & H & G \ H & B & F \ G & F & C \end{vmatrix} = 0$.Substituting the values:$\begin{vmatrix} 0 & \frac{h}{2} & \frac{g}{2} \ \frac{h}{2} & 0 & \frac{f}{2} \ \frac{g}{2} & \frac{f}{2} & c \end{vmatrix} = 0$.Expanding along the first row:$0 - \frac{h}{2} \left( \frac{ch}{2} - \frac{gf}{4} ight) + \frac{g}{2} \left( \frac{hf}{4} - 0 ight) = 0$.$-\frac{ch^2}{4} + \frac{hgf}{8} + \frac{ghf}{8} = 0$.$-\frac{ch^2}{4} + \frac{hgf}{4} = 0$.Multiplying by $4$,we get $-ch^2 + hgf = 0$,which implies $hgf = ch^2$.Dividing by $h$ (assuming $h 
eq 0$),we get $gf = ch$.

If an equation $hxy + gx + fy + c = 0$ represents a pair of lines,then

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