Tag: distances and midpoints

As $\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0 $ represent a pair of line $\displaystyle \therefore \begin{vmatrix}a &h  &g \h  &b  &f \g  &f  &c \end{vmatrix}=0$
or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$ Now say point ofintersection onY axis be $\displaystyle (0,y _{1} $ and  point of intersection of pair of line be obtained by solving the equations $\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$ $\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$ $\displaystyle \begin{matrix}\Rightarrow  \ \Rightarrow  \end{matrix} \left{\begin{matrix}hy _{1}+g=0 \by _{1}+f=0 \end{matrix}\right.>  ()$ and $\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$  On compairing the equation given in () we get $\displaystyle bg=fh $ and  $\displaystyle bg^{2}=fgh ....(2) $ Again $\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ meet at y-axis $\displaystyle \therefore x=0$ $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$ Now using (2) and (3) in equation (I) we have $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ $\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$ $\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4) $ Now adding (2) and (4) $\displaystyle 2fgh=ch^{2}+bg^{2}$

The equation $\displaystyle ax^{2}+2hxy+by^{2}=0$