If the equation ax^2+2hxy+by^2+2gx+2fy+c=0 re | vedclass

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(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ is given by the solution of the partial derivati

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$\frac{c(a+b)-f^2-g^2}{ab-h^2}$

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(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ is given by the solution of the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$. $2ax+2hy+2g=0 \implies ax+hy+g=0$ $2hx+2by+2f=0 \implies hx+by+f=0$ Solving these equations using Cramer's rule: $x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$. The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2$. Using the condition for a pair of lines $abc+2fgh-af^2-bg^2-ch^2=0$,we can simplify the expression for $x_0^2+y_0^2$. Substituting the values,we get $D^2 = \frac{(hf-bg)^2+(gh-af)^2}{(ab-h^2)^2} = \frac{h^2f^2+b^2g^2-2hfbg+g^2h^2+a^2f^2-2ghaf}{(ab-h^2)^2}$. Using $abc+2fgh=af^2+bg^2+ch^2$,this simplifies to $\frac{c(a+b)-f^2-g^2}{ab-h^2}$.

If the equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines,then the square of the distance of their point of intersection from the origin is

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