If the equation ax^2 + 2hxy + by^2 + 2gx + 2f | vedclass

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(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is found by solving $\frac{

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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 given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is found by solving $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$. These partial derivatives are $2ax + 2hy + 2g = 0$ and $2hx + 2by + 2f = 0$. Solving these,we get $x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$. The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2 = \left(\frac{hf-bg}{ab-h^2}\right)^2 + \left(\frac{gh-af}{ab-h^2}\right)^2$. Using the condition for a pair of lines $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$,we simplify the expression to $D^2 = \frac{c(a+b)-f^2-g^2}{ab-h^2}$.

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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