• # question_answer The square of distance between the point of intersection of the lines represented by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ and origin, is A)            $\frac{c(a+b)-{{f}^{2}}-{{g}^{2}}}{ab-{{h}^{2}}}$    B)            $\frac{c(a-b)+{{f}^{2}}+{{g}^{2}}}{\sqrt{ab-{{h}^{2}}}}$ C)            $\frac{c(a+b)-{{f}^{2}}-{{g}^{2}}}{ab+{{h}^{2}}}$   D)            None of these

Let the lines represented by given equation be            $y={{m}_{1}}x+{{c}_{1}}$and$y={{m}_{2}}x+{{c}_{2}}$.            Then $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$            $=b(y-{{m}_{1}}x-{{c}_{1}})(y-{{m}_{2}}x-{{c}_{2}})=0$            Comparing the coefficients of ${{x}^{2}},\ xy,\ x,\ y$ and constant term, we get  ${{m}_{1}}{{m}_{2}}=\frac{a}{b},\ {{m}_{1}}+{{m}_{2}}=\frac{-2h}{b},\ {{m}_{1}}{{c}_{2}}+{{m}_{2}}{{c}_{1}}=\frac{2g}{b},$            ${{c}_{1}}+{{c}_{2}}=-\frac{2f}{b}$and ${{c}_{1}}{{c}_{2}}=\frac{c}{b}$            Also the point of intersection of $y={{m}_{1}}x+{{c}_{1}}$and $y={{m}_{2}}x+{{c}_{2}}$ is $\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}},\frac{{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)$            Therefore, the square of distance of this point from origin is ${{\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)}^{2}}+\frac{{{({{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}})}^{2}}}{{{({{m}_{1}}-{{m}_{2}})}^{2}}}$            $=\frac{[{{({{c}_{1}}+{{c}_{2}})}^{2}}-4{{c}_{1}}{{c}_{2}}]+[{{({{m}_{1}}{{c}_{2}}+{{m}_{2}}{{c}_{1}})}^{2}}-4{{m}_{1}}{{m}_{2}}{{c}_{1}}{{c}_{2}}]}{{{({{m}_{1}}+{{m}_{2}})}^{2}}-4{{m}_{1}}{{m}_{2}}}$            Now putting the value defined above, we get the required distance i.e., $\frac{-c(a+b)+{{f}^{2}}+{{g}^{2}}}{{{h}^{2}}-ab}$.