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
Correct Answer: A
Solution :
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., \[\frac{-c(a+b)+{{f}^{2}}+{{g}^{2}}}{{{h}^{2}}-ab}\].
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