A) \[{{(x{{y}_{1}}-y{{x}_{1}})}^{2}}={{d}^{2}}({{x}^{2}}+{{y}^{2}})\]
B) \[{{({{x}_{1}}{{y}_{1}}-xy)}^{2}}=({{x}^{2}}+{{y}^{2}})\]
C) \[{{(x{{y}_{1}}+y{{x}_{1}})}^{2}}=({{x}^{2}}-{{y}^{2}})\]
D) \[({{x}^{2}}-{{y}^{2}})=2({{x}_{1}}+{{y}_{1}})\]
Correct Answer: A
Solution :
If the equation of line is \[y=mx\] and the length of perpendicular drawn on it from the point \[({{x}_{1}},{{y}_{1}})\] is d, then \[\frac{{{y}_{1}}-m{{x}_{1}}}{\sqrt{1+{{m}^{2}}}}=\pm d\,\,\Rightarrow \,{{({{y}_{1}}-m{{x}_{1}})}^{2}}={{d}^{2}}(1+{{m}^{2}}).\] But \[m=\frac{y}{x},\] therefore on eliminating 'm' , the required equation is \[{{(x{{y}_{1}}-y{{x}_{1}})}^{2}}={{d}^{2}}({{x}^{2}}+{{y}^{2}})\].
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