A) \[\frac{x}{{{x}_{1}}+{{x}_{2}}}+\frac{y}{{{y}_{1}}+{{y}_{2}}}=1\]
B) \[\frac{x}{{{x}_{1}}-{{x}_{2}}}+\frac{y}{{{y}_{1}}-{{y}_{2}}}=1\]
C) \[\frac{x}{{{y}_{1}}+{{y}_{2}}}+\frac{y}{{{x}_{1}}+{{x}_{2}}}=1\]
D) \[\frac{x}{{{y}_{1}}-{{y}_{2}}}+\frac{y}{{{x}_{1}}-{{x}_{2}}}=1\]
Correct Answer: A
Solution :
The mid-point of the chord is\[\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]. The equation of the chord in terms of its mid point is \[T={{S}_{1}}\]. \[\therefore x\left( \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)+y\left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right)\] \[=2\left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right)\left( \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] \[\Rightarrow x({{y}_{1}}+{{y}_{2}})+y({{x}_{1}}+{{x}_{2}})=({{x}_{1}}+{{x}_{2}})({{y}_{1}}+{{y}_{2}})\] \[\Rightarrow \] \[\frac{x}{{{x}_{1}}+{{x}_{2}}}=\frac{y}{{{y}_{1}}+{{y}_{2}}}=1\]
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