Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Option 1: $\frac{11}{13}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{16}{9}$
Option 4: $\frac{9}{16}$
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Correct Answer: $\frac{13}{11}$
Solution :
Given:
$(x-\frac{1}{3})^2+(y-4)^2=0$
$(x-\frac{1}{3})^2=0$ , $(y-4)^2=0$
So, $x=\frac{1}{3}, y=4$
Now, $\frac{y+x}{y-x}=\frac{4+\frac{1}{3}}{4-\frac{1}{3}}=\frac{13}{11}$
Hence, the correct answer is $\frac{13}{11}$.
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