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Id [(1/25, 0), (x,1/25)]=[(5, 0), (-a, 5...

Id `[(1/25, 0), (x,1/25)]=[(5, 0), (-a, 5)]^(-2)` , then the value of `x` is
a. `a/125`
b. `(2a)/125`
c. `(2a)/25`
d. none of these

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