What is the distance (13/3 , 1/7),(1/3 , 8/7) ?

The distance (13/3 , 1/7),(1/3 , 8/7) is square root of 17

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distance (13/3 , 1/7),(1/3 , 8/7)

distance

(\frac{13}{3}, \frac{1}{7}), (\frac{1}{3}, \frac{8}{7})

17

Solution steps

The distance between

(x_{1}, y_{1}), (x_{2}, y_{2})

is:

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The distance between

(\frac{13}{3}, \frac{1}{7})

and

(\frac{1}{3}, \frac{8}{7})

= (\frac{1}{3} - \frac{13}{3})^{2} + (\frac{8}{7} - \frac{1}{7})^{2}

Simplify

= 17

y = \frac{1}{x ^{2} + 1}

f (x) = \frac{( x - 1 )}{x + 1}

x^{3} - 5 x

f (x) = \frac{( 3 x - 1 )}{2 x + 5}

y = 3 x^{6} - x^{8}

What is the distance (13/3 , 1/7),(1/3 , 8/7) ?
The distance (13/3 , 1/7),(1/3 , 8/7) is square root of 17