What is the distance (4/5 , 13/5),(8/5 , 11/5) ?

The distance (4/5 , 13/5),(8/5 , 11/5) is 2/(sqrt(5))

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distance (4/5 , 13/5),(8/5 , 11/5)

distance

(\frac{4}{5}, \frac{13}{5}), (\frac{8}{5}, \frac{11}{5})

\frac{2}{5}

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{4}{5}, \frac{13}{5})

and

(\frac{8}{5}, \frac{11}{5})

= (\frac{8}{5} - \frac{4}{5})^{2} + (\frac{11}{5} - \frac{13}{5})^{2}

Simplify

= \frac{2}{5}

2 x^{4} - 5

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

f (x) = 8 x - 10

f (x) = \frac{x ^{2} - 144}{x + 12}

y = - \frac{x}{6} - 2, (9, - 2)

What is the distance (4/5 , 13/5),(8/5 , 11/5) ?
The distance (4/5 , 13/5),(8/5 , 11/5) is 2/(sqrt(5))