What is the midpoint (sqrt(32),3),(sqrt(2),-3) ?

The midpoint (sqrt(32),3),(sqrt(2),-3) is (5/(sqrt(2)),0)

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Find the Midpoint ( square root of 32,3) , ( square root of 2,-3)

midpoint

(32, 3), (2, - 3)

(\frac{5}{2}, 0)

Solution steps

Midpoint of

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

(\frac{x _{2} + x _{1}}{2}, \frac{y _{2} + y _{1}}{2})

(x_{1}, y_{1}) = (32, 3), (x_{2}, y_{2}) = (2, - 3)

= (\frac{2 + 32}{2}, \frac{- 3 + 3}{2})

Refine

= (\frac{5}{2}, 0)

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

y = 3 (x - 2)^{2} + 1

F (x) = 2 x - 1

g (x) (f (x)) f (x) = 3 - 2 x, g (x) = 3 x^{2} + 2

f (c) = 85 c

What is the midpoint (sqrt(32),3),(sqrt(2),-3) ?
The midpoint (sqrt(32),3),(sqrt(2),-3) is (5/(sqrt(2)),0)