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Popular >

midpoint (12,2),(8,-4)

Solution

midpoint

(12, 2), (8, - 4)

Solution

(10, - 1)

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}) = (12, 2), (x_{2}, y_{2}) = (8, - 4)

= (\frac{8 + 12}{2}, \frac{- 4 + 2}{2})

Refine

= (10, - 1)

Popular Examples

y = cos (x)

critical (-x^3)/4+2x^2+8/3-4x critical points

\frac{- x ^{3}}{4} + 2 x^{2} + \frac{8}{3} - 4 x

monotone f(x)=x^2-5x monotone intervals

f (x) = x^{2} - 5 x

range of-sqrt(2x-3)+6 range

- 2 x - 3 + 6

domain of x^3+2x^2-3x+1 domain

x^{3} + 2 x^{2} - 3 x + 1

Frequently Asked Questions (FAQ)

What is the midpoint (12,2),(8,-4) ?
The midpoint (12,2),(8,-4) is (10,-1)