How do you find the midpoint of (-4,-3),(4,-8)?

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Answer 1 · 2016-03-03T13:57:06

#P_("mid") -> (0,-11/2)#

Let mid point be #P_("mid")#

Let first point be #P_1 ->(x_1,y_1)->(-4,-3)#

Let second point be #P_2->(x_2,y_2)->(4,-8)#

Then#" "P_("mid") = (P_2+P_1)/2#..... ( mean value)

#=> " "P_("mid")=((x_2+x_1)/2 , (y_2+y_1)/2 ) #

#=> " "P_("mid")= ((4+(-4))/2 ,((-8)+(-3))/2) = (0,-11/2)#

Tony B

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Something to think about")#

Using the mean value is more straightforward than using

#P_1+ (P_2-P_1)/2# which is the alternative.

Permit me to demonstrate just for the mid x value

Distance between is #" "P_2-P_1 ->(x_2-x_1) = (4-(-4))=+8#
Count from #P_1" to "x_("mid") = +8/2 = 4#

Actual #x# value is #P_1+4 = -4+4=0#