How do you find the midpoint of the line segment with endpoints #(sqrt50,1)# and #(sqrt2,- 1)#?

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Answer 1 · 2016-12-17T16:50:54

#(3sqrt2,0)#

If the end points are #(x_1, y_1) and (x_2, y_2)#

then the midpoint is = #[x_1 + x_2]/2, [y_1 + y_2}/2#

Here #(x_1,y_1) = (sqrt50, 1) and (x_2,y_2) = (sqrt2 , -1)#

So midpoint = #[sqrt50 + sqrt2]/2, [1 +(-1)]/2#

or, #[sqrt(5*5*2) +sqrt2]/2, [1 - 1]/2#

or, #[5sqrt2 + sqrt2]/2, 0/2#

or,#[sqrt2(5+1)]/2, 0#

or, #[6sqrt2]/2, 0#

or, #3sqrt2, 0#

Answer 2 · 2016-12-17T16:56:35

Mid point #P_("mean") ->(x_("mean"),y_("mean"))=(3sqrt(2),0)#

The mid point is the mean values

Let point 1 be #P_1=(x_1,y_1) = (sqrt(50),1)#
Let point 2 be #P_2=(x_2,y_2)=(sqrt(2),-1))#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider #x_1 = sqrt(50)#

Note that #2xx25=50# but #25=5^2# so we have:

#x_1=sqrt(2xx5^2) = 5sqrt(2)#

Thus #P_1->(x_1,y_1)=(5sqrt(2),1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the mean values")#

#x_("mean") ->barx = (5sqrt(2)+sqrt(2))/2= 3sqrt(2)#

#y_("mean")->bary = (1-1)/2=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mid point #P_("mean") ->(x_("mean"),y_("mean"))=(3sqrt(2),0)#