I need help finding the midpoint of this problem please?

Question

(sqrt18, sqrt12) , (sqrt8, sqrt27)

I know how to find the midpoint, but the square roots in this problem are throwing me off. The answer is already given in the back of my book so you don't need to work it out for me..just please explain what I should do with the squareroots.

Answer 1

To avoid the confusion, just rename the points

(x1,y1) = (sqrt18, sqrt12)
(x2,y2) = (sqrt8, sqrt27)

mid point = [x2+x1]/2 , [y2+y1]/2

substitute your quadratic expressions and voila...the answer should be there waiting for you. Good luck!

Answer 2

So you want to know how to find

(sqrt18 + sqrt8)/2, (sqrt12+sqrt27)/2

sqrt18 = sqrt(9*2)= 3sqrt2
sqrt8 = sqrt(4*2)= 2sqrt2
sqrt12 = sqrt(4*3)= 2sqrt3
sqrt27 = sqrt(9*3)= 3sqrt3

sqrt18+sqrt8 = 5sqrt2
sqrt12+sqrt27=5sqrt3
midpoint = 5/2(sqrt2, sqrt3)

Answer 3

mid pt of (sqrt18, sqrt12) , (sqrt8, sqrt27)
is the average =
1/2 ( sqrt(18) +sqrt(8), sqrt(12) + sqrt(27) )
= (sqrt(4.5)+sqrt(2), 3/2(sqrt(4/3)+sqrt(3) )

Answer 4

{ [sqrt(18)+sqrt(8)]/2, [sqrt(12)+sqrt(27)]/2 }

sqrt(18) = sqrt(9*2) = 3*sqrt(2)
sqrt(8) = sqrt(4*2) = 2*sqrt(2)
sqrt(12) = sqrt(4*3) = 2*sqrt(3)
sqrt(27) = sqrt(9*3) = 3*sqrt(3)

And substitute:

{ [3*sqrt(2)+2*sqrt(2)]/2, [2*sqrt(3)+3*sqrt(3)]/2 }
{ 5*sqrt(2)/2, 5*sqrt(3)/2 }