Find the midpoint between (a/2, sqrt{3a}/2) and (x/3, sqrt{3y}/3)
2007-11-22 08:35:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Just find the average of the x's and the average of the y's.
1/2 *(a/2 + x/3) = a/4 + x/6 = (3a + 2x)/12
1/2 * (√[3a]/2 + √[3y]/3) = √[3a]/4 + √[3y]/6 = √3(3√a+2√y)/12
So midpoint is:
((3a + 2x)/12 , √3(3√a+2√y)/12 )
2007-11-22 08:44:20 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x cordinate of midpoint = [(a/2) + (x/3)]/2
=>(3a + 2x)12
y - coordinate of midpoint = [sqrt(3a)/2 + sqrt(3y)/3]/2
=>[3sqrt(3a) + 2sqrt(3y)]12
2007-11-22 08:46:22 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
(a/2 + x/3)/2 , (sqrt3a /2 + sqrt3y /3)/2 =
(3a + 2x)/12 , (3sqrt{3a} +2sqrt{3y})/12
2007-11-22 08:42:56 · answer #3 · answered by Steve A 7 · 0⤊ 0⤋
[(a/2+x/3)/2, (sqrt{3a}/2+sqrt{3y}/3)/2]
2007-11-22 08:43:08 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋