SOLUTION: Two vertices of an equilateral triangle are (3,4)and (-2,3) the third vertex can be

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Question 938101: Two vertices of an equilateral triangle are (3,4)and (-2,3) the third vertex can be
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Two vertices of an equilateral triangle are (3,4)and (-2,3) the third vertex can be
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The distance between the given points is sqrt(26)
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Draw circles or radius sqrt(26) with the given points as centers.
The 2 intersections of the circles are the 2 possible 3rd vertices.
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email via the TY note for help or to check your work.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Two vertices of an equilateral triangle are (3,4)and (-2,3) the third vertex can be
There are obviously two solutions:

 and 

First we find the distance between them to find out the
lengths of the three sides of the equilateral triangle.
d+=+sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29
d+=+sqrt%28%28-2-3%29%5E2%2B%283-4%29%5E2%29
d+=+sqrt%28%28-5%29%5E2%2B%28-1%29%5E2%29
d+=+sqrt%2825%2B1%29
d+=+sqrt%2826%29

Let the third vertex be the point (u,v)
 
Then the distance from (u,v) to (3,4) must
equal to the distance from (u,v) to (-2,3),
and each must equal to sqrt%2826%29.
sqrt%28%28u-3%29%5E2%2B%28v-4%29%5E2%29%22%22=%22%22sqrt%28%28u-%28-2%29%5E%22%22%29%5E2%2B%28v-3%29%5E2%29%22%22=%22%22sqrt%2826%29
Squaring all three sides:
%28u-3%29%5E2%2B%28v-4%29%5E2%22%22=%22%22%28u%2B2%29%5E2%2B%28v-3%29%5E2%22%22=%22%2226
%28u-3%29%5E2-%28u%2B2%29%5E2%22%22=%22%22%28v-3%29%5E2-%28v-4%29%5E2
%28%28u-3%29%5E%22%22-%28u%2B2%29%29%28%28u-3%29%5E%22%22%2B%28u%2B2%29%29%22%22=%22%22%28%28v-3%29%5E%22%22-%28v-4%29%29%28%28v-3%29%2B%28v-4%29%29   
%28u-3-u-2%29%28u-3%2Bu%2B2%29%22%22=%22%22%28v-3-v%2B4%29%28v-3%2Bv-4%29 
%28-5%29%282u-1%29%22%22=%22%22%281%29%282v-7%29
-10u%2B5%22%22=%22%222v-7
-10u-2v%22%22=%22%22-12
5u%2Bv%22%22=%22%226
v=6-5u
Substitute in
%28u%2B2%29%5E2%2B%28v-3%29%5E2%22%22=%22%2226
%28u%2B2%29%5E2%2B%286-5u-3%29%5E2%22%22=%22%2226
%28u%2B2%29%5E2%2B%283-5u%29%5E2%22%22=%22%2226
u%5E2%2B4u%2B4%2B9-30u%2B25u%5E2=26
26u%5E2-26u%2B13=26
26u%5E2-26u-13=0
2u%5E2-2u-1=0  
u+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+ 
u+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%282%29%28-1%29+%29%29%2F%282%2A2%29+
u+=+%282+%2B-+sqrt%284%2B8%29%29%2F4+
u+=+%282+%2B-+sqrt%2812%29%29%2F4+
u+=+%282+%2B-+sqrt%284%2A3%29%29%2F4+
u+=+%282+%2B-+2sqrt%283%29%29%2F4+
u+=+%282%281+%2B-+sqrt%283%29%29%29%2F4+
u+=+%281+%2B-+sqrt%283%29%29%2F2+
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Using the +,
u+=+%281+%2B+sqrt%283%29%29%2F2+
v=6-5u
v=6-5%28%281+%2B+sqrt%283%29%29%2F2%29
v=12%2F2-%285+%2B+5sqrt%283%29%29%2F2%29
v=%2812-%285+%2B+5sqrt%283%29%29%29%2F2%29
v=%2812-5+-+5sqrt%283%29%29%2F2
v=%287+-+5sqrt%283%29%29%2F2,

so one solution is , which is about (-0.4,7.8) 

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Using the -,
u+=+%281+-+sqrt%283%29%29%2F2+
v=6-5u
v=6-5%28%281+-+sqrt%283%29%29%2F2%29
v=12%2F2-%285+-+5sqrt%283%29%29%2F2%29
v=%2812-%285+-+5sqrt%283%29%29%29%2F2%29
v=%2812-5+%2B+5sqrt%283%29%29%2F2
v=%287+%2B+5sqrt%283%29%29%2F2,

so the other solution is , which is about (1.4,-0.8) 
 
Edwin