Question

Given points A (1,1) and B (5,5), find all possible coordinates of point C that will make triangle ABC an equilateral.

Answer 1

So far, I tried finding the distance between point A & B which I found out is radical 32... and then I don't know what to do after it

Answer 2

I tried doing something like \[\sqrt{(x-1)^{2}+(y-1)^{2}} = \sqrt{32}\]
\[\sqrt{(x-5)^{2}+(y-5)^{2}} = \sqrt{32}\]
and then \[\sqrt{(x-1)^{2}+(y-1)^{2}} =\sqrt{(x-5)^{2}+(y-5)^{2}} \]

Answer 3

good start
Now square both sides
Then expand and add like terms
solve for y in terms of x

Answer 4

If done correctly, I got y=-x+6...? Is this correct?

Answer 5

yes it is, however i think i was wrong
this will give you coordinates such that AC = BC but they won't equal sqrt(32) thus giving a isosceles triangle not an equilateral triangle
hold on i think have the correct solution
instead of setting them equal to each other you have to use substitution to solve system of 2 equations.

Answer 6

i need your help dumb cow to check joe math's work

Answer 7

What are the 2 equations?

Answer 8

Solve 1st equation for y:
\[(x-1)^{2} + (y-1)^{2} = 32\]
\[y = \pm \sqrt{32-(x-1)^{2}} +1\]
substitute in for y in 2nd equation:
\[(x-5)^{2}+(\sqrt{32-(x-1)^{2}}-4)^{2} = 32\]
simplify
\[\rightarrow \sqrt{32-(x-1)^{2}} = 5-x\]
\[\rightarrow x^{2} - 6x -3 =0\]
Use quadratic formula
\[x = 3\pm2\sqrt{3}\]

Answer 9

Sorry I got lost when you simplified \[\sqrt{32-(x-1)^{2}} \]
How did it become 5-x?