<s>[INST] Find the coordinate of third point of equilateral triangle. I have two points A and B whose coordinates are $(3,4)$ and $(-2,3)$ The third point is C. We need to calculate its coordinates.
I think there will be two possible answers, as the point C could be on the either side of line joining A and B.
Now I put AB = AC = BC.
We calculate AB by distance formula : $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
= $ \sqrt{5^2 + 1^2} = \sqrt{26}$
I plug this value into the distance of AC and BC and finally equate to get :
$5x + y = 6$
Now what can I do? There are two variables, I am getting equation of a line! How can I solve this?
 [/INST] Call the position of point $C$ by the coords $(a, b)$. The equations for $C$ are then 
$$
\sqrt{(a-3)^2 + (b - 4)^2} = \sqrt{26} \\
\sqrt{(a+2)^2 + (b - 3)^2} = \sqrt{26} 
$$
Squaring both, we get
$$
(a-3)^2 + (b - 4)^2 = 26 \\
(a+2)^2 + (b - 3)^2 = 26
$$
$$
a^2 - 6a + 9 + b^2 - 8b + 16= 26 \\
a^2 + 4a + 4 + b^2 - 6b + 9= 26
$$
Subtracting these two gives
$$
-10a + 5 - 2b + 7 = 0
$$
or 
$$
6 = 5a + b
$$
which is a line both points must lie on. Writing this as 
$$
b = 6 - 5a
$$
we can substitute in either equation. Let's got with the second:
$$
a^2 + 4a + 4 + b^2 - 6b + 9= 26
$$
becomes
$$
a^2 + 4a + 4 + (6-5a)^2 - 6(6-5a) + 9= 26
$$
which is a quadratic that can now be solved for the two possible values of $a$. 
(Once you do so, you use $b = 6 - 5a$ to find the corresponding $b$-values.)
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