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?
 A: 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.)
A: If you like vector approach:
Displace (shift/translate) point A by vector(-3,-4) to come to the origin.
Multiply the new radius vector  $ab$  with $ e^ {i \pi/3} , e^ {-i \pi/3}$ (once clockwise and once anticlockwise) to obtain new points $ C_1 $ and $ C_2 $. The multiplying factor is $ (1/2 \pm i \sqrt 3/2) $.
Displace these points back to original positions translating by $ (3,4) $. 
If you multiply thrice, all points of a hexagon would also be reached.
A: The solution using a quadratic equation is a useful tool that will work for any
non-degenerate triangle with known sides.
In this particular case, however, you can exploit a known property of the
equilateral triangle:
if $M$ is the midpoint of side $AB$ then $\triangle AMC$ 
is a right triangle with
a $60$-degree angle at $A$, and $CM = \sqrt3 \, AM$.
We can combine this with some techniques that borrowed from other useful
notions of coordinate geometry.
To find $M$, take the arithmetic means of the coordinates of $A =(3,4)$ and $B =(-2,3)$:
$$x_M = \frac12(x_A + x_B) = \frac12(3-2) = \frac12.$$
$$y_M = \frac12(y_A + y_B) = \frac12(4+3) = \frac72.$$
So $M=\left(\frac12, \frac72\right)$.
Now to get from $M$ to $A$ you increase $x$ by $\frac52$ and $y$ by $\frac12$,
so that line has slope $\frac15$.
The line $CM$ must have slope $-5$ to make a right angle with $AM$,
that is, for a $\frac12$ change in $x$ we get a $-\frac52$ change in $y$.
(In fact, the line with slope $-5$ through $\left(\frac12, \frac72\right)$
is the line $5x + y = 6$ that you found by other means.)
So the point $D = \left(x_m + \frac12, y_M - \frac52\right) =
\left(\frac12 + \frac12, \frac72 - \frac52\right)  = (1,1)$
is somewhere on the line $CM$ (not necessarily between $C$ and $M$)
and $DM = AM$.
But $C$ is just $\sqrt3$ times as far from $M$, so scale up the changes in $x$ and $y$
accordingly:
$$C = \left(\frac12 + \frac12\sqrt3, \frac72 - \frac52\sqrt3\right).$$
And of course we could also go an equal distance from $M$ in the opposite direction:
$$C' = \left(\frac12 - \frac12\sqrt3, \frac72 + \frac52\sqrt3\right).$$
Not surprisingly, these are the same results you get by solving the quadratic equation.
We could be a bit more sophisticated about all this (and save some words)
by treating the directed segment $MA$ as a vector and performing transformations
on it that rotate it by $90$ degrees (in either direction) and scale it by $\sqrt3$.
But if we know how to write a general rotation-and-scaling transformation
as a $2\times2$ matrix then we might as well take $AB$ as a vector and 
rotate it $60$ degrees with a scale factor of $1$ to determine the vector $AC$.
So this method also generalizes in a useful way when you know an applicable
angle of the triangle.
A: Calculate slope of AB (m), distance AB = s
Point A is at (h,k)
tan-1 m = angle N
s x sin (60 + angle N) = y
s x cos (60 + angle N) = x
Third point is at x+h, y+k
Use (180 + 60 + angle N) for second answer
and h,k is now at point B
