Question 705616: Two of the vertices of an equilateral triangle are (2,1) and (6,5). Find the possible coordinates of the remaining vertex.
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! Drawing this would take time and planning, but here is a very useful description:
*NOTE: Please see POSSIBLE EASIER WAY further down!
Use two circles, one centered at (2,1) and the other centered at (6,5). The radius of each circle will be the size of the distance from point (2,1) to point (6,5). These two circles intersect at two points. Either of those intersections may be chosen as your unknown vertex of your equilateral triangle.
Symbolically developing the circles---
The two circles become
AND
Reminder: The points of intersection will be vertices for the other point of the equilateral triangle.
*POSSIBLE EASIER WAY:
The line perpendicular to the two-point defined one, and containing the midpoint of (2,1) and (6,5) will contain both possible unknown vertices of the equilateral triangle. Use this new found line and its intersection with one of the circles.
Slope of line defined by (2,1) & (6,5) is +1.
Slope of new target line is then -1.
Midpoint of (2,1) & (6,5) is the point (4,3).
New target line still needs known y intercept, "b". y=-x+b, b=y+x, b=3+4=7.
New target line is therefore, y=-x+7
Look for solution to the system
Solving this should be easier than dealing with two circles.
...in fact, it's very fast in comparison.
Substitute the expression for y from the line equation into the circle equation. Simplification gives,  . General solution to quadratic equation gives  or  .
Solution for y then is  or  .
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