The points $B = (3, 1)$ and $C = (-2, 5)$ are the vertices of an equilateral triangle $ABC$. Calculate the area and perimeter of this triangle.

Question

I encountered a problem while solving this task. It is an equilateral triangle so $\vert AB \vert = \vert AC \vert = \vert BC \vert$. I then calculated the length of the segment $\vert BC \vert$

$$\vert AB \vert = \sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}$$

So my equation looks like this:

$$X_A = 3, Y_A = 1, X_B = -2, Y_B = 5$$ $$\vert BC \vert = \sqrt{(-2-3)^2+(5-1)^2} = \sqrt{(-5)^2+4^2} = \sqrt{41}$$

In the solution, the correct answer to this subsection is $\sqrt{61}$ which means it comes out that $X_A = -2, Y_A = 5, X_B = 3, Y_B = 1$. How is it that suddenly one segment length of a triangle has two solutions? As far as I know, there is no rule according to which $X_A, X_B, Y_A, Y_B$ should be given, because in the end a segment can have only one length.

EDIT: Sorry, I rewrote the task on the sheet wrong.... Point $B = (3,-1)$, not $(3, 1)$. Already always comes out one result

Answer 1

Consider line segment AB. You can construct a circle centered at A that passes through B and a circle at B that passes through A. These circles will intersect at two points (C and D) on opposite sides of the line segment.

You are essentially calculating the two points of intersection using the distance formula.

Answer 2

First solve for |BC|, which is $\sqrt{41}$. I suppose your book must have an error. Then since we know that the triangle is equilateral, the perimeter must be $3a$ units and the area $\frac{a^2\sqrt 3}{4}$ squared units, where $a$ is each side length.

If you wish to explicitly solve for A, you can via the distance formula. You get two different points, for a reason explained by @TurlocTheRed quite well. Algebraically solving here will lead to quadratic equations, giving you two values for each of abscissae and ordinates.