I'm given a square in the coordinate plane, the y co-ordinates of the square vertices are $y1$, $y2$, $y3$, and $y4$. How should I calculate the area of the square?
I tried to have $x1$, $x2$, $x3$, and $x4$ set, and then
$(x1-x2)^2 + (y1-y2)^2 = (x2-x3)^2 + (y2-y3)^2$
$(x2-x3)^2 + (y2-y3)^2 = (x3-x4)^2 + (y3-y4)^2$
$(x3-x4)^2 + (y3-y4)^2 = (x4-x1)^2 + (y4-y1)^2$
$(x1-x2)^2 + (y1-y2)^2 = (x4-x1)^2 + (y4-y1)^2$
$(x1-x3)^2 + (y1-y3)^2 = 2(x2-x3)^2 + 2(y2-y3)^2$
$2(x1-x2)^2 + 2(y1-y2)^2 = (x2-x4)^2 + (y2-y4)^2$
Trying to solve x is difficult. Is there a better way?
Best Answer
Order the y-axis coordinates in increasing order so that $y_1 \leq y_2 \leq y_3 \leq y_4$. Let $a = y_2 - y_1$ and $b = y_3 - y_1$, and notice that in following figure the inscribed square is the square you are trying to find the area. By the Pythagorean theorem, the area of that square is $a^2 + b^2$.