In the coordinate plane, one of the vertices of a square is the point \((-3, -4).\) If the diagonals of that square

Let the length of the side of the square be s.

Note that diagonals of a square intersect at the center of the square.
Therefore, by Pythagoras theorem (s/2) $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2$$ = square of distance between vertex and center.

So $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2\ =\ \left(3\ -\ \left(-3\right)\right)^2\ +\ \left(2\ -\ \left(-4\right)\right)^2\ =\ 72$$

$$\left(\frac{s}{2}\right)^2\ =\ 36$$

$$\left(\frac{s}{2}\right)\ =\ 6$$

$$s\ =\ 12$$

Area = $$s\ \cdot\ s\ =\ 144$$ D