Show that the area of a triangle having vertices $(t, t-2)$,$(t+2, t+2)$,and $(t+3, t)$ does not depend on the value of $t$.

(N/A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(x_1, y_1) = (t, t-2)$,$(x_2, y_2) = (t+2, t+2)$,and $(x_3, y_3) = (t+3, t)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |t((t+2) - t) + (t+2)(t - (t-2)) + (t+3)((t-2) - (t+2))|$
Area $= \frac{1}{2} |t(2) + (t+2)(2) + (t+3)(-4)|$
Area $= \frac{1}{2} |2t + 2t + 4 - 4t - 12|$
Area $= \frac{1}{2} |(2t + 2t - 4t) + (4 - 12)|$
Area $= \frac{1}{2} |0 - 8| = \frac{1}{2} |-8| = 4$ square units.
Since the result is $4$,which is a constant,the area of the triangle does not depend on the value of $t$.