Show that,for every $a \in R$,the area of the triangle having the vertices $A(5, a)$,$B(2, 5)$,and $C(2, 3)$ is $3$ square units.

(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 $A(5, a)$,$B(2, 5)$,and $C(2, 3)$.
Substituting these values into the formula:
Area $= \frac{1}{2} |5(5 - 3) + 2(3 - a) + 2(a - 5)|$
Area $= \frac{1}{2} |5(2) + 6 - 2a + 2a - 10|$
Area $= \frac{1}{2} |10 + 6 - 10|$
Area $= \frac{1}{2} |6|$
Area $= 3$ square units.
Since the result is independent of $a$,the area of the triangle is $3$ square units for every $a \in R$.