SOLUTION: Find the value(s) of y such that the triangle with the given vertices has an area of 8 square units. (Enter your answers as a comma-separated list.) (−4, 5), (−3, 3), (−3, y

Let  A = (-4,5)

     B = (-3,3)

     C = (-3,y)


Notice that the side  BC  is vertical line  x= -3  of the length of  |y-3|     (absolute value).


Also notice that the distance of the point A from this line x= -3  is  -3 - (-4) = -3 + 4 = 1 unit.



    You may consider the side  BC  of the length  |y-3|  units as a base of the triangle  ABC,
    and consider the distance of the point A from the line x= 3 as the length of the altitude
    of this triangle drawn to the base BC.



So, your equation for the triangle area is


     = 8,

or

    |y - 3| = 16.


This equation has two solutions  y= 3+16 = 19  and  y= 3-16 = -13.


ANSWER.  There are two solutions for y:  y= -13  or  y= 19.