How would you set up and solve this word problem?

Heureka probably knows how to do this better with vectors.......and maybe even come up with more values for y than I did......but, here's my attempt  :

The first value for y is pretty easily derived.......let the base be defined the two points (-5,1) and (-4,y).........if we let y = 1.........the length of the base = 1 and the height of the oblique triangle [ with  (0,5) as the "apex" point ] will be 4.....so......the area of this triangle =  (1/2)(b)(h)  = (1/2)(1)(4)  = 2 sq units

To get a second point....let the base be defined by the two points  (-5,1) and (0,5).....and the length of this base = sqrt (41)

And the equation of the line going through these two points is

y   = (4/5)x + 5   which we can write as    (4/5)x - y + 5  = 0

Using the area formula for a triangle, we can calculate the needed height  of our second triangle

2 = sqrt(41)/2 *h    →  h  = 4/sqrt(41)

So........we need to find a point with an x coordinate of -4 that is 4/sqrt(41)   units from the line

y = (4/5)x + 5

Using the equation for the perpendicular distance from a point to a given line, we have

abs [ (4/5)(-4)  -1y + 5 ] / ( sqrt [ 16/25 + 1 ] )  =  4/sqrt(41)

abs [ -16/5 -1y + 5 ] / sqrt (41)/ 5  = 4/sqrt(41)

abs [ 9/5 - y] * 5  = 4

abs [9/5 -y] = 4/5

And we have two equations here

9/5 - y  = 4/5                 or                  9/5 - y   = - 4/5

y = 9/5  - 4/5  = 1                                 y =  9/5 + 4/5  =  13/5   = 2.6

Notice that this gave us not only a second point, but also my original "guess" as well

So.......the two values of y are :     1, 2.6

Here's an  image of both triangles that we found......the first is triangle ABC   and the second is triangle ABD

Find the value(s) of y such that the triangle with the given vertices has an area of 2 square units. (Enter your answers as a comma-separated list.)

(−5, 1), (0, 5), (−4, y)

y= ?

$\begin{array}{lcll} \vec{A} = \binom{a_x}{a_y} = \binom{-5}{1} \qquad \vec{B} = \binom{b_x}{b_y} = \binom{0}{5} \qquad \vec{C} = \binom{c_x}{c_y} = \binom{-4}{y} \\\\ \vec{B}-\vec{A} = \binom{b_x}{b_y}-\binom{a_x}{a_y} = \binom{b_x-a_x}{b_y-a_y} = \binom{0-(-5)}{5-1} = \binom{5}{4} \\ \vec{C}-\vec{A} = \binom{c_x}{c_y}-\binom{a_x}{a_y} = \binom{c_x-a_x}{c_y-a_y} = \binom{-4-(-5)}{y-1} = \binom{1}{y-1} \\ \\ \end{array}\\ \begin{array}{rcll} 2\cdot A_{\text{triangle}} &=& | (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A}) | \\ &=& | \binom{5}{4} \times \binom{1}{y-1} | \\ &=& | 5\cdot(y-1) - 4\cdot 1 | \\ &=& | 5\cdot y -5 - 4 | \\ &=& | 5\cdot y -9 | \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& + ( 5\cdot y_2 -9 ) \\ 5\cdot y_2 &=& 9+ 2\cdot A_{\text{triangle}} \\ y_2 &=& \frac{9+ 2\cdot A_{\text{triangle}}}{5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_2 &=& \frac{9+ 2\cdot 2}{5}\\ y_2 &=& \frac{13}{5}\\ \mathbf{y_2} & \mathbf{=} & \mathbf{2.6} \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& - ( 5\cdot y_1 -9 ) \\ 2\cdot A_{\text{triangle}} &=& - 5\cdot y_1 +9 \\ 5\cdot y_1 &=& 9 - 2\cdot A_{\text{triangle}} \\ y_1 &=& \frac{9 - 2\cdot A_{\text{triangle}}} {5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_1 &=& \frac{9 - 2\cdot2} {5} \\ y_1 &=& \frac{5} {5} \\ \mathbf{y_1} & \mathbf{=} & \mathbf{1} \\ \end{array}$