Atriangle with vertices located at (−2, −2) and (4, −2) has an area of 24 square units. which is one possible location of the other vertex?

Actually, the answer is (4, 6).

Step-by-step explanation:

I did this problem on a test and got it correct. Also, I know that the formula for a triangles area is bh/2 or base*height divided by 2.The absolute value of the x-values combined is the base which is 6 because I -2 + 4 I = 6 units. To find a height that would give you 24 units squared, you would have to multiply the height that you would put for a rectangle ( 4) by 2 which is 8 units. If you add the absolute value of the y-value in the points ( -2, -2) or ( 4, -2) to the absolute value of the y- value in (4, 6) which gives you the equation I -2 + 6 I = 8 units It will prove that (4, 6) is the missing vertices for the right triangle.

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