Find a value of y such that the triangle with the given vertices has an area of 40 square units.
(7,0),(7,-5),(-9,y)

A)y = -9 or y = -10
B)y = 7 or y = -3
C)y = 9 or y = 0
D)y = 8 or y = 3
E)y = 10 or y = 8

Question

Find a value of y such that the triangle with the given vertices has an area of 40 square units. ​
​(7,0),​(7,-5),​(-9,y)
​
A)​y = -9 or y = -10
B)​​y = 7 or y = -3
C)​y = 9 or y = 0
D)​y = 8 or y = 3
E)​y = 10 or y = 8

Quizplus · Accepted Answer

One way to find the area of a triangle is by using the formula A = 1/2 * base * height. We can choose the point (7,0) or (7,-5) as one of the vertices of the triangle and use that as the base.
If we choose (7,0) as the base, then the height of the triangle is the distance between the point (-9,y) and the line x = 7. The perpendicular line from (-9,y) to x = 7 would have an x-coordinate of 7, so the distance between (-9,y) and (7,0) is:
distance = sqrt[(7 - (-9))^2 + (0 - y)^2] = sqrt[256 + y^2]
Therefore, the area of the triangle is:
A = 1/2 * base * height = 1/2 * (7 - 7) * sqrt[256 + y^2] = 0
This equation simplifies to:
0 = 1/2 * sqrt[256 + y^2]
0 = sqrt[256 + y^2]
256 = y^2
y = 16 or y = -16
However, y = 16 is not a possible solution because it would give us a point (-9,16) that is not on the coordinate plane. Therefore, y = -16 is the only solution if we choose (7,0) as the base. 
If we choose (7,-5) as the base, then the height of the triangle is the distance between the point (-9,y) and the line x = 7. Again, the perpendicular line from (-9,y) to x = 7 would have an x-coordinate of 7, so the distance between (-9,y) and (7,-5) is:
distance = sqrt[(7 - (-9))^2 + (-5 - y)^2] = sqrt[256 + (y + 5)^2]
Therefore, the area of the triangle is:
A = 1/2 * base * height = 1/2 * (7 - 7) * sqrt[256 + (y + 5)^2] = 0
This equation simplifies to:
0 = 1/2 * sqrt[256 + (y + 5)^2]
0 = sqrt[256 + (y + 5)^2]
256 = (y + 5)^2
y + 5 = 16 or y + 5 = -16
y = 11 or y = -21
However, y = -21 is not a possible solution because it would give us a point (-9,-21) that is not on the coordinate plane. Therefore, y = 11 is the only solution if we choose (7,-5) as the base. 
So the possible values of y are -16 and 11, and the only option that includes either of those values is C, y = 9 or y = 0. However, y cannot be 0 because that would give us a line passing through two of the vertices of the triangle, which is not a triangle at all. Therefore, the only possible value of y is -16, which means that the answer is C, y = 9.

Find a Value of Y Such That the Triangle with the Given