Answer #1: There are infinite possibilities for the third vertex. If it were isosceles, there are 6 possibilities.

Explanation: Since the distance between the two points is 6, the height must be 4 in order to have an area of 12. Therefore, the 3rd vertex could be anywhere on the y = 3 line. Therefore, there are infinite possibilities.

If it were isosceles, then I knew the only places there could be a vertex is if the two segments not shown are equal, one segment and the one shown are equal, or the one shown and the other segment are equal. That is 3. There are also the negatives of those, and 3 + 3 = 6.

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Answer #2: The third vertex is anywhere on the 2 or -2 y-axis. If the triangle is isoceles, the third vertex is either (2,-10) or (-2,-10).

Explanation: First I found the length of the base of the triangle from (0,6) to (0,12) to be six units (obviously, 12-6 = 6). The problem states that the area of the triangle must be 12. Remembering that the area of a triangle is found by using one half the base times the height (1/2b*h), I set up the problem:
1/2(6)*h = 12
Reduced:
3*h = 12
Divide both sides by 3:
h=4 The height of the triangle must be four. The third vertex of the triangle can be anywhere on the 2 or -2 y-axis and the height of the triangle will still be four units. For the triangle to be isoceles, the third point must be on a perpendicular bisector to the base. The line x = -10 is the perpendicular bisector to the base, which lies on y = 0. The intersection of x = -10 and y = 2 or y = -2 is (2,-10) and (-2,-10). Therefore, the third points for an isoceles triangle are (2,-10) and (-2,-10). The Descartes triangle is named after the famous mathemetician/philosopher Rene Descartes.

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Answer#3: The possibilities are:(4,12), (4,11), (4,10), (4,9), (4,8), (4,7), (4,6), (-4,6), (-4,7), (-4,8), (-4,9), (-4,10), (-4,11), and (-4,12).
If the triangle is isosceles, then the third coordinate would be (4,9) or (-4,9).

Explanation: First, we drew the coordinates that were already given to us, then we found the height of the triangle by using the guess and check method.
We concluded that the height of the triangle would be 4 since 6 times 4 divided by 2 is 12, which is how many units the area of the triangle is. We knew that the x coordinate would always be either 4 or -4 and the y coordinates were in between 12 and 6, so we came up with 14 different possibilities.

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Answer #4: {6,4; 7,4; 8,4; 9,4 (if it's isosceles); 10,4; 11,4; 12,4}

Explanation: I started by solving trying to find the hight of the triangle. I used the formula A = 1/2 bh. this took me to the equation
12 = 1/2(6)(x)
12 = 1/2(6x)
12 = 3x
x = 4.
This means the height of the triangle is 4 units. I then set the height in all possible positions that would make the height perpendicular to the base.

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Answer #5: The third vertex can be positioned on (2,12), (3,8), (4,6), and (6,4) and the triangle can not be isosceles and have an area of 12 units squared at the same time. Bonus: It is called "Descartes' Triangle" because Descartes is the man who came up with the triangle.

Explanation: I went through each of the coordinates and found the area of each triangle using the formula A = 1/2bh. When I did that I came up with all of the triangle that had an area of 12 units squared. I decided that you could not have an isosceles triangle because none of those numbers squared then divided by 2 equaled 12.

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Answer #6: all the possible answer for the third vertex is x = 4 and x = -4 and the possible answer of the isosceles is y = 4 and y = -4

Explanation: The area of the triangle is 0.5 time b h. if we know the two vertices of a triangle are located at (0,6) and (0,12) then the possible answer just can be 4 and -4. and the isosceles tiangle is 4 and -4 too, because we know isosceles traingle have two line is equal and one triangle can be (0,6), (0,12) and (4,9) and the next triangle can be (0,12), (0,6) and (-4,6.5)

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Answer #7: The possible points for the other vertex are all points on lines parallel to the x axis that go through the points (0, 4) and (0, -4).

Explanation: To get my answer i first found the formula for the area of a triangle (1/2 base times height). Then i plugged in my numbers (1/2 6 times h = 12). Then I worked the problem and got that the height is 4. After that i drew some of the points and realized that all of the points on that line would work. I also moticed that points on the line that go through -4 also work.

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Answer #8: The point can either be (-4, 9) or (4, 9) if the triangle was supposed to be isosceles. OR the third vertex can be on the x-axis.

Explanation: From point (0, 6) to (0, 12)they're 6 units apart. So i determined 6 units as the base for the triangle. Since they gave us 12 has the area I would plug it in the area for triangles formula. 12 = 1/2(6)h so h = 4. So to gave it become isoscles then you would find the midpoint of the given point which equals to (0, 9). Then we can plug in either -4 or 4 as the x-axis to make it isosceles. OR you can have it anywhere on the x-axis to make it isosceles.

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Answer #9: The other vertex could of been either (4,9) or (14,9)

Explanation: If the area of the triangle is 12units^2
And the formula for a triangle is 1/2 Base*Height
The base is already given to me and that is the difference of y2 to y1 which is 6
so it is 1/2*6*Height = 12
or 3*Height = 12
divide 3 to both sides and you get height = 4
so you move x right 4 on the graph and move it to the middle of the first 2 points which is 9 or move x left 4 and move it to the middle of the first 2 points

If the triangle was isosceles the answer would remain the same? The reason they call this triangle the Descartes' Triangle is because thay named it after the mathmetician who discovered it Rene' Descartes'