math question

[Dencology]

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.Guys this should be easy but i have a hard time with it. i did use the formula distance to find the two bases and once you divide it by half you can get the area. but for some reason the distance formula is not working. can you help me?
in a two dimentional coordinate system point A=(3,-6) point B=(1,1) and point C=(7,6). find the area of triangle ABC.
a. 26
b. 24
c. 20
d. 18
e. 42

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[Streetwolf]

.Guys this should be easy but i have a hard time with it. i did use the formula distance to find the two bases and once you divide it by half you can get the area. but for some reason the distance formula is not working. can you help me?
in a two dimentional coordinate system point A=(3,-6) point B=(1,1) and point C=(7,6). find the area of triangle ABC.
a. 26
b. 24
c. 20
d. 18
e. 42

.

Well you need the base and height, not just two sides. You can make one side a base but then you need to find the height. Graph it and you'll see that there is no clear cut height in this triangle. You'd have to divide it into a couple more triangles and figure out the height.

Instead I suggest using the determinant. To do this you need to set one of these coordinates to (0, 0) and use the other two. I suggest setting B to (0, 0) and so A = (2, -7) and C = (6, 5).

Put A in the top row and C in the bottom row of the 2x2 matrix. Now you have (2, -7) on top and (6, 5) on bottom. So 2 is above 6, and -7 is above 5. What you do is take the (top left * bottom right) - (top right * bottom left). Take the absolute value of that and that's your area.

In this case you have (2*5) - (-7*6) = 10 - (-42) = +52. Don't forget though that this is a triangle. Determinant finds the area of a parallelogram. Divide by 2 and get 26, choice (A).

 

[vvvv]

Well you need the base and height, not just two sides. You can make one side a base but then you need to find the height. Graph it and you'll see that there is no clear cut height in this triangle. You'd have to divide it into a couple more triangles and figure out the height.

Instead I suggest using the determinant. To do this you need to set one of these coordinates to (0, 0) and use the other two. I suggest setting B to (0, 0) and so A = (2, -7) and C = (6, 5).

Put A in the top row and C in the bottom row of the 2x2 matrix. Now you have (2, -7) on top and (6, 5) on bottom. So 2 is above 6, and -7 is above 5. What you do is take the (top left * bottom right) - (top right * bottom left). Take the absolute value of that and that's your area.

In this case you have (2*5) - (-7*6) = 10 - (-42) = +52. Don't forget though that this is a triangle. Determinant finds the area of a parallelogram. Divide by 2 and get 26, choice (A).

good way to solve

 

[smile101]

Well you need the base and height, not just two sides. You can make one side a base but then you need to find the height. Graph it and you'll see that there is no clear cut height in this triangle. You'd have to divide it into a couple more triangles and figure out the height.

Instead I suggest using the determinant. To do this you need to set one of these coordinates to (0, 0) and use the other two. I suggest setting B to (0, 0) and so A = (2, -7) and C = (6, 5).

Put A in the top row and C in the bottom row of the 2x2 matrix. Now you have (2, -7) on top and (6, 5) on bottom. So 2 is above 6, and -7 is above 5. What you do is take the (top left * bottom right) - (top right * bottom left). Take the absolute value of that and that's your area.

In this case you have (2*5) - (-7*6) = 10 - (-42) = +52. Don't forget though that this is a triangle. Determinant finds the area of a parallelogram. Divide by 2 and get 26, choice (A).

I have been using this formula:
A= 1/2 [(x1y2+ x2y3 + x3y1)-(y1x2+y2x3+y3x1)]

but streetwolf's way is really simple and less time consuming.

 

[Streetwolf]

I have been using this formula:
A= 1/2 [(x1y2+ x2y3 + x3y1)-(y1x2+y2x3+y3x1)]

but streetwolf's way is really simple and less time consuming.

Actually I use the same formula believe it or not! The only difference is that yours works for all 3 points and mine only needs 2. Since I relocate one of the points to (0,0), and in the above case I chose point B which was (x2, y2), all terms in that equation with an x2 or a y2 drop out. This leaves us with the formula I used.

 

[theedaddy77]

that was amazing that formula for determinant. i had a quick question. for this to work you need to set one of the point to 0,0 so whatever point you use how do you know it affects the other point? in this case u just minus the x and y corrdinates from all the rest, do you always do that are did you do that because 1,1 is in the first quadrant? thanks for the help!

 

[Streetwolf]

Ya just do whatever you did to that one point, to all the other points. Since I used (1,1) I subtracted 1 from the x and 1 from the y. Or put another way, I moved the origin up 1 and to the right 1.

You need to put a point to (0,0) to use the 2 point determinant formula. Otherwise you'll need to use the 3 point equation in one of the posts above.