Calculate Area of Triangle ABC

[mathdad]

Point A(3, 4), point B(8, 5) and point C(7, 8) are located in quadrant 1 and form Triangle ABC.

Note:

Point A(a, b)
Point B(c, d)
Point C(e, f)

Find the area of Triangle ABC using the formula below.

A = (1/2)(a*d - c*b + c*e - e*d + e*b - a*e)

I think this is just a plug and chug problem. It looks tricky but in reality, it's not that bad.

I see it this way:

a = 3
b = 4
c = 8
d = 5
e = 7
f = 8

I plug the values of a through e (not including f) into the formula to calculate the area of Triangle ABC. Am I right?

 

[MarkFL]

As you can find derived here:

http://mathhelpboards.com/math-notes-49/finding-area-triangle-formed-3-points-plane-2954.html

we have:

\(\displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|\)

and putting this in terms of the given coordinates, we have:

\(\displaystyle A=\frac{1}{2}\left|(e-a)(d-b)-(c-a)(f-b) \right|\)

\(\displaystyle A=\frac{1}{2}\left|-ad+af+bc-be-cf+de \right|\)

Change the signs to match your formula:

\(\displaystyle A=\frac{1}{2}\left|ad-af-bc+be+cf-de \right|\)

Thus, I would posit that the formula you have been given has a few typos in it. :D

 

[mathdad]

As you can find derived here:

http://mathhelpboards.com/math-notes-49/finding-area-triangle-formed-3-points-plane-2954.html

we have:

\(\displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|\)

and putting this in terms of the given coordinates, we have:

\(\displaystyle A=\frac{1}{2}\left|(e-a)(d-b)-(c-a)(f-b) \right|\)

\(\displaystyle A=\frac{1}{2}\left|-ad+af+bc-be-cf+de \right|\)

Change the signs to match your formula:

\(\displaystyle A=\frac{1}{2}\left|ad-af-bc+be+cf-de \right|\)

Thus, I would posit that the formula you have been given has a few typos in it. :D

I made a few typos. Thanks.