Find The Area of a Triangle Given Three Vertices

[llaunyc]

Is there a way to find the area of a triangle if you are given three vertices on the coordinate plane? Assume that the three points do not form a right triangle...

Let's say I can calculate the length between the 2 closer points, B and C, and that's my base, and segment BC. How can I find the point at which A creates a perpendicular line with BC, my height?

[hardykarim]

Sure you can.

Let the point D=middle of segment BC with coordinates:

Xd=(Xb+Xc)/2
Yd=(Yb+Yc)/2

Then we can calculate the distance between A and D, your height! ;)

[jnelson0612]

Sure you can.

Let the point D=middle of segment BC with coordinates:

Xd=(Xb+Xc)/2
Yd=(Yb+Yc)/2

Then we can calculate the distance between A and D, your height! ;)

Excellent job!

[upasana]

use the determinant method of matrix to evaluate the area
(x1, y1),
(x2, y2),
(x3, y3)

Area will be 0.5* |x1 y1 1|
|x2 y2 1|
|x3 y3 1|

[kaushik.pal.calling]

Sure you can.

Let the point D=middle of segment BC with coordinates:

Xd=(Xb+Xc)/2
Yd=(Yb+Yc)/2

Then we can calculate the distance between A and D, your height! ;)

Excellent job!

Distance between mid-point of a side and the opposite vertex need not be the altitude. Let B=(-2,0) & C=(2,0). Mid-point of BC, D=(0,0). AD will be altitude of triangle ABC only if A is on y-axis.

Given 3 vertices of a triangle, area is given by:
sqrt(s(s-a)(s-b)(s-c)) where S=(a+b+c)/2 and a, b and c are the length of 3 sides

[tim]

hardykarim, your approach doesn't work. it will give you a median and not an altitude. thanks kaushik for pointing this out..

upasana, your method works but is probably overkill on the GMAT..

kaushik's Heron's Formula method will also work but is probably also overkill..

remember, the GMAT does not expect you to know less common formulas such as these. if you must resort to one of these formulas, you have almost surely missed the trick..