# Calculate the area of the triangle with vertices (-2,-2),(3-7), and (4,4)? Please do it without using length and angles.

embizze | High School Teacher | (Level 1) Educator Emeritus

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Compute the area of the triangle with vertices at (-2,-2),(3,-7) and (4,4) without computing lengths or measures of angles:

(1) The area of a triangle with vertices `(x_1,y_1),(x_2,y_2),(x_3,y_3)` can be found by computing `1/2` of the determinant of the matrix:

`[[x_1,y_1,1],[x_2,y_2,1],[x_3,y_3,1]]` , making the sign positive if necessary.

Thus the area is:

`+- 1/2 [[-2,-2,1],[3,-7,1],[4,4,1]]`

`=+-1/2[(14-8+12)-(-28-8-6)]=+-1/2[18-(-42)]=30`

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The area of the triangle is 30 square units.

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(2) Alternatively you could use Pick's theorem: A=`I+1/2 B-1` where I is the number of interior lattice points, B the number of lattice points on the boundary. So area is `25+1/2(12)-1=31-1=30`

Most of the other ways to find the area require finding a length(s) or angle -- vectors compute the length (if you use the dot product formula `A=1/2sqrt((AB*BC)(AC*AC)-(AB*AC)^2)=1/2sqrt(|AB|^2|AC|^2-(AB*AC)^2)` while the cross product formula requires an angle `Area=1/2|AB"x"AC|=1/2[|AB||AC|sin theta n]`

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