Can someone please help me with this Geometry problem?

Question

Find the area of △ABC with vertices at A (2,-3), B (0,1), and C (0,-2).

For this I got 3, am I correct?

Answer 1

(2,3), (0,1), (0,-2)

it helps to at least plot the points and do a rough sketch

connect the dots

the area = the area of a right triangle, 4 by 2 = A = bh/2 = 4(2)/2 = 4

minus the area of a smaller right triangle 2 by 1 = A= bh/2 = 2(1)/2 = 1

Area = 4- 1 = 3

You're right.

Or just take the base as the distance from (0,1) to (0,-2) = 3 rotating the graph 90 degrees

and the height as 0 - - 2 = 2

then area = 3(2)/2 = 3

Or calculate the perimeter, divide by 2 = (3sqr5+3)/2

calculate each side length: 2sqr5, 3 and sqr5

Area by Heron's Formula = square root of product of 4 terms (3sqr5+3)/2 and the difference between that and each side. It's tedious, but it also comes out to A= 3