Question

Find the area of the triangle with vertices at (1,0), (0,-1), and (2,-2)

Answer 1

A(1,0), B (0,-1) and C(2,-2) are the vertices .
Area of the Triangle = Absolute value of[1/2(Xc-Xa)(Yb-Ya) - (Xb-Xa)(Yc-Ya)]
Use this formula

Answer 2

Let's call A(1,0), B(0,-1), C(2,-2).

AB = sqrt(2) long, while BC=AC = sqrt(5) long.

Thus the triangle is isosceles. We can find the length of AD, where D is the midpoint of AB, using Pythagoras' theorem: AD = sqrt((sqrt(5)^2 - sqrt(2)^2)) = sqrt(5-1) = 2.

Thus area is 1/2 base x height = 1/2 x sqrt(2) x 2 = sqrt(2)

Answer 3

but if the choices were 1, 1.25, 1.5 and 1.75.. which one is it?

Answer 4

Actually it's none of those, since the answer is sqrt(2) = 1.41.