Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 30

Given the coordinates (2, 0), (0, 2), and (-1, 1).
Let A=(2, 0), B(0, 2), and C(-1, 1).

Find the equation of line AB using A(2, 0) and B(0, 2).
The slope of line AB is (2-0)/(0-2)=-1
The equation of line AB is y=-1x+2
Find the equation of line BC using B(0, 2) and C(-1 1).
The slope of line BC is (1-2)/(-1-0)=(-1)/-1=1
The equation of line BC is y=1x+2
Find the equation of line AC using A(2, 0) and C(-1, 1).
The slope of line AC is (1-0)/(-1-2)=-1/3
The equation of line AC is y-0=(-1/3)(x-2)=>y=-1/3x+2/3
Set up the intervals for integration.
int_-1^0(x+2)-(-1/3x+2/3)dx+int_0^2(-x+2)-(-1/3x+2/3)dx
=int_-1^0(4/3x+4/3)dx+int_0^2(-2/3x+4/3)dx
=[4/3*x^2/2+4/3x] from x=-1 to x=0 + [-2/3*x^2/2+4/3x] from x=0 to x=2
=[2/3x^2+4/3x] from x=-1 to x=0 + [-1/3x^2+4/3x] from x=0 to x=2
=[0-(2/3-4/3)]+[(-4/3+8/3)-0]
=(-2/3+4/3)+(-4/3+8/3)
=2/3+4/3
=6/3
=2
The area of the triangle with vertices (2, 0), (0, 2), and (-1, 1) is 2 units squared.