Can anyone help me out with this problem? Find the area of the triangle with the vertices A(0,1), B(-3,-2), and C(-4,4). Any pointers would be greatly appreciated.
The triangle you are finding is made up of three line sections,
You need to make equations for each line and then use integrals to get the area.
For example you can integrate from C to A and then subtract out the green triangles using 1/2 * base * height
There are many ways to solve for the area of the mentioned triangle.
For me, an easy one is by coordinate geometry or by analyzing the triangle on the xy plane.
The trick is to enclose the triangle in a rectangular, and then the area of the said triangle is equal to the area of the rectangle minus the areas of the 3 right triangles outside of the said triangle.
The enclosing rectangle:
>>>The highest y-coordinate is that of point C, which is 4. The lowest y is that of B, which is -2. So, the height of the rectangle is 4 +2 = 6.
>>>The leftmost x is that of C, which is -4. The rightmost x is the 0 of A. So, the width of the rectangle is 4+0 = 4.
>>>Therefore, the enclosing rectangle is 6 by 4.
Area of rectangle = 6*4 = 24 sq.units.
area of right triangle to the left of side BC is
A1 = (1/2)(6)(-3 -(-4)) = (1/2)(6)(1) = 3 sq.units
area of right triangle above side AC is
A2 = (1/2)(4)(4-1) = (1/2)(4)(3) = 6 sq.units
area of right triangle below side AB is
A3 = (1/2)(0 -(-3))(1 -(-2)) = (1/2)(3)(3) = 4.5 sq.units
Therefore, area of triangle ABC = 24 -3 -6 -4.5 = 10.5 sq.units ...answer.
Now, since you post your question under Calculus, I think you want to solve it by calculus or by integration.
It will be a long, long solution.
Are you interested?