What is the area of a triangle with vertices at (0, 0), (6, 0), and (3, 7)?
To find the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), you can use the following formula:

Area = |(x1(y2-y3) + x2(y3-y1) + x3(y1-y2))/2|

In this case, the vertices are (0, 0), (6, 0) and (3, 7), so x1=0, y1=0, x2=6, y2=0, x3=3 and y3=7.

Substituting these values into the above formula, the area of the triangle is:

|(0(0-7) + 6(7-0) + 3(0-0))/2|

= |(-42)/2|

= 21 square units.

So, the area of the triangle with vertices at (0, 0), (6, 0), and (3, 7) is 21 square units.
Complete the following task based on the instruction and optional input provided.### Instruction:
What is the area of a triangle with vertices at (0, 0), (6, 0), and (3, 7)?

### Output:
To find the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), you can use the following formula:

Area = |(x1(y2-y3) + x2(y3-y1) + x3(y1-y2))/2|

In this case, the vertices are (0, 0), (6, 0) and (3, 7), so x1=0, y1=0, x2=6, y2=0, x3=3 and y3=7.

Substituting these values into the above formula, the area of the triangle is:

|(0(0-7) + 6(7-0) + 3(0-0))/2|

= |(-42)/2|

= 21 square units.

So, the area of the triangle with vertices at (0, 0), (6, 0), and (3, 7) is 21 square units.