Answer:
The area of this triangle is about 21.2132 square units.
Step-by-step explanation:
First, find the lengths of the legs AB and BC.
Length of AB ===
Find the difference in position vertically:
![-2-4=-6](https://tex.z-dn.net/?f=-2-4%3D-6)
The points are 6 units apart vertically.
Find the difference in position horizontally:
![3-0=3](https://tex.z-dn.net/?f=3-0%3D3)
The points are 3 units apart horizontally.
These lengths form a right triangle with the distance between the points as the hypotenuse, so you can use the pythagorean theorem to solve:
![a^2+b^2=c^2\\3^2+6^2=c^2\\9+36=c^2\\45=c^2\\c\approx6.7082](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3Dc%5E2%5C%5C3%5E2%2B6%5E2%3Dc%5E2%5C%5C9%2B36%3Dc%5E2%5C%5C45%3Dc%5E2%5C%5Cc%5Capprox6.7082)
AB is about 6.7082 units long.
Length of BC ===
Same process as above.
Find the vertical distance:
![-4--2=-2](https://tex.z-dn.net/?f=-4--2%3D-2)
2 units apart vertically.
Find the horizontal distance:
![-3-3=-6](https://tex.z-dn.net/?f=-3-3%3D-6)
6 units apart horizontally.
Use the pythagorean theorem:
![2^2+6^2=c^2\\4+36=c^2\\40=c^2\\c=6.3246](https://tex.z-dn.net/?f=2%5E2%2B6%5E2%3Dc%5E2%5C%5C4%2B36%3Dc%5E2%5C%5C40%3Dc%5E2%5C%5Cc%3D6.3246)
BC is about 6.3246 units long.
Area ===
Finally, you can use these to find the area of the triangle. The area of a right triangle is just half the area of a rectangle with the same base and height:
![A=\frac{b\times h}{2}\\\\A=\frac{6.7082\times6.3246}{2}\\\\A=\frac{42.4264}{2}\\\\A=21.2132](https://tex.z-dn.net/?f=A%3D%5Cfrac%7Bb%5Ctimes%20h%7D%7B2%7D%5C%5C%5C%5CA%3D%5Cfrac%7B6.7082%5Ctimes6.3246%7D%7B2%7D%5C%5C%5C%5CA%3D%5Cfrac%7B42.4264%7D%7B2%7D%5C%5C%5C%5CA%3D21.2132)
The area of this triangle is about 21.2132 square units.