<u>ANSWER: </u>
The area of the triangle with vertices at (0, – 2), (8, – 2), and (9, 1) is 12 square units.
<u>SOLUTION:
</u>
Given, vertices of the triangle are A(0, -2), B(8, -2) and C(9, 1).
We have to find the area of the given triangle.
We know that,
![\text { Area of triangle }=\frac{1}{2}\left(x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right)](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Area%20of%20triangle%20%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%28x_%7B1%7D%5Cleft%28y_%7B2%7D-y_%7B3%7D%5Cright%29%2Bx_%7B2%7D%5Cleft%28y_%7B3%7D-y_%7B1%7D%5Cright%29%2Bx_%7B3%7D%5Cleft%28y_%7B1%7D-y_%7B2%7D%5Cright%29%5Cright%29)
![\text { Where, }\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \text { are vertices of the triangle. }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Where%2C%20%7D%5Cleft%28x_%7B1%7D%2C%20y_%7B1%7D%5Cright%29%2C%5Cleft%28x_%7B2%7D%2C%20y_%7B2%7D%5Cright%29%2C%5Cleft%28x_%7B3%7D%2C%20y_%7B3%7D%5Cright%29%20%5Ctext%20%7B%20are%20vertices%20of%20the%20triangle.%20%7D)
Here in our problem, ![\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,-2),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(8,-2) \text { and }\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)=(9,1)](https://tex.z-dn.net/?f=%5Cleft%28%5Cmathrm%7Bx%7D_%7B1%7D%2C%20%5Cmathrm%7By%7D_%7B1%7D%5Cright%29%3D%280%2C-2%29%2C%5Cleft%28%5Cmathrm%7Bx%7D_%7B2%7D%2C%20%5Cmathrm%7By%7D_%7B2%7D%5Cright%29%3D%288%2C-2%29%20%5Ctext%20%7B%20and%20%7D%5Cleft%28%5Cmathrm%7Bx%7D_%7B3%7D%2C%20%5Cmathrm%7By%7D_%7B3%7D%5Cright%29%3D%289%2C1%29)
Now, substitute the above values in the formula.
![\text { Area of triangle }=\frac{1}{2}\left(x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right)](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Area%20of%20triangle%20%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%28x_%7B1%7D%5Cleft%28y_%7B2%7D-y_%7B3%7D%5Cright%29%2Bx_%7B2%7D%5Cleft%28y_%7B3%7D-y_%7B1%7D%5Cright%29%2Bx_%7B3%7D%5Cleft%28y_%7B1%7D-y_%7B2%7D%5Cright%29%5Cright%29)
![=\frac{1}{2}(0(-2-1)+8(1-(-2))+9(-2-(-2))](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%280%28-2-1%29%2B8%281-%28-2%29%29%2B9%28-2-%28-2%29%29)
![=\frac{1}{2}(0+8(1+2)+9(2-2))](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%280%2B8%281%2B2%29%2B9%282-2%29%29)
![\frac{1}{2}(0 + 0 +24) = \frac{24}{12} = 2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%280%20%2B%200%20%2B24%29%20%3D%20%5Cfrac%7B24%7D%7B12%7D%20%3D%202)
Hence, the area of the triangle is 12 sq units.
Answer:
20
Step-by-step explanation:
Plug in -4 for n:
(-4)^2 - (-4) = 16 + 4 = 20
Answer:
50+30z=250+10z
z=10
Step-by-step explanation:
50+30z=250+10z
Substract 50 on both sides
50+30z-50=250+10z-50
Simplify
30z=10z+200
Substract 10z on both sides
30z-10z=10z+200-10z
20z=200
200/20 = 10
Answer z=10
Answer:
=(-1,-5)
Step-by-step explanation:
y=X1=4
y=-2x-5-2