What is the area of a triangle with vertices at (0, −2) ,(8, −2) , and ​ (9, 1) ?Enter your answer in the box( there is no provided image )

The area of a triangle with vertices at  (0, −2) ,(8, −2) and ​(9, 1) is 12 square unitsSolution: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. The area of triangle when vertices are given is:[tex]\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|[/tex][tex]\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. }[/tex]Here in our problem,[tex]\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)[/tex]Now, substitute the above values in the formula:[tex]\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|[/tex][tex]\begin{array}{l}{=\frac{1}{2}|0(-2-1)+8(1-(-2))+9(-2-(-2))|} \\\\ {=\frac{1}{2}|0+8(1+2)+9(2-2)|} \\\\ {=\frac{1}{2}|0+24+0|} \\\\ {=\frac{24}{2}} \\\\ {=12 \text { square units }}\end{array}[/tex]Hence, the area of the triangle is 12 square units.