Question

find the area of triangle with vertices A (at₁² ,2at₁) B(at²​₂,2at₂) C(2at²_​3 , 2at₂)

Answer 1

$$\begin{gathered} \mathrm{Using}\mathrm{the}\mathrm{formula}, \\ \mathrm{Area}=\frac{1}{2}\left|\left[{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right]\right| \\ =\frac{1}{2}\left|\left[{{\mathrm{at}}_1}^2\left(2{\mathrm{at}}_2-2{\mathrm{at}}_3\right)+{{\mathrm{at}}_2}^2\left(2{\mathrm{at}}_3-2{\mathrm{at}}_1\right)+{{\mathrm{at}}_3}^2\left(2{\mathrm{at}}_1-2{\mathrm{at}}_1\right)\right]\right| \\ =\mathrm{a}\left[{{\mathrm{t}}_1}^2\left({\mathrm{t}}_2-{\mathrm{t}}_3\right)+{{\mathrm{t}}_2}^2\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)+{{\mathrm{t}}_3}^2\left({\mathrm{t}}_1-{\mathrm{t}}_1\right)\right] \\ \mathrm{Now}\mathrm{you}\mathrm{can}\mathrm{simplify}\mathrm{more}\mathrm{for}\mathrm{your}\mathrm{own}\mathrm{practice}. \end{gathered}$$