SOLUTION
(57) We want to find the length of each side of the triangle with sides
![P_1(2,1),P_2(-4,1),P_3(-4,-3)](https://img.qammunity.org/2023/formulas/mathematics/college/eff45mtthex6dtokyuet6n6ao9cyuvw7b9.png)
Using the distance formula we have
![d = √((x_2 - x_1)^2 + (y_2-y_1)^2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/u032ksae2ro1cpvb938kks45mkvnbkmxi0.png)
Between P1 and P2, we have
![\begin{gathered} d=√((-4-2)^2+(1-1)^2) \\ d=√((-6)^2) \\ P_1P_2=√(36)=6\text{ units } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vlb92tdwopztuwtij1vzyka1nvkzuum3ys.png)
Between P2 and P3, we have
![\begin{gathered} P_2P_3=√((x_2 - x_1)^2 + (y_2-y_1)^2) \\ P_2P_3=√(\left(-4-(-4)\right)^2+\left(-3-1\right)^2) \\ =√(0+(-4)^2) \\ =√(16)=4\text{ units } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9mgi0m259ogjoaugx35ldpo2fyr6ny5mtm.png)
Between P1 and P3, we have
![\begin{gathered} P_1P_3=√(\left(-4-2\right)^2+\left(-3-1\right)^2) \\ =√((-6)^2+(-4)^2) \\ =√(36+16) \\ =√(52)\text{ units } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sgsa4nr27dp6wotdn2gpe6fbb4wcbh7610.png)
Now looking at sides P1P2 = 6 units and P2P3 = 4 units we can see that
![\begin{gathered} √(6^2+4^2) \\ =√(36+16) \\ √(52)=P_1P_3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cvciunz2iml4ytdv2c38vh3iovxbgjav2m.png)
Hence, this is a right-angle triangle because it is in accordance with Pythagorean theorem. P1P3 is the hypotenuse