SOLUTION
(57) We want to find the length of each side of the triangle with sides
[tex]P_1(2,1),P_2(-4,1),P_3(-4,-3)[/tex]
Using the distance formula we have
[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}[/tex]
Between P1 and P2, we have
[tex]\begin{gathered} d=\sqrt{(-4-2)^2+(1-1)^2} \\ d=\sqrt{(-6)^2} \\ P_1P_2=\sqrt{36}=6\text{ units } \end{gathered}[/tex]
Between P2 and P3, we have
[tex]\begin{gathered} P_2P_3=\sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2} \\ P_2P_3=\sqrt{\left(-4-(-4)\right)^2+\left(-3-1\right)^2} \\ =\sqrt{0+(-4)^2} \\ =\sqrt{16}=4\text{ units } \end{gathered}[/tex]
Between P1 and P3, we have
[tex]\begin{gathered} P_1P_3=\sqrt{\left(-4-2\right)^2+\left(-3-1\right)^2} \\ =\sqrt{(-6)^2+(-4)^2} \\ =\sqrt{36+16} \\ =\sqrt{52}\text{ units } \end{gathered}[/tex]
Now looking at sides P1P2 = 6 units and P2P3 = 4 units we can see that
[tex]\begin{gathered} \sqrt{6^2+4^2} \\ =\sqrt{36+16} \\ \sqrt{52}=P_1P_3 \end{gathered}[/tex]
Hence, this is a right-angle triangle because it is in accordance with Pythagorean theorem. P1P3 is the hypotenuse
