Question

(04.01 mc)

Answer 1

Part A

By the distance formula,

$rs=\sqrt{(4-2)^2 + (4-3)^2}=\sqrt{5}\\\\st=\sqrt{(4-5)^2 + (4-0)^2}=\sqrt{17}\\\\rt=\sqrt{(2-5)^2 + (3-0)^2}=3\sqrt{2}$

Part B

$m_{rs}=\frac{4-3}{4-2}=\frac{1}{2}\\\\m_{rt}=\frac{3-0}{2-5}=-1\\\\m_{st}=\frac{4-0}{4-5}=-4$

Part C

The triangle is scalene because each side has a different length.