解：?$(1)$?∵該函數(shù)是正比例函數(shù)，∴常數(shù)項(xiàng)必須為?$0$?且一次項(xiàng)系數(shù)不為?$0$?
即?$2k - 2 = 0$?且?$1 - 3k \neq 0，$?∴?$k = 1$?
?$(2)$?當(dāng)?$k = 0$?時(shí)，函數(shù)為?$y = (1 - 0)x + 0 - 2 = x - 2$?
∵一次項(xiàng)系數(shù)?$1 ＞ 0，$?∴?$y$?隨?$x$?的增大而增大
又∵?$x_{1} ＜ x_{2}，$?∴?$y_{1} ＜ y_{2}$?
