(1) 解:
由于點(diǎn)P在過點(diǎn)A(-2,3)且與x軸平行的直線上,所以點(diǎn)P的縱坐標(biāo)與點(diǎn)A的縱坐標(biāo)相同,即
$m - 1 = 3$
解得:
$m = 4$
將$m=4$代入$2m+4$得:
$2m + 4 = 12$
所以,點(diǎn)P的坐標(biāo)為$(12,3)。$
(2) 解:
由于點(diǎn)P到x軸的距離是1,所以
$|m - 1| = 1$
解得:
$m = 2$ 或 $m = 0$
當(dāng)$m=2$時,代入$2m+4$得$2m + 4 = 8,$此時點(diǎn)P的坐標(biāo)為$(8,1);$
當(dāng)$m=0$時,代入$2m+4$得$2m + 4 = 4,$此時點(diǎn)P的坐標(biāo)為$(4,-1)。$
所以,點(diǎn)P的坐標(biāo)為$(8,1)$或$(4,-1)。$
(3) 解:
由于點(diǎn)P到x軸、y軸的距離相等,所以
$|2m + 4| = |m - 1|$
解得:
$2m + 4 = m - 1$ 或 $2m + 4 = -(m - 1)$
即:
$m = -5$ 或 $m = -1$
當(dāng)$m=-5$時,代入$2m+4$和$m-1$得$2m + 4 = -6,$$m - 1 = -6,$此時點(diǎn)P的坐標(biāo)為$(-6,-6);$
當(dāng)$m=-1$時,代入$2m+4$和$m-1$得$2m + 4 = 2,$$m - 1 = -2,$此時點(diǎn)P的坐標(biāo)為$(2,-2)。$
所以,點(diǎn)P的坐標(biāo)為$(-6,-6)$或$(2,-2)。$