直线y=2x+12分别交x轴、y轴于A(-6,0)、B(0,12),
P(m,6),△PAB是等腰直角三角形,
1）PA=PB,PA⊥PB,
(m+6)^2+36=m^2+36,且(m+6)^2+36+m^2+36=36+144,
整理得12m+36=0,且2m^2+12m-108=0,
m=-3,不满足m^2+6m-54=0.
2）AP=AB,AP⊥AB,
(m+6)^2+36=36+144,且(m+6)^2+36+36+144=36+m^2,
化简得 m+6=土12,且12m=-216,
解得m=-18.P(-18,6).
3)BP=BA,BP⊥BA,留给您练习.