Question

直线y=-4/3-4与x轴交于点a,与y轴交于点b,点p在直线ab上且△poa为等腰三角形,求点p坐标

Answer 1

易得A(-3,0),B(0,-4),OA=3,OB=4,∴AB=5,

过P作PQ⊥X轴于Q,则ΔAPQ∽ΔABO,

⑴PO=PA,则OQ=AQ=3/2,

∴PQ是ΔOAB的中位线,∴PQ=1/2OB=2,∴P1(-3/2,-2),

⑵AO=AP=3,PQ/OB=AP/AB=AQ/AO,PQ=3×4/5=12/5,AQ=9/5,

∴OQ=6/5,∴P2(-6/5,-12/5),

⑶OA=OP=3,∵P在直线Y=-4/3X-4上,设P(m,-4/3m-4),

则m^2+(-4/3m-4)^2=3^2,

25m^2+96m+63=0,

m=(-96±54)/50,m=-3(就是A点舍去),m=-21/25,

∴P3(-21/25,-93/25).