試求x軸上求一點P,使其到兩定點A(-3,2),B(4,-5)距離相等 Sol 設P(x.0) AP^2=(x+3)^2+2^2=x^2+6x+13 BP^2=(x-4)^2+5^2=X^2-8x+41 AP^2=BP^2 x^2+6x+13=x^2-8x+41 14x=28 x=2 P(2,0) or L1:AB直線方程式:(y-2)/(x+3)/(-5-2)/(4+3)=-1 x+3=-y+2 x+y=-1 AB中點C(0.5,-1.5) L2過C垂直AB直線 L2直線方程式:x-y=2 L3與x軸交於(2,0) P(2,0)
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