建立平面直角坐标系,如图所示,则B(4,0),E(4,2),F(2,4),A(0,0).
设直线AE,BF的斜率分别为kAE,kBF,
则kAE=2/4=1/2,kBF=(4"-" 0)/(2"-" 4)=-2.
于是kAE·kBF=1/2×(-2)=-1,
故BF⊥AE.
11.导学号57084076在x轴上求一点P,使得
(1)P到A(4,1)和B(0,4)的距离之差最大,并求出最大值;
(2)P到A(4,1)和C(3,4)的距离之和最小,并求出最小值.
解(1)如图,设直线BA与x轴交于点P,此时P为所求点,
且|PB|-|PA|=|AB|=√("(" 0"-" 4")" ^2+"(" 4"-" 1")" ^2 )=5.∵直线BA的斜率kBA=(1"-" 4)/4=-3/4,
∴直线BA的方程为y=-3/4x+4.
令y=0,得x=16/3,即P(16/3 "," 0).故距离之差最大值为5,此时P点的坐标为(16/3 "," 0).
(2)作A关于x轴的对称点A',则A'(4,-1),连接CA',则|CA'|为所求最小值,直线CA'与x轴交点为所求点.
又|CA'|=√("(" 4"-" 3")" ^2+"(-" 1"-" 4")" ^2 )=√26,
直线CA'的斜率kCA'=("-" 1"-" 4)/(4"-" 3)=-5,
则直线CA'的方程为y-4=-5(x-3).
令y=0,得x=19/5,即P(19/5 "," 0).
故距离之和最小值为√26,此时P点的坐标为(19/5 "," 0).