Question

已知A(1,0)B(3,2)C(0,4),点D满足AB⊥CD,且AD平行BC,试求点D的坐标？要过程~

Answer 1

用斜率解
# 两线平行当且仅当其斜率相等；
# 两线垂直当且仅当其斜率之积为-1；

设点D(x,y)

AB⊥CD:
[(2-0)/(3-1)]*[(y-4)/(x-0)]=-1
(2/2)*[(y-4)/x]=-1
(y-4)/x=-1
y-4=-x
y+x=4

AD平行BC:
[(y-0)/(x-1)]=[(4-2)/(0-3)]
y/(x-1)=(2/-3)
y=(2/-3)*(x-1)
-3y=2(x-1)
-3y=2x-2
2x+3y=2

解二元一次方程联立
y+x=4...........①
2x+3y=2.......②

①移项得
x=4-y
代入②得
2*(4-y)+3y=2
8-2y+3y=2
y=-6
代入x=4-y得
x=4-(-6)
x=10

故点D=(10,-6)