Question

已知A（-2,3）和B（1,4）,在y轴上找一点C,使三角形ABC为直角三角形.

Answer 1

因为AB,BC,AC都可能为斜边,所以分三种情况来解答:
设C点的坐标为C(0,Y),AB,BC及AC边可求,
AB^2=(-2-1)^2+(3-4)^2=10
BC^2=(1-0)^2+(4-Y)^2=Y^-8Y+17
AC^2=(-2-0)^2+(3-Y)^2=Y^2-6Y+13
若AB为直角三角形的斜边,
AB^2=(-2-1)^2+(3-4)^2=10
(-2-0)^2+(3-Y)^2+(1-0)^2+(4-Y)^2=10
化简得,
Y^2-7Y+10=0
即
(Y-2)(Y-5)=0
Y=2或Y=5
故当AB为斜边时,C点坐标为C(0,2)或C(0,5)
若BC为斜边,则:
AC^2+AB^2=BC^2
Y^2-6Y+13+10=Y^2-8Y+17
Y=-3
故当BC为斜边是C点坐标为C(0,-3)
若AC为斜边,则
AB^2+BC^2=AC^2
10+Y^2-8Y+17=Y^2-6Y+13
Y=7
故当BC为斜边是C点坐标为C(0,7)
所以C点的坐标为:
C(0,2),C(0,5),C(0,-3),C(0,7)