答案
解:A(0,-2),B(-2,1),C(3,2),
(1)由勾股定理得:
AB=
=,
BC=
=,
AC=
=5;
(2)由已知得A′(0,-4),B′(-4,2),C′(6,4),
由勾股定理得:
A′B′=
=2,
B′C′=
=2,
A′C′=
=10;
(3)∵
===,
∴这六条线段成比例;
(4)根据三角形三边长可以判定△ABC∽△A′B′C′,
所以形状相同.
解:A(0,-2),B(-2,1),C(3,2),
(1)由勾股定理得:
AB=
=,
BC=
=,
AC=
=5;
(2)由已知得A′(0,-4),B′(-4,2),C′(6,4),
由勾股定理得:
A′B′=
=2,
B′C′=
=2,
A′C′=
=10;
(3)∵
===,
∴这六条线段成比例;
(4)根据三角形三边长可以判定△ABC∽△A′B′C′,
所以形状相同.