题目内容
在直角坐标系中,已知两点A
、B
以及动点C
、D
,则当以点A、 B、C 、D为顶点的四边形的周长最小时,比值
为 .
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试题分析:先根据两点间的距离公式求出AB的值,再过点B作关于y轴的对称点B′,过点A作关于x轴的对称点A′,连接A′B′分别交x、y轴于点D、C,由两点之间线段最短可知线段A′B′即为四边形ABCD的周长最小值,用待定系数法求出过A′B′两点的直线解析式,即可求出C、D的坐标.
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∴四边形ABCD周长=AB+BC+CD+AD=2
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∴求其周长最小值,就是求BC+CD+AD的最小值.过B作y轴对称点B′(4,5),
则BC=B′C,
过A作x轴对称点A′(-8,-3),则AD=A′D
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∴BC+CD+AD=B′C+CD+A′D≥A′B′
即A′、D、C、B′四点共线时取等号
可求出相应的C、D坐标,
设直线A′B′的方程是y=kx+b(k≠0),
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点评:根据对称的性质作出A、B的对称点A′、B′及求出其坐标是解答此题的关键.
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