题目内容
12.设m,n∈N,若A(m,0),B(0,n),C(1,3)三点共线,则m,n的值是$\left\{\begin{array}{l}{n=6}\\{m=2}\end{array}\right.$,$\left\{\begin{array}{l}{n=4}\\{m=4}\end{array}\right.$.
分析 对m,n分类讨论,利用三点共线与斜率的关系即可得出.
解答 解:①当m=0时,A,B,C三点不共线,舍去;
②当m=1时,A,B,C三点不共线,舍去;
③当m≠0,1时,
kAB=$\frac{n}{-m}$,kCA=$\frac{3}{1-m}$.
∵A,B,C三点共线,
∴$\frac{n}{-m}$=$\frac{3}{1-m}$,
当n=3时,m不存在,舍去.
∴n≠3,
化为$m=\frac{n}{n-3}$=1+$\frac{3}{n-3}$为自然数,
可得n=0,4,2,6.
而n=0,2,时,舍去.
∴$\left\{\begin{array}{l}{n=6}\\{m=2}\end{array}\right.$,$\left\{\begin{array}{l}{n=4}\\{m=4}\end{array}\right.$.
综上可得:$\left\{\begin{array}{l}{n=6}\\{m=2}\end{array}\right.$,$\left\{\begin{array}{l}{n=4}\\{m=4}\end{array}\right.$.
故答案为:$\left\{\begin{array}{l}{n=6}\\{m=2}\end{array}\right.$,$\left\{\begin{array}{l}{n=4}\\{m=4}\end{array}\right.$.
点评 本题考查了三点共线与斜率的关系、分类讨论方法,考查了推理能力与计算能力,属于中档题.
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