【高校数学Ⅱ】3-1-1 直線上の点｜問題集

\(AB=|1-(-3)|\)
\(\ \ \ \ \ \ =4\)

\(BC=|5-1|\)
\(\ \ \ \ \ \ =4\)

\(\displaystyle \frac{-3+1}{2}\)
\(\displaystyle =\frac{-2}{2}\)
\(=-1\)
よって、\(M(-1)\)

\(\displaystyle \frac{1・(-3)+2・5}{2+1}\)
\(\displaystyle =\frac{7}{3}\)
よって、\(\displaystyle P\left(\frac{7}{3}\right)\)

\(\displaystyle \frac{(-1)・1+3・5}{3-1}\)
\(\displaystyle =\frac{14}{2}\)
\(=7\)
よって、\(Q(7)\)

\(\displaystyle \frac{2・(-5)+1・2}{1+2}\)
\(\displaystyle =\frac{-8}{3}\)
よって、\(\displaystyle P\left(-\frac{8}{3}\right)\)

\(\displaystyle \frac{-5+2}{2}\)
\(\displaystyle =\frac{-3}{2}\)
よって、\(\displaystyle M\left(-\frac{3}{2}\right)\)

\(\displaystyle \frac{(-2)・(-5)+1・2}{1-2}\)
\(\displaystyle =\frac{12}{-1}\)
\(=-12\)
よって、\(Q(-12)\)

\(\displaystyle \frac{1・(-10)+2・(-1)}{2+1}\)
\(\displaystyle =\frac{-12}{3}\)
\(=-4\)
よって、\(P(-4)\)

\(\displaystyle \frac{-10+(-1)}{2}\)
\(\displaystyle =\frac{-11}{2}\)
よって、\(\displaystyle M\left(-\frac{11}{2}\right)\)

\(\displaystyle \frac{(-1)・(-10)+2・(-1)}{2-1}\)
\(\displaystyle =\frac{8}{1}\)
\(=8\)
よって、\(Q(8)\)