【高校数学Ⅱ】3-1-2 平面上の点｜問題集

\(AB=\sqrt{(4-1)^2+(6-2)^2}\)
\(\ \ \ \ \ \ =\sqrt{3^2+4^2}\)
\(\ \ \ \ \ \ =\sqrt{25}\)
\(\ \ \ \ \ \ =5\)

\(AB=\sqrt{(2-(-3))^2+(-4-1)^2}\)
\(\ \ \ \ \ \ =\sqrt{5^2+5^2}\)
\(\ \ \ \ \ \ =\sqrt{50}\)
\(\ \ \ \ \ \ =5\sqrt{2}\)

\(AB=\sqrt{(5-3)^2+(-2-(-2))^2}\)
\(\ \ \ \ \ \ =\sqrt{2^2+0^2}\)
\(\ \ \ \ \ \ =\sqrt{4}\)
\(\ \ \ \ \ \ =2\)

\(AB=\sqrt{2^2+(-3)^2}\)
\(\ \ \ \ \ \ =\sqrt{13}\)

\(P(0,y)\)とおく。
\(AP=\sqrt{(-4)^2+(y-2)^2}\)
\(BP=\sqrt{1^2+(y-(-1))^2}\)
\(AP=BP\)より、
\(16+(y-2)^2=1+(y+1)^2\)
\(6y=18\)
\(y=3\)
よって、
\(P(0,3)\)

\(\displaystyle \left(\frac{0+7}{2},\frac{-6+0}{2}\right)\)
\(\displaystyle =\left(\frac{7}{2},\frac{-6}{2}\right)\)
よって、\(\displaystyle M\left(\frac{7}{2},-3\right)\)

\(\displaystyle \left(\frac{4・0+3・7}{3+4},\frac{4・(-6)+3・0}{3+4}\right)\)
\(\displaystyle =\left(\frac{21}{7},\frac{-24}{7}\right)\)
よって、\(\displaystyle P\left(3,-\frac{24}{7}\right)\)

\(\displaystyle \left(\frac{4・0-3・7}{-3+4},\frac{4・(-6)-3・0}{-3+4}\right)\)
\(\displaystyle =\left(\frac{-21}{1},\frac{-24}{1}\right)\)
よって、\(Q(-21,-24)\)

\(Q(x,y)\)とおくと、点\(A\)は
\(\displaystyle \left(\frac{x}{2},\frac{y-4}{2}\right)=(-3,2)\)
\(\left\{\begin{array}{l}\displaystyle \frac{x}{2}=3 \\ \displaystyle \frac{y-4}{2}=2\end{array}\right.\)
これを解くと、
\(x=-6,y=8\)
よって、
\(Q(-6,8)\)

\(\displaystyle \left(\frac{1+5+3}{3},\frac{1+2+4}{3}\right)\)
\(\displaystyle =\left(3,\frac{7}{3}\right)\)
よって、\(\displaystyle G\left(3,\frac{7}{3}\right)\)

\(\displaystyle \left(\frac{-1+2+5}{3},\frac{3-1+0}{3}\right)\)
\(\displaystyle =\left(2,\frac{2}{3}\right)\)
よって、\(\displaystyle G\left(2,\frac{2}{3}\right)\)