【高校数学Ⅱ】3-1-4 2直線の関係｜要点まとめ

\(y=-3x+9\)は傾きが\(-3\)なので、平行な直線の傾きは\(-3\)
\(y-1=-3(x+2)\)
\(y=-3x-5\)

\(\displaystyle y=\frac{1}{5}x+\frac{1}{5}\)
傾きが\(\displaystyle \frac{1}{5}\)なので、平行な直線の傾きは\(\displaystyle \frac{1}{5}\) \(\displaystyle y-3=\frac{1}{5}(x-2)\)
\(\displaystyle y=\frac{1}{5}x+\frac{13}{5}\)

\(y=-3x+9\)は傾きが\(-3\)なので、垂直な直線の傾きは\(\displaystyle \frac{1}{3}\)
\(\displaystyle y-1=\frac{1}{3}(x+2)\)
\(\displaystyle y=\frac{1}{3}x+\frac{5}{3}\)

\(\displaystyle y=\frac{1}{5}x+\frac{1}{5}\)
傾きが\(\displaystyle \frac{1}{5}\)なので、垂直な直線の傾きは\(-5\) \(y-3=-5(x-2)\)
\(y=-5x+13\)

点\((-2,3)\)と直線\(3x+y-2=0\)の距離\(h\)は
\(\displaystyle h=\frac{|3・(-2)+1・3-2|}{\sqrt{3^2+1^2}}\)
\(\displaystyle \ \ =\frac{5}{\sqrt{10}}\)
\(\displaystyle \ \ =\frac{\sqrt{10}}{2}\)

点\((1,2)\)と直線\(4x-3y-5=0\)の距離\(h\)は
\(\displaystyle h=\frac{|4・1-3・2-5|}{\sqrt{4^2+(-3)^2}}\)
\(\displaystyle \ \ =\frac{7}{\sqrt{25}}\)
\(\displaystyle \ \ =\frac{7}{5}\)

点\((0,0)\)と直線\(2x+y-4=0\)の距離\(h\)は
\(\displaystyle h=\frac{|2・0+0-4|}{\sqrt{2^2+1^2}}\)
\(\displaystyle \ \ =\frac{4}{\sqrt{5}}\)
\(\displaystyle \ \ =\frac{4\sqrt{5}}{5}\)