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== 点と直線の距離の公式 ==
【点と直線の距離の公式】
 点(x0,y0) と直線ax+by+c=0の距離は
ax0+by0+ca2+b2
(式の見方)
#初心者が陥りやすい落とし穴(その1)
 ax+by+c=0だから,分子は0だろうと考えると,それは違う!
 図を見れば分かるように,点(x0,y0) は直線 上にないから,方程式ax0+by0+c=0を満たさない.
 この公式の分子に書いてあるのは,方程式でなく式です.正確に言えば「式の値」です.
 例えば,x+y+1=0 は,直線の方程式で,その直線上にある点(1,2) はこの直線の方程式を満たしますが,直線外の1点(1,0) は,1+0+1=2 となって,直線の方程式を満たしません.それなら,この「式」1+0+1 の「値」2 は,何の役に立つのかと言うと,これが「求める距離」に関係のある数字になっています.ただ,そのままでは距離にはならないので,距離になるように尺度を調整する方法があるのです.
#初心者が陥りやすい落とし穴(その2)
 試験の前の日に,公式の形をぼんやりと覚えただけの生徒がよくやる間違いは,分母に2つの数字があったということから,次のように2つの数字を当てはめる傾向です.
(3,4) と直線2x+y+3=0の距離は
2×3+4+332+42 ←分母が間違い!
 公式をよく見ると,点の座標(x0,y0)ではなく,a,b,c の3つある係数の内で,cの面目を丸つぶれにしてa, bの2つだけ使うようになっているのです.

(具体例とイラストによる解説)
 点(1, 2)と直線x+y−1=0の距離を考えてみます.
 直線x+y−1=0上の点(1, 0)は直線x+y−1=0上にあるから,x+y−1の値は,当然0になります.
 直線x+y−2=0上の点(2, 0)の座標をx+y−1に代入すると,2+0−1=1になります.これは,x+y−2=0 ⇒ x+y−1=1となることからも分かります.この事情は,直線x+y−2=0上の点(1, 1)(0, 2)についても同様で,直線x+y−2=0上の点は,すべてx+y−1の式の値が1になります.
 直線x+y−3=0上の点(3, 0)の座標をx+y−1に代入すると,3+0−1=2になります.これは,x+y−3=0 ⇒ x+y−1=2となることからも分かります. この事情は,直線x+y−3=0上の点(2, 1)(1, 2)についても同様で,直線x+y−3=0上の点は,すべてx+y−1の式の値が2になります.
 直線x+y=0上の点(0, 0)の座標をx+y−1に代入すると,0+0−1=−1になります.これは, x+y=0 ⇒ x+y−1=−1となることからも分かります.この事情は,直線x+y=0上の点(−1, 1)(1, −1)についても同様で,直線x+y=0上の点は,すべてx+y−1の式の値が−1になります.

 以上の考察から,直線x+y−1=0の「上にない」点の座標(x0, y0)を「式」x+y−1に代入しても0にはならないが,直線x+y−1=0からの距離に応じて「平行線の縞模様になる」ことが分かります.そこで,点(x0, y0)と直線x+y−1=0との距離を求めるには,これら平行線の縞模様x+y−1=0, x+y−1=1, x+y−1=2, ... の1目盛り当たりの間隔を掛ければよいことになります.
 右図において点(1, 0)(0, 1)の距離は,1辺の長さが1の正方形の対角線の長さだから,2,茶色で示した1目盛りの間隔は22=12になります.
 そこで,初めに考えた問題:「点(1, 2)と直線x+y−1=0の距離」を求めるには,
 まず,点の座標(1, 2)を直線の方程式の左辺だけを切り出した式x+y−1に代入して「式の値」を求める.
1+2−1=2
次に,この式の値2に縞模様1目盛り当たりの間隔12を掛けて 22=2…(答)


(公式に絶対値が必要な理由)
 前の図において,平面全体は
x+y−1=0, x+y−1=1, x+y−1=2, ...
x+y−1=−1, x+y−1=−2, x+y−1=−3, ...のように,「式」x+y−1の「値」..., −2, −1, 0, 1, 2, 3, ...を持った平行線の縞模様に分けられるが,点と直線の距離は「正または0」の数で考えるので,負の数..., −3, −2, −1などになるところは,向きが逆になっているだけで距離は正の値3, 2, 1に読み替えなければならない.このようにして,ax0+by0+cの値が負になる場合は正の数に直し,正の数の場合はそのまま使うように絶対値を付けると,どんな場合でも使えるようになる.
|ax0+by0+c|
(1目盛りの間隔が1a2+b2となる理由)
 右図のように,ax+byの値が1だけ違うとき,平行な2つの直線の間隔は,次のように求めることができる.
 ax+by=1 (a, b≠0)の直線について,y=0のときx=1ax=0のときy=1bだから,水色の直角三角形の面積は,S=12ab
 これは,斜辺の長さ(1a)2+(1b)2=a2+b2abと茶色で示した間隔Hで求めても等しいはずだから
12ab=S=a2+b22ab×H
H=1a2+b2

(公式のまとめ)
(x0, y0)と直線ax+by+c=0 (a2+b2≠0)の距離は
ax0+by0+ca2+b2

【例題1.1】
 点(1, 2)と直線3x+4y−5=0の距離を求めてください.
(解答)
3×1+4×2532+42=65…(答)
【例題1.2】
 点(2, 0)と直線y=2x+1の距離を求めてください.
(解答)
 直線の方程式を2x−y+1=0の形に変形しておく
2×2+0+122+(1)2=55=5…(答)
【例題1.3】
 原点(0, 0)と直線4x−3y=25の距離を求めてください.
(解答)
 直線の方程式を4x−3y−25=0の形に変形しておく
4×03×02542+(3)2=255=5…(答)
【例題1.4】
 点(4, −5)と直線x=−1の距離を求めてください.
(解答)
 直線の方程式をx+0y+1=0の形に変形しておく
4+0×(5)+112+02=5…(答)

【問題1】 (選択肢の中から正しいものを1つクリック)
※解答すれば,解説が出ます.
(1)
 点(3, −4)と直線2x−y−5=0の距離を求めてください.
15 1 5 5
(2)
 原点(0, 0)と直線3x+4y=5の距離を求めてください.
15 1 5 5

(3)
 点(14,13)と直線x3y4=1の距離を求めてください.
15 1 2 5
(4)
 点(1, 0)と直線y=12x+2の距離を求めてください.
15 1 5 5

(平行な2直線間の距離)
【例題2.1】
 平行な2直線3x+4y+1=03x+4y+2=0の距離を求めてください.
(解答)
 直線3x+4y+1=0上の1点(1, −1)3x+4y+2=0の距離を求める.
3×1+4×(1)+232+42=15 …(答)
一般に,平行な2直線ax+by+c=0ax+by+d=0の距離は
dca2+b2…(*)
となることが証明できます.
(証明)
 直線ax+by+c=0上の1点を(x0, y0)とおくと
ax0+by0+c=0…(1)
が成り立つ.
 このとき,点(x0, y0)と直線ax+by+d=0の距離は
ax0+by0+da2+b2…(2)
ここで,(1)により,ax0+by0=−cを(2)に代入すると
dca2+b2(証明終わり)
 以上のように証明できますが,例題2.1のように一方の直線上の点を適当に見つけて,点と直線の距離の公式を適用すれば解けるので,この(*)は覚えるまでもないでしょう.
 一般に「教科書に書かれている公式は黙って使ってもよい」が「教科書に書かれていない公式を黙って使うと減点されることがある」「使いたければ証明してから使う方がよい」と考えられるので,(*)は結果が出てから目で確認する程度に使うとよいでしょう.

【問題2】 (選択肢の中から正しいものを1つクリック)
※解答すれば,解説が出ます.
(1)
 平行な2直線3x−4y−2=03x−4y+1=0の距離を求めてください.
15 35 5 3
(2)
a, b≠0のとき,平行な2直線xa+yb=1xa+yb=0の距離を求めてください.

(3)
 3点A(−1, 1), B(2, −2), C(3, 3)を頂点とする△ABCの面積を求めてください.
32 23 9 18
(4)
 3点A(−1, 2), B(1, −3), C(2, 4)を頂点とする△ABCの面積を求めてください.

(数学Bのベクトルを使った証明)
 ※各々の教科書には,数学Ⅱで習う範囲を逸脱しないように気を付けながら,無駄なく正確な証明が書かれていますが,結構長い証明になっており,それらの数式変形を目で追っていくのは苦痛を伴います.
 特に,どこに連れて行くのか,今どこにいるのか,その変数は何を表すのかが必ずしも明らかでないとつらい.(生徒も)
 ここでは数学Bのベクトルを使って,もっと視覚的に分かりやすい方法で「点と直線の距離の公式」を証明することを考える.その延長でさらに,3次元空間における「点と平面の距離の公式」も示すことを目指す.
 直線ax+by+c=0の法線ベクトル(垂直なベクトル)n=(a,b)をその大きさa2+b2で割ると単位ベクトル(長さが1のベクトル)になる.
nn=(aa2+b2,ba2+b2)
 そこで,点(x0, y0)から,その単位ベクトルの何倍を継ぎ足せば直線ax+by+c=0上にあるかを調べたらよい.この倍率tが求める距離になる
x=x0+aa2+b2t
y=y0+ba2+b2t
(x, y)が,直線ax+by+c=0上にあるためには
a(x0+aa2+b2t)+b(y0+ba2+b2t)+c=0
a2a2+b2t+b2a2+b2t+ax0+by0+c=0
(a2+b2)t+(ax0+by0+c)a2+b2=0
t=(ax0+by0+c)a2+b2a2+b2=ax0+by0+ca2+b2
 距離は正の数だから
ax0+by0+ca2+b2

【3次元空間における点と平面の距離】
 3次元空間における点(x0, y0, z0)と平面ax+by+cz+d=0の距離は
ax0+by0+cz0+da2+b2+c2
(証明)
 平面ax+by+cz+d=0の法線ベクトルn=(a,b,c)をその大きさa2+b2+c2で割ると単位ベクトルになる.
nn=(aa2+b2+c2,ba2+b2+c2,ca2+b2+c2)
 そこで,点(x0, y0, z0)から,その単位ベクトルの何倍を継ぎ足せば平面ax+by+cz+d=0上にあるかを調べたらよい.この倍率tが求める距離になる
x=x0+aa2+b2+c2t
y=y0+ba2+b2+c2t
z=z0+ca2+b2+c2t
(x, y, z)が,平面ax+by+cz+d=0上にあるためには
a(x0+aa2+b2+c2t)+b(y0+ba2+b2+c2t)
+c(z0+ca2+b2+c2t)+d=0
a2a2+b2+c2t+b2a2+b2+c2t+c2a2+b2+c2t
+ax0+by0+cz0+d=0
(a2+b2+c2)t+(ax0+by0+cz0+d)a2+b2+c2=0
t=(ax0+by0+cz0+d)a2+b2+c2a2+b2+c2
=ax0+by0+cz0+da2+b2+c2
 距離は正の数だから
ax0+by0+cz0+da2+b2+c2

【問題3】 (選択肢の中から正しいものを1つクリック)
※解答すれば,解説が出ます.
(1)
 3次元空間において,点(4, 1, 2)と平面2x−2y+z+4=0の距離(点から平面に引いた垂線の長さ)を求めてください.
1 2 3 4
(2)
 3次元空間において,原点(0, 0, 0)と平面2x−3y+6z+7=0の距離(原点から平面に引いた垂線の長さ)を求めてください.
1 2 3 4

 平面上の平行の2直線の場合と同様に,3次元空間の平行な2平面の間の距離について,次の式を証明することができます.
ただし,直線の場合と同様で,丸暗記して使うよりは一方の平面上の点と他方の平面との距離を求める方が,答案としての好感度は上でしょう.
 平行な2平面ax+by+cz+d=0ax+by+cz+e=0の距離は
eda2+b2+c2…(*)
(証明)
 平面ax+by+cz+d=0上の1点を(x0, y0, z0)とおくと
ax0+by0+cz0+d=0…(1)
が成り立つ.
 このとき,点(x0, y0, z0)と平面ax+by+cz+e=0の距離は
ax0+by0+cz0+ea2+b2+c2…(2)
ここで,(1)により,ax0+by0+cz0=−dを(2)に代入すると
eda2+b2+c2(証明終わり)

【問題4】 (選択肢の中から正しいものを1つクリック)
※解答すれば,解説が出ます.
(1)
 3次元空間において,平行な2つの平面4x+2y5z5=04x+2y5z+5=0の距離を求めてください.
1 2 3 4
(2)
 3次元空間において,平行な2つの平面2x+3y3z+4=02x+3y3z8=0の距離を求めてください.
1 2 3 4

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