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== 2点を通る直線 ==

■解説
 2つの点の座標が与えられているとき,これら2点を通る直線の式(1次関数の式)は,次の方法で求めることができる.

 (I) 求める1次関数の式(直線の式)を y=ax+b とおく.
 (II) この式の x , y に2つの点の座標を代入して,a , b の連立方程式を作る.
 (III) a , b を求めて,y=ax+b の形で答える.


※  y=2x+1 のような式を,「1次関数の式」「1次関数を表わす式」「直線の式」「直線の方程式」という.

※ 次の例1,2,3において,x , y に値を代入すると, a , b の式になることに注意.( x , y の連立方程式になるのではない. )  
【例1】
 yx の1次関数で,そのグラフが2点 A(−1 ,−3) , B(2 , 3) を通るとき,この1次関数の式を求めなさい.

(答案)
 求める1次関数の式を y=ax+b とおく.
 点 A(−1 ,−3) を通るから x=−1 , y=−3 を代入して,
__________−3=−a+b …(1)
 点 B(2 , 3) を通るから x=2 , y=3 を代入して,
__________3=2a+b …(2)
(1)−(2) より −6=−3a
__________a=2
これを(1)に代入して b=−1
ゆえに,y=2x−1 …(答)
【例2】
 yx の1次関数で,そのグラフが2点 A(1 , 2) , B(4 ,−4) を通るとき,この1次関数の式を求めなさい.

(答案)
 求める1次関数の式を y=ax+b とおく.
A(1 , 2) , B(4 ,−4) を通るから
2=a+b …(1)
−4=4a+b …(2)
(1)−(2) より 6=−3a
______________a=−2
これを(1)に代入して 2=−2+b
______________b=4
ゆえに,y=−2x+4 …(答)  
【例3】  
 yx の1次関数で,そのグラフが2点 A(−1 , 2) , B(2 ,−3) を通るとき,この1次関数の式を求めなさい.

(答案)
 求める1次関数の式を y=ax+b とおく.
A(−1 , 2) , B(2 ,−3) を通るから
2=−a+b …(1)
−3=2a+b …(2)
(1)−(2) より 5=−3a
______________a=−.53n
これを(1)に代入して 2= .53n+b
______________b= .13n
ゆえに,y=−.53nx+.13n …(答)

※暗算ではできないので,計算用紙が必要
(各々2~3問あり、計7問なので注意)
問題1 
[ 第1問 / 全3問中 ] [ 採点する ] [ Help ] [ 次の問題 ]
(1) yx の1次関数で,そのグラフが2点 A(−1 , 1) , B(2 ,−5) を通るとき,この1次関数の式を求めなさい.

_______y=−x−
問題2 
[ 第1問 / 全2問中 ] [ 採点する ] [ Help ] [ 次の問題 ]
(1) 2点 A(1 , 4), B(3 , 2) を通る直線と x 軸との交点の座標を求めなさい.

____________________________( , )
問題3 
[ 第1問 / 全2問中 ] [ 採点する ] [ Help ] [ 次の問題 ]
(1) 3点 A(3 , 4) , B(0 ,−2) , C(5 , a) が同一直線上にあるように定数 a の値を定めなさい.


____________________________a=

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