Grade

Type of questions

Mathematics Senior High

(2)の問いについてです。 定点となるMを右の写真の解のような形で表してはいけないのでしょうか。ダメな理由も教えていただけるとありがたいです

Check 例題 360 直線のベクトル方程式(1)円3 07*** (1) 異なる2点A(a),B() に対して, p=(1-t)+t6 (1) 表される図形はどのような図形か. (2) 3点A(a),B(b),C(c) を頂点とする △ABC がある. 辺ACを 21 に内分する点M () を通り,辺ABに平行な直線のベクトル 方程式をa, 6, こと媒介変数を用いて表せ 考え方 (1) ja+(-a) と変形すると,点P(j) は点Aを通り, ABに平行な直線上にあ ることがわかる (2)M(m)を通り、ABに平行な直線のベクトル方程式は,p=m+tAB と表せる。 解答 (1) = (1-1)+16=a+1(-a) 点P()は,点Aを通り b=a+1(6-9) 1 変化する 定点 A1=0 6-d=ABに平行な直線, すなわち直線AB上を動き, b-a a t=0 のとき, = より, 点Aの位置 t=1 のとき, = より,点Bの位置 t=1 B tが0から1まで変 わるとき、点Pは点 にある。 よって、求める図形は, 線分AB である. AからABの向き (2) 求める直線上の任意の点をP() とする.点M(㎡) に, Bまで動く。 a+2c は,辺ACを2:1 に内分する点だから, m= 3 求める直線は辺AB と平行だから,その方向ベクト ルは, AB (S-C A(a) よって,=m+tAB=+2c+(-a) P(p) (M(m) 3 すなわち, = (1/31) a1+1+1/2/30 B(b) c(c) AB JS Focus 点A(a)を通り, d に平行な直線のベクトル方程式は, p=a+td 2点A(a),B(b) を通る直線のベクトル方程式は, b=(1-t)a+tb とくに, t のとき, 線分AB を表す 足して1

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English Junior High

1.(1)②、(2)②、(3)①、(4)③④⑧⑩、(5)③④⑤、(6)③④、(7)①④⑥、(8)①②③⑥、(9)の解説をして欲しいです。3枚目が答えです

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