Bài 2. PHƯƠNG TRÌNH ĐƯỜNG TRÒN

Hikari Dorabase
Xem chi tiết
Trung Nguyen
28 tháng 3 2020 lúc 21:29

gt->\(A\left(\frac{9}{4};-\frac{1}{4}\right)\)

Giả sử B(b;2-b); C(c;\(\frac{3-2c}{6}\))

Có:\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AM}\)

\(\overrightarrow{AB}\left(b-\frac{9}{4};2-b+\frac{1}{4}\right)\rightarrow\overrightarrow{AB}\left(b-\frac{9}{4};\frac{5}{4}-b\right)\)

\(\overrightarrow{AC}\left(c-\frac{9}{4};\frac{3-2c}{6}+\frac{1}{4}\right)\rightarrow\overrightarrow{AC}\left(c-\frac{9}{4};\frac{3}{4}-\frac{1}{3}c\right)\)

\(\overrightarrow{AM}\left(-1-\frac{9}{4};1+\frac{9}{4}\right)\rightarrow\overrightarrow{AM}\left(-\frac{13}{4};\frac{13}{4}\right)\)

\(\rightarrow\left\{{}\begin{matrix}b-\frac{9}{4}+c-\frac{9}{4}=2.\frac{-13}{4}\\\frac{5}{4}-b+\frac{3}{4}-\frac{1}{3}c=2.\frac{13}{4}\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}b+c=-2\\-b-\frac{1}{3}c=\frac{9}{2}\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}b=-\frac{23}{4}\\c=\frac{15}{4}\end{matrix}\right.\)

Vậy \(B\left(-\frac{23}{4};\frac{31}{4}\right);C\left(\frac{15}{4};-\frac{3}{4}\right)\)

Bình luận (0)