Question

[Hoangg Minh]

Cho tam giác ABC có A(3;1) B(2,6) C(4;-1) a.Tính chứ vi tam giác ABC B.Tính góc A C.Tìm toạ độ trực tâm H của tam giác ABC.

Answer 1

[Nguyễn Lê Phước Thịnh]

a: A(3;1); B(2;6); C(4;-1)

\(AB=\sqrt{\left(2-3\right)^2+\left(6-1\right)^2}=\sqrt{5^2+1^2}=\sqrt{26}\)

\(AC=\sqrt{\left(4-3\right)^2+\left(-1-1\right)^2}=\sqrt{2^2+1^2}=\sqrt{5}\)

\(BC=\sqrt{\left(4-2\right)^2+\left(-1-6\right)^2}=\sqrt{2^2+7^2}=\sqrt{53}\)

Chu vi tam giác ABC là:

\(C_{ABC}=\sqrt{26}+\sqrt{5}+\sqrt{53}\left(đvđd\right)\)

b: Xét ΔABC có 

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{26+5-53}{2\cdot\sqrt{26\cdot5}}\simeq-0,96\)

=>\(\widehat{A}\simeq165^0\)

c: Gọi H(x,y) là trực tâm của ΔABC

\(\overrightarrow{AH}=\left(x-3;y-1\right)\)

\(\overrightarrow{BH}=\left(x-2;y-6\right)\)

\(\overrightarrow{BC}=\left(2;-7\right);\overrightarrow{AC}=\left(1;-2\right)\)

H là trực tâm nên ta có: AH\(\perp\)BC và BH\(\perp\)AC

=>\(\left\{{}\begin{matrix}\overrightarrow{AH}\cdot\overrightarrow{BC}=0\\\overrightarrow{BH}\cdot\overrightarrow{AC}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left(x-3\right)+\left(-7\right)\left(y-1\right)=0\\1\left(x-2\right)+\left(-2\right)\left(y-6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-6-7y+7=0\\x-2-2y+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-7y=-1\\x-2y=-10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-7y=-1\\2x-4y=-20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3y=-1+20=19\\x-2y=-10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{19}{3}\\x=-10+2y=-10-\dfrac{38}{3}=-\dfrac{68}{3}\end{matrix}\right.\)