Question

Cho tam giác ABC cân tại A với A (2; - 2) , B (-1 ; -2) và C (a;b)Tìm a,b biết cosA = \(\dfrac{3}{5}\)

Answer 1

\(\overrightarrow{BA}=\left(3;0\right)\Rightarrow AB=3=AC\) ; \(\overrightarrow{AC}=\left(a-2;b+2\right)\) ; \(\overrightarrow{BC}=\left(a+1;b+2\right)\)

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=\dfrac{6\sqrt{5}}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-2\right)^2+\left(b+2\right)^2=9\\\left(a+1\right)^2+\left(b+2\right)^2=\dfrac{36}{5}\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(\dfrac{1}{5};-\dfrac{22}{5}\right)\\\left(a;b\right)=\left(\dfrac{1}{5};\dfrac{2}{5}\right)\end{matrix}\right.\)