Question

Exercițiul 14 a) și b) dau coroana

Answer 1

$\it a)\ \ BC^2=(x_C-x_B)^2+(y_C-y_B)^2 =(2-1)^2+(5-2)^2=1+9=10\\ \\ BC=\sqrt{10}\\ \\ \\ A,\ B,\ C\ coliniare\ dac\breve a\ \ \begin{vmatrix}0&-1&1\\ 1&2&1\\ 2&5&1\end{vmatrix}=0\\ \\ \\ \ell_2-\ell_1\ \d si\ \ell_3-\ell_1\ \Rightarrow \ \begin{vmatrix}0&-1&1\\ 1&3&0\\ 2&6&0\end{vmatrix}=0\Rightarrow\begin{vmatrix}1&3\\ 2&6\end{vmatrix}=0 \Rightarrow\\ \\ \\ \Rightarrow1\cdot6=2\cdot3 \Rightarrow 6-6=0\ (A) \Rightarrow A,\ B,\ C\ sunt\ coliniare\ .$

$\it b)\ \ Dac\breve a \ M(x,\ y)\ este\ mijlocul\ lui\ [AB],\ atunci:\\ \\ \begin{cases} \it x=\dfrac{x_A+x_B}{2}=\dfrac{0+1}{2}=\dfrac{1}{2}\\ \\ \\ \it y=\dfrac{y_A+y_B}{2}=\dfrac{-1+2}{2}=\dfrac{1}{2} \end{cases} \ \Rightarrow M\Big(\dfrac{1}{2},\ \dfrac{1}{2}\Big)$

$\it\ Fie\ A'(x,\ y)\ simetricul\ lui\ A\ fa\c{\it t}\breve a\ de\ B \Rightarrow B\ este\ mijlocul\ lui\ [AA'],\ deci:\\ \\ \\ x_B=\dfrac{x_A+x_{A'}}{2}\Rightarrow 2x_B=x_A+x_{A'}\Rightarrow x_{A'}=2x_B-x_A=2\cdot1-0=2\\ \\ \\ Analog,\ y_{A'}=2y_B-y_A=2\cdot2-(-1)=4+1=5\\ \\ \\ Deci,\ avem\ A'(2,\ 5)$