5)
[tex]\it\ \ AB:\ \dfrac{y-y_A}{y_B-y_A} = \dfrac{x-x_A}{x_B-x_A} \Rightarrow\ \dfrac{y-3}{-2-3}=\dfrac{x-2}{-3-2} \Rightarrow y-3=x-2|_{+3} \Rightarrow \\ \\ \\ \Rightarrow y=x+1 \Rightarrow panta\ dreptei\ AB\ este\ m=1 \Rightarrow panta\ perpendicularei\ pe\ AB\\ \\ este\ m'=-1[/tex]
[tex]\it\ Ecua\c{\it t}ia\ perpendicularei\ pe\ AB\ este\ \ y=-x+n\\ \\ Mijlocul\ segmentului\ [AB]\ este\ M(x_M,\ \ y_M).\\ \\ \\ x_M=\dfrac{x_A+x_B}{2}=\dfrac{2+(-3)}{2}=-\dfrac{1}{2} \\ \\ \\ y_M=\dfrac{y_A+y_B}{2}=\dfrac{3+(-2)}{2}=\dfrac{1}{2}[/tex]
[tex]\it M\Big(-\dfrac{1}{2},\ \ \dfrac{1}{2}\Big)\ apar\c{\it t}ine\ perpendicularei\ pe\ AB \Rightarrow \dfrac{1}{2}=\dfrac{1}{2}+n \Rightarrow n=0\\ \\ \\ Deci,\ ecua\c{\it t}ia\ mediatoarei\ segmentului\ [AB]\ este:\ \ y=-x[/tex]
6)
[tex]\it \overline{u}\cdot\overline{v}=|\overline{u}|\cdot|\overline{v}|cos(\widehat{\overline{u},\overline{v}}) \Rightarrow cos(\widehat{\overline{u},\overline{v}})=\dfrac{\overline{u}\cdot\overline{v}}{|\overline{u}|\cdot|\overline{v}|}=\dfrac{5}{2\cdot3}=\dfrac{5}{6}[/tex]
