A(-10,-13),B(-2,3),C(2,1)
Notăm mediana cu CM, M ∈ AB, BN ⊥ CM, N ∈ CM
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Coordonatele punctului M, mijlocului segmentului AB, sunt:
[tex]\boxed{\boldsymbol{M(x_{M}, y_{M}): \ \ x_{M} = \dfrac{x_{A} + x_{B}}{2} ; \ \ y_{M} = \dfrac{y_{A} + y_{B}}{2} }}[/tex]
[tex]x_{M} = \dfrac{-10-2}{2} = -6; \ \ \ y_{M} = \dfrac{-13+3}{2} = -5[/tex]
[tex]\implies \boldsymbol{M(-6,-5)}[/tex]
Ecuația dreptei CM
[tex]\dfrac{y - y_{C}}{y_{M} - y_{C}} = \dfrac{x - x_{C}}{x_{M} - x_{C}} \implies \dfrac{y - 1}{-5 - 1} = \dfrac{x - 2}{-6 - 2} \implies \dfrac{y - 1}{-6} = \dfrac{x - 2}{-8}[/tex]
[tex]\implies \boldsymbol{3x - 4y - 2 = 0}[/tex]
Distanța de la un punct M la o dreaptă d:
[tex]M \big(x_{M}; y_{M}\big); \ \ \ \ \ ax + by + c = 0[/tex]
[tex]\boxed{\boldsymbol{d(M;d) = \dfrac{\big|a \cdot x_{M} + b \cdot y_{M} + c\big|}{\sqrt{a^{2} + b^{2} }} }}[/tex]
[tex]B(-2,3); \ \ \ 3x - 4y - 2 = 0 \implies a = 3, b = - 4, c = - 2[/tex]
[tex]d(B;CM) = \dfrac{\big|a \cdot x_{B} + b \cdot y_{B} + c\big|}{\sqrt{a^{2} + b^{2} }} = \dfrac{\big|3 \cdot (-2) - 4 \cdot 3 - 2\big|}{\sqrt{4^{2} + 3^{2} }} = \dfrac{\big|- 6 - 12 - 2\big|}{\sqrt{25}}[/tex]
[tex]= \dfrac{\big|- 20\big|}{5} = \dfrac{20}{5} = 4[/tex]
[tex]\Rightarrow \boldsymbol{ d(B;CM) = 4 }[/tex]
