Question

Se dau punctele A(-3;1) B(2;0) C(1;4). Determinati ecuatia medianei duse din varful A al triunghiului ABC.

Answer 1

$A(-3,1),B(2,0),C(1,4)$

$Fie\:M(x_{M},y_{M}) \: mijlocul \: lui \: BC$

$x_{M}=\frac{x_{B}+x_{C}}{2} = \frac{2 + 1}{2} = \frac{3}{2}$

$y_{M}=\frac{y_{B}+y_{C}}{2} = \frac{0 + 4}{2} = \frac{4}{2} = 2$

$= > M(\frac{3}{2},2)$

$Fie \: AM \: mediana \: dusa \: din \: A$

$AM:\frac{y-y_{M}}{y_{A}-y_{M}}=\frac{x-x_{M}}{x_{A}-x_{M}}$

$AM: \frac{y - 2}{1 - 2} = \frac{x - \frac{3}{2} }{ - 3 - \frac{3}{2} }$

$AM: \frac{y - 2}{ - 1} = \frac{x - \frac{3}{2} }{ - \frac{9}{2} }$

$AM: - \frac{9}{2} (y - 2) = - (x - \frac{3}{2} )$

$AM: - \frac{9}{2} y + 9 = - x + \frac{3}{2}$

$AM: - \frac{9}{2} y + x + 9 - \frac{3}{2} = 0$

$AM: - \frac{9}{2} y + x + \frac{15}{2} = 0\:|\times2$

$AM : -9y+2x+15=0$

$AM :-9y=-2x-15 \:|:(-9)$

$AM:y=\frac{2}{9}x+\frac{15}{9}$

$AM:y=\frac{2}{9}x+\frac{5}{3}$