Respostas

2013-10-17T12:37:41-03:00
  \left[\begin{array}{ccc}1&-3&1\\-2&1&1\\x&y&1\end{array}\right] =0
1-3x-2y-x-6-y=0
-4x-3y-5=0
y= \frac{4x}{3} -  \frac{-5}{3}

AC
AC= \sqrt{}[(x-1) ^{2}+ (y+3) ^{2} ]

AB
AB= \sqrt{}[(-2-1) ^{2}+(1+3)^{2}]
AB= \sqrt{}(9+16)=5

AC que é AC=3AB/5
AC=  \frac{ \sqrt{}[(x-1)^2+(y+3)^2]=3 }{5}
 \sqrt{}[(x-1)^2+(y+3)^2]=3
(x-1)^2+ (y+3)^2=9
y= \frac{-4x}{3}- \frac{5}{3}
(x-1)^2+( \frac{-4x}{3} \frac{-5}{3}+3)^2 = 9
(x-1)^2+( \frac{-4x}{3}+ \frac{4}{3})^2=9

x'= \frac{-4}{5}
x''= \frac{14}{5}

Para x= \frac{-4}{5} dá y= \frac{-3}{5}
Para y=  \frac{14}{5} dá y=  \frac{-9}{5}

Resposta= d)y=-3/5
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