Question

os pontos A(1,-3) B(-2,1) e C(X,Y) sao colineares o ponto c esta
entre A e B e AC=3/5AB nessa condiçao
a)x=4/5
b)y=3/5
C)x=2/5
d)y=-3/5
e)x=-6/5

Answer 1

$\left[\begin{array}{ccc}1&-3&1\\-2&1&1\\x&y&1\end{array}\right] =0$
$1-3x-2y-x-6-y=0$
$-4x-3y-5=0$
$y= \frac{4x}{3} - \frac{-5}{3}$

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