Question #50f9b

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Answer 1 · 2016-02-23T18:42:06

$(a) k->y=(-4x+25)/3, l->y=(-x+10)/2; (b) D(4,3), (x-4)^2+(y-3)^2=50, r=5sqrt(2)$

Repeating the points
$A(5,10)$
$B(-3,4)$
$C(-1,-2)$
So the midpoints of interest are
$M_(AB)(1,7)$
$M_(AC)(2,4)$

(a)
Finding the slopes ( $k_n=(Delta y)/(Delta x) and p_n=-1/(k_n)$ )

$AB-> k_1=(4-10)/(-3-5)=(-6)/-8=3/4$ => $p_1=-4/3$
$AC-> k_2=(-2-10)/(-1-5)=(-12)/(-6)=2$ => $p_2=-1/2$

Equations of lines

$k-> (y-7)=(-4/3)(x-1)$ => $y=(-4x+4)/3+7$ => $y=(-4x+25)/3$
$l->(y-4)=(-1/2)(x-2)$ => $y=(-x+2)/2+4$ => $y=(-x+10)/2$

(b)
$D=knnl$

Finding D

$(-4x+25)/3=(-x+10)/2$ => $-8x+50=-3x+30$ => $5x=20$ => $x=4$
$->y=(-4+10)/2=6/2$ => $y=3$
$-> D(4,3)$

Since D is the circumcenter then
$AD=BD=CD=r$
(choosing CD) $=> r=sqrt((4+1)^2+(3+2)^2)=sqrt(25+25)=sqrt(50)=5sqrt(2)$

Circle's equation
$(x-x_D)^2+(y-y_D)^2=r^2$
$(x-4)^2+(y-3)^2=50$