# Question 50f9b

Feb 23, 2016

(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)#

#### Explanation:

Repeating the points
$A \left(5 , 10\right)$
$B \left(- 3 , 4\right)$
$C \left(- 1 , - 2\right)$
So the midpoints of interest are
${M}_{A B} \left(1 , 7\right)$
${M}_{A C} \left(2 , 4\right)$

(a)
Finding the slopes (${k}_{n} = \frac{\Delta y}{\Delta x} \mathmr{and} {p}_{n} = - \frac{1}{{k}_{n}}$)

$A B \to {k}_{1} = \frac{4 - 10}{- 3 - 5} = \frac{- 6}{-} 8 = \frac{3}{4}$ => ${p}_{1} = - \frac{4}{3}$
$A C \to {k}_{2} = \frac{- 2 - 10}{- 1 - 5} = \frac{- 12}{- 6} = 2$ => ${p}_{2} = - \frac{1}{2}$

Equations of lines

$k \to \left(y - 7\right) = \left(- \frac{4}{3}\right) \left(x - 1\right)$ => $y = \frac{- 4 x + 4}{3} + 7$ => $y = \frac{- 4 x + 25}{3}$
$l \to \left(y - 4\right) = \left(- \frac{1}{2}\right) \left(x - 2\right)$ => $y = \frac{- x + 2}{2} + 4$ => $y = \frac{- x + 10}{2}$

(b)
$D = k \cap l$

Finding D

$\frac{- 4 x + 25}{3} = \frac{- x + 10}{2}$ => $- 8 x + 50 = - 3 x + 30$ => $5 x = 20$ => $x = 4$
$\to y = \frac{- 4 + 10}{2} = \frac{6}{2}$ => $y = 3$
$\to D \left(4 , 3\right)$

Since D is the circumcenter then
$A D = B D = C D = r$
(choosing CD)$\implies r = \sqrt{{\left(4 + 1\right)}^{2} + {\left(3 + 2\right)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2}$

Circle's equation
${\left(x - {x}_{D}\right)}^{2} + {\left(y - {y}_{D}\right)}^{2} = {r}^{2}$
${\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} = 50$