contestada

D is the midpoint of CE.E has coordinates (-3,-2), and D has coordinates (2 1/2, 1). Find the coordinate of C.

Respuesta :

midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
(-3,-2)....x1 = -3 and y1 = -2
(x,y)......x2 = x and y2 = y
now we sub
(-3 + x) / 2, (-2 + y) / 2

(-3 + x) / 2 = 2 1/2
-3 + x = 5/2 * 2
-3 + x = 5
x = 5 + 3
x = 8

(-2 + y) / 2 = 1
-2 + y = 1 * 2
-2 + y = 2
y = 2 + 2
y = 4

so the coordinates of point C are : (8,4)
to the risk of sounding redundant.

[tex]\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \qquad \qquad D\left(2\frac{1}{2},1 \right)\implies D\left(\frac{5}{2},1 \right)\\\\ -------------------------------\\\\ \textit{middle point of 2 points }[/tex]

[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({{ x}}\quad ,&{{ y}})\quad % (c,d) E&({{ -3}}\quad ,&{{ -2}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{-3+x}{2}~~,~~\cfrac{-2+y}{2} \right)=\stackrel{midpoint}{\left( \frac{5}{2}~,~1 \right)}\implies \begin{cases} \cfrac{-3+x}{2}=\cfrac{5}{2}\\\\ -3+x=5\\ \boxed{x=8}\\ ----------\\ \cfrac{-2+y}{2}=1\\\\ -2+y=2\\ \boxed{y=4} \end{cases}[/tex]