MATH SOLVE

5 months ago

Q:
# M is the midpoint of CF for the points C(4, 10) and F(8, 8). Find MF.A. 5B. 2 (Square root) 5C.(Square root) 5D. 10

Accepted Solution

A:

hmmm ok... we know C and F, what is M anyway?

[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &C&(~ 4 &,& 10~) % (c,d) &F&(~ 8 &,& 8~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+4}{2}~,~\cfrac{8+10}{2} \right)\implies \stackrel{M}{(6,9)}[/tex]

ok, what is the length of the segment MF then?

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &M&(~ 6 &,& 9~) % (c,d) &F&(~ 8 &,& 8~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ MF=\sqrt{(8-6)^2+(8-9)^2}\\\\\\ MF=\sqrt{2^2+(-1)^2}\implies MF=\sqrt{5}[/tex]

[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &C&(~ 4 &,& 10~) % (c,d) &F&(~ 8 &,& 8~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+4}{2}~,~\cfrac{8+10}{2} \right)\implies \stackrel{M}{(6,9)}[/tex]

ok, what is the length of the segment MF then?

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &M&(~ 6 &,& 9~) % (c,d) &F&(~ 8 &,& 8~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ MF=\sqrt{(8-6)^2+(8-9)^2}\\\\\\ MF=\sqrt{2^2+(-1)^2}\implies MF=\sqrt{5}[/tex]