+0  
 
0
115
1
avatar+2769 

The sum of the x-coordinates of the vertices of a triangle in the Cartesian plane equals \(\sqrt13\). Let equal the sum of the x-coordinates of the midpoints of the sides of the triangle. Find \(S^2\).

tertre  Mar 22, 2018
 #1
avatar+92933 
+2

The sum of the x-coordinates of the vertices of a triangle in the Cartesian plane equals\(\sqrt{13}\)

Let S equal the sum of the x-coordinates of the midpoints of the sides of the triangle. Find S^2

 

Let the x vaues of the vertices be a,b,and c

\(a+b+c=\sqrt{13}\)

 

the x values of the midpoints of the sides are 

    \(\frac{a+b}{2},\:\;\frac{b+c}{2},\;\;\frac{a+c}{2}\\ sum=\frac{a+b}{2}+\frac{b+c}{2}+\frac{a+c}{2}\\ sum=\frac{a+b+b+c+a+c}{2}\\ sum=\frac{2(a+b+c)}{2}\\ S=a+b+c = \sqrt{13}\\ S^2=13\)

Melody  Mar 22, 2018

9 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.