+0  
 
0
55
1
avatar+2347 

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
Sort: 

1+0 Answers

 #1
avatar+92221 
+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

30 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details