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**+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