Chapter 13.5, Problem 16WE

To determine

To find: the length of the longest median of the triangle

Answer to Problem 16WE

The length of the longest median of the triangle is $M_1{M}_3=\sqrt{26}$

Explanation of Solution

Given information: The vertices are $X\left(-2,3\right),Y\left(6,-3\right)\&Z\left(4,7\right)$

Calculation:

  $M_1,M_2,M_3$ are the midpoints

  $\begin{array}{l}{M}_1=\left(\frac{-2+6}{2},\frac{3-3}{2}\right)\\ M_1=\left(\frac{4}{2},\frac{0}{2}\right)=\left(2,0\right)\\ M_2=\left(\frac{4+6}{2},\frac{7-3}{2}\right)\\ M_2=\left(\frac{10}{2},\frac{4}{2}\right)=\left(5,2\right)\\ M_3=\left(\frac{4-2}{2},\frac{7+3}{2}\right)\\ M_3=\left(\frac{2}{2},\frac{10}{2}\right)=\left(1.5\right)\\ M_1{M}_2=\sqrt{{\left(5-2\right)}^2+{\left(2-0\right)}^2}=\sqrt{13}\\ M_2{M}_3=\sqrt{{\left(5-1\right)}^2+{\left(2-5\right)}^2}=\sqrt{25}\\ M_1{M}_3=\sqrt{{\left(2-1\right)}^2+{\left(5-0\right)}^2}=\sqrt{26}\end{array}$

The longest median of the triangle is $M_1{M}_3$