Chapter 1.8, Problem 44E

To determine

To calculate: The lengths of the medians of the triangle with the vertices $A\left(1,0\right),B\left(3,6\right)\text{ and }C\left(8,2\right)$ .

Answer to Problem 44E

The lengths of the medians of the triangle with vertices $A\left(1,0\right),B\left(3,6\right)\text{ and }C\left(8,2\right)$ are $\frac{\sqrt{145}}{2},\frac{\sqrt{109}}{2}\text{ and }\sqrt{37}$ .

Explanation of Solution

Given information:

The three vertices of the triangle $A\left(1,0\right),B\left(3,6\right)\text{ and }C\left(8,2\right)$ .

Formula used:

Median of a triangle is defined as the line segment joining two points from a vertex to the midpoint of the opposite side.

Mid-point formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is mathematically expressed as,

  $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2}\right)$

Distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

Calculation:

Consider the provided vertices of the triangle $A\left(1,0\right),B\left(3,6\right)\text{ and }C\left(8,2\right)$ .

Recall the mid-point formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is mathematically expressed as,

  $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2}\right)$

Let D, E and F be the mid-points of line segments AB, BC and CA respectively.

So, mid-point of line segment $AB$ is calculated as,

  $\begin{array}{c}\text{D}=\left(\frac{1+3}{2},\frac{0+6}{2}\right)\\ =\left(\frac{4}{2},\frac{6}{2}\right)\\ =\left(2,3\right)\end{array}$

Now, mid-point of line segment $BC$ is calculated as,

  $\begin{array}{c}\text{E}=\left(\frac{3+8}{2},\frac{6+2}{2}\right)\\ =\left(\frac{11}{2},\frac{8}{2}\right)\\ =\left(\frac{11}{2},4\right)\end{array}$

Mid-point of line segment $AC$ is calculated as,

  $\begin{array}{c}\text{F}=\left(\frac{1+8}{2},\frac{0+2}{2}\right)\\ =\left(\frac{9}{2},\frac{2}{2}\right)\\ =\left(\frac{9}{2},1\right)\end{array}$

By plotting all the points on the coordinate plane, we get the following figure,

  

In the above figure, AE, BF and CD represents the three medians of the triangle ABC.

Recall the distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

So, length of median AE is calculated as,

  $\begin{array}{c}d\left(\text{AE}\right)=\sqrt{{\left(\frac{11}{2}-1\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{{\left(\frac{9}{2}\right)}^2+4^2}\\ =\sqrt{\frac{81}{4}+16}\\ =\frac{\sqrt{145}}{2}\end{array}$

Length of median BF is calculated as,

  $\begin{array}{c}d\left(\text{BF}\right)=\sqrt{{\left(\frac{9}{2}-3\right)}^2+{\left(1-6\right)}^2}\\ =\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{\frac{9}{4}+25}\\ =\frac{\sqrt{109}}{2}\end{array}$

Now, the length of median CD is calculated as,

  $\begin{array}{c}d\left(\text{CD}\right)=\sqrt{{\left(8-2\right)}^2+{\left(2-3\right)}^2}\\ =\sqrt{{\left(6\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{36+1}\\ =\sqrt{37}\end{array}$

Thus, the lengths of the medians of the triangle with vertices $A\left(1,0\right),B\left(3,6\right)\text{ and }C\left(8,2\right)$ are $\frac{\sqrt{145}}{2},\frac{\sqrt{109}}{2}\text{ and }\sqrt{37}$ .