Chapter 1.8, Problem 48E

To determine

To verify: The point M is equidistant from the vertices of triangle ABC.

Explanation of Solution

Given information:

The point M is the midpoint of the line segment AB and figure,

  

Formula used:

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,

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

Proof:

Consider the given figure,

  

In the above figure, M is the mid-point of the line segment AB.

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,

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

So, coordinates of point M will be calculated as,

  $\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

Now, to show that M is equidistant from the vertices of triangle ABC, prove that the distance from point M to the vertices are equal, i.e.,

  $d\left(A,M\right)=d\left(B,M\right)=d\left(C,M\right)$

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, distance between A and M is calculated as,

  $\begin{array}{c}d\left(A,M\right)=\sqrt{{\left(\frac{a}{2}-a\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\\ =\sqrt{\frac{a^2+b^2}{4}}\\ =\frac{\sqrt{a^2+b^2}}{2}\end{array}$

Distance between B and M is calculated as,

  $\begin{array}{c}d\left(B,M\right)=\sqrt{{\left(\frac{a}{2}-0\right)}^2+{\left(\frac{b}{2}-b\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\\ =\sqrt{\frac{a^2+b^2}{4}}\\ =\frac{\sqrt{a^2+b^2}}{2}\end{array}$

Distance between C and M is calculated as,

  $\begin{array}{c}d\left(C,M\right)=\sqrt{{\left(\frac{a}{2}-0\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\\ =\sqrt{\frac{a^2+b^2}{4}}\\ =\frac{\sqrt{a^2+b^2}}{2}\end{array}$

Since, $d\left(A,M\right)=d\left(B,M\right)=d\left(C,M\right)$ , so, it is proved that M is equidistant from the vertices of triangle ABC.

Thus, it is proved that point M is equidistant from the vertices of triangle ABC.