College Algebra, Chapter 2, 2.1, Section 2.1, Problem 56

If point $M$ is the midpoint of the segment $AB$. Show that $M$ is equidistant from the vertices of triangle $ABC$




If $M$ is the midpoint of segment $AB$, then its coordinates is $\displaystyle \left( \frac{a}{2}, \frac{b}{2} \right)$
Thus, by using distance formula,

$
\begin{equation}
\begin{aligned}
d_{MA} &= \sqrt{\left( a- \frac{a}{2} \right)^2 + \left( 0 - \frac{b}{2} \right)^2 }\\
\\
d_{MA} &= \sqrt{\left( \frac{a}{2} \right)^2 + \left( - \frac{b}{2} \right)^2 }\\
\\
d_{MA} &= \sqrt{\frac{a^2}{4} + \frac{b^2}{4} }\\
\\
d_{MA} &= \frac{\sqrt{a^2 + b^2}}{2} \text{ units}
\end{aligned}
\end{equation}
$


Similarly,

$
\begin{equation}
\begin{aligned}
d_{MB} &= \sqrt{\left( 0- \frac{a}{2} \right)^2 + \left( b - \frac{b}{2} \right)^2 }\\
\\
&= \sqrt{\left( -\frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2 }\\
\\
&= \sqrt{\frac{a^2}{4} + \frac{b^2}{4} }\\
\\
&= \frac{\sqrt{a^2 + b^2}}{2} \text{ units}
\end{aligned}
\end{equation}
$


And,

$
\begin{equation}
\begin{aligned}
d_{MC} &= \sqrt{\left( 0- \frac{a}{2} \right)^2 + \left( b - \frac{b}{2} \right)^2 }\\
\\
d_{MC} &= \sqrt{\left( -\frac{a}{2} \right)^2 + \left( - \frac{b}{2} \right)^2 }\\
\\
d_{MC} &= \sqrt{\frac{a^2}{4} + \frac{b^2}{4} }\\
\\
d_{MC} &= \frac{\sqrt{a^2 + b^2}}{2} \text{ units}
\end{aligned}
\end{equation}
$


It shows that $M$ is equidistant from the vertices of triangle $ABC$.