College Algebra, Chapter 2, 2.1, Section 2.1, Problem 48

Prove that the points $A(-1,3), B(3,11)$ and $C(5,15)$ are collinear by showing that $d(A,B) + d(B,C) = d(A,C)$
By using distance formula,

$
\begin{equation}
\begin{aligned}
d_{(A,B)} &= \sqrt{(11-3)^2 + (3-(-1))^2}\\
\\
&= \sqrt{8^2 + 4^2}\\
\\
&= \sqrt{64+16}\\
\\
&= \sqrt{80} \text{ units }
\end{aligned}
\end{equation}
$


Then,

$
\begin{equation}
\begin{aligned}
d_{(B,C)} &= \sqrt{(15-11)^2 + (5-3)^2}\\
\\
&= \sqrt{4^2 + 2^2}\\
\\
&= \sqrt{16+4}\\
\\
&= \sqrt{20} \text{ units}
\end{aligned}
\end{equation}
$

Lastly,

$
\begin{equation}
\begin{aligned}
d_{(A,C)} &= \sqrt{(15-3)^2 + (5-(-1))^2}\\
\\
&= \sqrt{12^2 + 6^2}\\
\\
&= \sqrt{144+36}\\
\\
&= \sqrt{180} \text{ units}
\end{aligned}
\end{equation}
$

Therefore,

$
\begin{equation}
\begin{aligned}
d(A,B) + d(B,C) = d(A,C) \\
\\
\sqrt{80} + \sqrt{20} &= \sqrt{180} && \text{Factor out the largest squares}\\
\\
\sqrt{(16)(5)} + \sqrt{(4)(5)} &= \sqrt{(36)(5)} && \text{Simplify}\\
\\
4\sqrt{5} + 2\sqrt{5} &= 6 \sqrt{5} && \text{Combine like terms}\\
\\
6 \sqrt{5} &= 6 \sqrt{5}
\end{aligned}
\end{equation}
$