College Algebra, Chapter 2, 2.4, Section 2.4, Problem 60

Use slopes to determine whether the given points are collinear.

a.) $(1, 1), (3,9), (6,21)$

b.) $(-1,3), (1,7), (4, 15)$

a.) Let $m_1$ be the slope through points $(1,1)$ and $(3, 9)$, $m_2$ be the slope through points $(1, 1)$ and $(6, 21)$, $m_3$ be the slope through $(3, 9)$ and $(6, 21)$.


$
\begin{equation}
\begin{aligned}

m_1 =& \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4
\\
\\
m_2 =& \frac{21 - 1}{6 - 1} = \frac{20}{5} = 4
\\
\\
m_3 =& \frac{21 - 9}{6 - 3} = \frac{12}{3} = 4

\end{aligned}
\end{equation}
$


Since $m_1 = m_2 = m_3$, all the points are collinear.

b.) Similarly, let $m_1$ be the slope through $(-1, 3)$ and $(1, 7)$

$\qquad m_2$ be the slope through $(1,7)$ and $(4,15)$

$\qquad m_3$ be the slope through $(-1,3)$ and $(4, 15)$


$
\begin{equation}
\begin{aligned}

m_1 =& \frac{7 - 3}{1 - (-1)} = \frac{4}{2} = 2
\\
\\
m_2 =& \frac{15 - 7}{4 - 1} = \frac{8}{3}
\\
\\
m_3 =& \frac{15 - 3}{4 - (-1)} = \frac{12}{5}

\end{aligned}
\end{equation}
$


Since $m_1 \neq m_2 \neq m_3$, the points do not lie on the same line.