College Algebra, Chapter 2, 2.1, Section 2.1, Problem 52

Plot the points $A(-2,1)$ and $B(12,-1)$ on a coordinate plane. Which (if either) of the points $C(5,-7)$ and $D(6,7)$ lies on the perpendicular bisector of the segment $AB$?

First we must get the slope of the segment $AB$ by using two point form
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1-1}{12-(-2)} = \frac{-2}{14} = \frac{-1}{7}$
Thus, the slope of the perpendicular bisector is equal to the negative reciprocal of the slope of segment $AB$. So,
$\displaystyle m_P = \frac{-1}{\frac{-1}{7}} = 7$
Then, since the segment is bisected, we must get the midpoint $AB$

$
\begin{equation}
\begin{aligned}
x &= \frac{-2+12}{2} = 5\\
\\
y &= \frac{1+(-1)}{2} = 0
\end{aligned}
\end{equation}
$

Thus, the midpoint of the segment $AB$ is $(5,0)$
Therefore, the equation of the perpendicular line can be obtained by using the point slope form $y = mx +b$

$
\begin{equation}
\begin{aligned}
y &= m_P x + b\\
\\
y &= 7x + b
\end{aligned}
\end{equation}
$

@ point $(5,0)$

$
\begin{equation}
\begin{aligned}
0 &= 7(5) + b\\
\\
b &= -35
\end{aligned}
\end{equation}
$

Then,
$y = 7x - 35$

Now, let's check which points $C(5,-7)$ and $D(6,7)$ lies on the line

$
\begin{equation}
\begin{aligned}
@\text{ point } & C &&& @ \text{ point } & D\\
\\
-7 &= 7(5) - 35 &&& 7 &= 7(6) - 35\\
\\
-7 &= 35 -35 &&& 7 &= 42-35\\
\\
-7 &\neq 0 &&& 7 &= 7
\end{aligned}
\end{equation}
$


It shows that $D(6,7)$ lies on the perpendicular bisector of the segment $PQ$

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