College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 2

Suppose that $P(2,0)$ and $Q(-4,8)$
a.) Plot $P$ and $Q$ on a coordinate plane.



b.) Find the distance from $P$ to $Q$.
By using the distance formula,

$
\begin{equation}
\begin{aligned}
d_{PQ} &= \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\\
\\
d_{PQ} &= \sqrt{(8-0)^2 + (-4-2)^2}\\
\\
d_{PQ} &= \sqrt{8^2 + (-6)^2}\\
\\
d_{PQ} &= 10 \text{units}
\end{aligned}
\end{equation}
$


c.) Find the midpoint of the segment $PQ$.

$
\begin{equation}
\begin{aligned}
m_{PQ}, \quad x &= \frac{2-4}{2} = -1\\
\\
y &= \frac{0+8}{2} = 4
\end{aligned}
\end{equation}
$


Therefore, $m_{PQ} (-1,4)$

d.) Find the slope of the line determined by $P$ and $Q$ and find equations for the line in point slope form and in slope intercept form. Then sketch a graph of the line.
$\displaystyle m = \frac{y_2-y_1}{x_2 - x_1} = \frac{8-0}{-4-2} = \frac{8}{-6} = - \frac{4}{3}$
Thus, by using point slope form,

$
\begin{equation}
\begin{aligned}
y - y_1 &= m(x - x_1)\\
\\
y - 0 &= \frac{-4}{3} (x-2)\\
\\
y &= \frac{-4}{3} x + \frac{8}{3}
\end{aligned}
\end{equation}
$


By using slope intercept form,

$
\begin{equation}
\begin{aligned}
y &= mx + b\\
\\
y &= \frac{-4}{3} x + b && \text{Solve for } b\\
\\
0 &= - \frac{4}{3} (2) + b\\
\\
b &= \frac{8}{3}
\end{aligned}
\end{equation}
$

Thus,$\displaystyle y = \frac{-4}{3} x + \frac{8}{3}$



e.) Sketch the circle that passes through $Q$ and has center $P$, and find the equation of that circle.
Recall that the general equation for the circle with center $(h,k)$ and radius $r$ is
$(x - h)^2 + (y- k)^2 = r^2$
Since the center is at $P(2,0)$
$(x-2)^2 + (y - 0)^2 = r^2$
And it pass through $Q(-4,8)$

$
\begin{equation}
\begin{aligned}
(-4-2)^2 + (8)^2 &= r^2\\
\\
(-6)^2 + (8)^2 &= r^2\\
\\
r^2 &= 100\\
\\
r &= 10 \text{ units}
\end{aligned}
\end{equation}
$

Thus, the equation of the circle is..
$(x-2)^2 + y^2 = 100$