Chapter 1, Problem 17T

(a)

To determine

To sketch: The points P and Q in the coordinate plane.

Explanation of Solution

The given points is $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

In point $P\left(-3,1\right)$, x-coordinate is negative and y-coordinate is positive, so it is lies on II Quadrant and point $Q\left(5,6\right)$ both coordinates are positive so it is lies in I Quadrant.

The coordinate plane of two given points is shown below,

Figure (1)

Figure (1) shows the points $P\left(-3,1\right)$ and $Q\left(5,6\right)$ in coordinate plane.

(b)

To determine

To find: The distance between $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

Answer to Problem 17T

The distance between P and Q is $\sqrt{89}$.

Explanation of Solution

Given:

The points is $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

Calculation:

The distance formula of two points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is,

$d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Where,

$d$ is the distance between two points.

$x_1$, $x_2$, $y_1$ and $y_2$ are coordinates of two points.

Substitute 5 for $x_2$, $-3$ for $x_1$, 6 for $y_2$ and 1 for $y_1$ from given points to the above formula to get distance,

$\begin{array}{c}d\left(P,Q\right)=\sqrt{{\left(5-\left(-3\right)\right)}^2+{\left(6-1\right)}^2}\\ =\sqrt{{\left(8\right)}^2+{\left(5\right)}^2}\\ =\sqrt{64+25}\\ =\sqrt{89}\end{array}$

Thus, the distance between P and Q is $\sqrt{89}$.

(c)

To determine

To find: The midpoint of line segment PQ.

Answer to Problem 17T

The midpoint of segment PQ is $\left(1,\frac{7}{2}\right)$.

Explanation of Solution

Given:

The points is $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

Calculation:

The formula of line segment from $A\left(x_1,y_1\right)$ to $B\left(x_2,y_2\right)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Substitute 5 for $x_2$, $-3$ for $x_1$, 6 for $y_2$ and 1 for $y_1$ from given points in the above formula to get midpoint,

$\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{\left(-3\right)+5}{2},\frac{1+6}{2}\right)\\ =\left(\frac{2}{2},\frac{7}{2}\right)\\ =\left(1,\frac{7}{2}\right)\end{array}$

Thus, the midpoint of segment PQ is $\left(1,\frac{7}{2}\right)$.

(d)

To determine

To find: The slope of line passes through two pints P and Q.

Answer to Problem 17T

The slope of line PQ is $\frac{5}{8}$.

Explanation of Solution

Given:

The points is $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

Calculation:

The formula of slope of line passes through two points is,

$m=\frac{y_2-y_1}{x_2-x_1}$

In above formula m is slope of line and $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$ are coordinates of two points.

Substitute 6 for $y_2$, 1 for $y_1$, 5 for $x_2$, and $-3$ for $x_1$ from given points to the above formula to get slope,

$\begin{array}{l}m=\frac{6-1}{5-\left(-3\right)}\\ m=\frac{5}{8}\end{array}$

Thus, the slope of line PQ is $\frac{5}{8}$.

(e)

To determine

To find: The perpendicular bisector of line PQ.

Answer to Problem 17T

The equation of perpendicular bisector is $y=-\frac{8}{5}x+\frac{51}{10}$.

Explanation of Solution

Given:

The points is $P\left(-3,1\right)$ and $Q\left(5,6\right)$.

Calculation:

The slope of line passes through points $P\left(-3,1\right)$ and $Q\left(5,6\right)$ is $m=\frac{5}{8}$ that obtained from part (d).

The lines are perpendicular if,

$m_1\cdot m_2=-1$

Where,

$m_1$ slope of line passes through two points.

$m_2$ is slope of perpendicular bisector.

Substitute $\frac{5}{8}$ for $m_1$ to get slope of perpendicular bisector,

$\begin{array}{c}\frac{5}{8}\cdot m_2=-1\\ m_2=\frac{-1}{\frac{5}{8}}\\ m_2=-\frac{8}{5}\end{array}$

The slope $m_2=-\frac{8}{5}$ is slope of perpendicular bisector of line PQ and perpendicular bisector is a line passes midpoint $\left(1,\frac{7}{2}\right)$ of line segment PQ obtained from part (c) with slope $-\frac{8}{5}$.

The formula of point-slope form of line is,

$y-y_1=m\left(x-x_1\right)$

Substitute $\frac{7}{2}$ for $y_1$, 1 for $x_1$ and $-\frac{8}{5}$ for min the above formula to get equation of perpendicular bisector,

$\begin{array}{c}y-\frac{7}{2}=-\frac{8}{5}\left(x-1\right)\\ y=-\frac{8}{5}x+\frac{8}{5}+\frac{7}{2}\\ y=-\frac{8}{5}x+\frac{16+35}{10}\\ y=-\frac{8}{5}x+\frac{51}{10}\end{array}$

Thus, the equation of perpendicular bisector is $y=-\frac{8}{5}x+\frac{51}{10}$.

(f)

To determine

To find: The equation of circle with diameter PQ.

Answer to Problem 17T

The equation of circle is ${\left(x-1\right)}^2+{\left(y-\frac{7}{2}\right)}^2=\frac{89}{4}$.

Explanation of Solution

Given:

The points is $P\left(-3,1\right)$, $Q\left(5,6\right)$ and the line segment PQ is a diameter of circle.

Calculation:

The equation of circle with center $\left(h,k\right)$ and radius r is,

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$. (1)

The distance between points $P\left(-3,1\right)$ and $Q\left(5,6\right)$ is $\sqrt{89}$ from part (b) that is a diameter of circle through given conditions.

The radius of circle is half of diameter that is,

$r=\frac{\sqrt{89}}{2}$

The line segment PQ is a diameter of circle and the midpoint of line segment PQ is center of circle that is $\left(1,\frac{7}{2}\right)$ obtained from part (c).

Substitute 1 for h, $\frac{7}{2}$ for k and $\frac{\sqrt{89}}{2}$ for r in equation (1) to get the equation of circle,

$\begin{array}{l}{\left(x-1\right)}^2+{\left(y-\frac{7}{2}\right)}^2={\left(\frac{\sqrt{89}}{2}\right)}^2\\ {\left(x-1\right)}^2+{\left(y-\frac{7}{2}\right)}^2=\frac{89}{4}\end{array}$

Thus, the equation of circle is ${\left(x-1\right)}^2+{\left(y-\frac{7}{2}\right)}^2=\frac{89}{4}$.