Chapter T, Problem 4BDT

(a)

To determine

To find: The slope of the line that contains $\left(5,-12\right)$ and $\left(-7,4\right)$.

Answer to Problem 4BDT

The slope of the line is $m=-\frac{4}{3}$.

Explanation of Solution

Formula used:

The slope of the line is $m=\frac{y_1-y_2}{x_1-x_2}$, where the line passes through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

Calculation:

It is given that the line that contains $\left(5,-12\right)$ and $\left(-7,4\right)$.

Substitute $x_1=5$, $y_1=-12$, $x_2=-7$, and $y_2=4$ in $m=\frac{y_1-y_2}{x_1-x_2}$ to find the slope of the line,

$\begin{array}{c}m=\frac{-12-4}{5-\left(-7\right)}\\ =\frac{-16}{12}\\ =-\frac{4}{3}\end{array}$

Thus, the slope of the line is $m=-\frac{4}{3}$.

(b)

To determine

To find: The equation and the intercept of the line that contains $\left(5,-12\right)$ and $\left(-7,4\right)$.

Answer to Problem 4BDT

The intercept of the line is $-\frac{16}{3}$ and the equation of the line is $y=-\frac{4}{3}x-\frac{16}{3}$.

Explanation of Solution

Given:

The slope of the line is $m=-\frac{4}{3}$.

Formula used:

The equation of the straight line is $y=mx+c$, where m is the slope of the line, c is the x-intercept.

Calculation:

Suppose, the equation of the straight line is $y=mx+c$.

It is given that, the slope of the line is $m=-\frac{4}{3}$.

Thus, substitute $m=-\frac{4}{3}$ in $y=mx+c$ to obtain the equation of the straight line,

$y=-\frac{4}{3}x+c$

It is given that, the line passes through the point $\left(-7,4\right)$.

Thus, substitute $x=-7$ and $y=4$ in $y=-\frac{4}{3}x+c$ to obtain the intercept of the straight line,

$\begin{array}{l}4=-\frac{4}{3}\left(-7\right)+c\\ c=4-\frac{28}{3}\\ c=-\frac{16}{3}\end{array}$

Thus, the value is $c=-\frac{16}{3}$, that is the intercept of the line.

Thus, substitute $c=-\frac{16}{3}$ in $y=-\frac{4}{3}x+c$ to obtain the equation of the straight line,

$y=-\frac{4}{3}x-\frac{16}{3}$

Thus, the intercept of the line is $-\frac{16}{3}$ and the equation of the line is $y=-\frac{4}{3}x-\frac{16}{3}$.

(c)

To determine

To find: The mid points of the points $\left(5,-12\right)$ and $\left(-7,4\right)$.

Answer to Problem 4BDT

The midpoint of the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is $\left(-1,-4\right)$.

Explanation of Solution

Formula used:

The midpoint of the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\left(\frac{x_1-x_2}{2},\frac{y_1-y_2}{2}\right)$.

Calculation:

It is given that the points are $\left(5,-12\right)$ and $\left(-7,4\right)$.

Substitute $x_1=5$, $y_1=-12$, $x_2=-7$, and $y_2=4$ in $\left(\frac{x_1-x_2}{2},\frac{y_1-y_2}{2}\right)$ to find the midpoints,

$\begin{array}{c}\left(\frac{-7+5}{2},\frac{4-12}{2}\right)=\left(\frac{-2}{2},\frac{-8}{2}\right)\\ =\left(-1,-4\right)\end{array}$

Thus, the midpoint of the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is $\left(-1,-4\right)$.

(d)

To determine

To find: The length of the segment between the points $\left(5,-12\right)$ and $\left(-7,4\right)$.

Answer to Problem 4BDT

The length of the segment between the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is 20.

Explanation of Solution

Formula used:

The length of the segment between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$.

Calculation:

It is given that the points are $\left(5,-12\right)$ and $\left(-7,4\right)$.

Substitute $x_1=5$, $y_1=-12$, $x_2=-7$, and $y_2=4$ in $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ to find the midpoints,

$\begin{array}{c}\sqrt{{\left(-7-5\right)}^2+{\left(4+12\right)}^2}=\sqrt{{\left(-12\right)}^2+{\left(16\right)}^2}\\ =\sqrt{144+256}\\ =\sqrt{400}\\ =\pm 20\end{array}$

Length cannot be any negative term. Thus, the length is 20.

Thus, the length of the segment between the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is 20.

(e)

To determine

To find: The equation of the perpendicular bisector of AB. $\left(5,-12\right)$ and $\left(-7,4\right)$.

Answer to Problem 4BDT

The equation of the perpendicular bisector is $y=\frac{3}{4}x-\frac{13}{4}$.

Explanation of Solution

Given:

The points are $A=\left(5,-12\right)$ and $B=\left(-7,4\right)$.

The slope of the line AB is $m=-\frac{4}{3}$.

The midpoint of the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is $\left(-1,-4\right)$.

Formula used:

The equation of the straight line is $y=mx+c$, where m is the slope of the line, c is the x-intercept.

Calculation:

Suppose, the equation of the straight line is $y=mx+c$.

It is given that, the slope of the line AB is $-\frac{4}{3}$.

Thus, slope of the perpendicular bisector of AB is $\frac{3}{4}$.

Thus, substitute $m=\frac{3}{4}$ in $y=mx+c$ to obtain the equation of the straight line,

$y=\frac{3}{4}x+c$

It is given that, the line passes through the midpoint, that is $\left(-1,-4\right)$.

Thus, substitute $x=-1$ and $y=-4$ in $y=\frac{3}{4}x+c$ to obtain the intercept of the straight line,

$\begin{array}{l}-4=\frac{3}{4}\left(-1\right)+c\\ c=-4+\frac{3}{4}\\ c=-\frac{13}{4}\end{array}$

Thus, the value is $c=-\frac{13}{4}$, that is the intercept of the line.

Thus, substitute $c=-\frac{13}{4}$ in $y=\frac{3}{4}x+c$ to obtain the equation of the straight line,

$y=\frac{3}{4}x-\frac{13}{4}$

Thus, the equation of the perpendicular bisector is $y=\frac{3}{4}x-\frac{13}{4}$.

(f)

To determine

To find: The equation of the circle that has diameter as AB.

Answer to Problem 4BDT

The equation of the circle is ${\left(x+1\right)}^2+{\left(y+4\right)}^2=100$.

Explanation of Solution

Given:

The points are $A=\left(5,-12\right)$ and $B=\left(-7,4\right)$.

The midpoint of the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is $\left(-1,-4\right)$.

The length of the segment between the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is 20.

Formula used:

The equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where the center of the circle is at the point $\left(h,k\right)$, r is the radius of the circle.

Calculation:

Suppose, the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

It is given that, the center of the circle is at the midpoint of the points $\left(5,-12\right)$ and $\left(-7,4\right)$, that is $\left(-1,-4\right)$.

Thus, substitute $h=-1$ and $k=-4$ in ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ to obtain the equation of the straight line,

${\left(x+1\right)}^2+{\left(y+4\right)}^2=r^2$

The radius is the half of the length of the segment AB.

It is given that the length of the segment between the points $\left(5,-12\right)$ and $\left(-7,4\right)$ is 20.

Thus, the radius is $\frac{20}{2}=10$.

Thus, substitute $r=10$ in ${\left(x+1\right)}^2+{\left(y+4\right)}^2=r^2$ to obtain the equation of the straight line,

$\begin{array}{l}{\left(x+1\right)}^2+{\left(y+4\right)}^2={10}^2\\ {\left(x+1\right)}^2+{\left(y+4\right)}^2=100\end{array}$

Thus, the equation of the circle is ${\left(x+1\right)}^2+{\left(y+4\right)}^2=100$.