Chapter 6.2, Problem 39E

Solution

a.

The $x$ -intercept and $y$ -intercept of the line $3x-4y=12$

The $x$ -intercept $A$ is $\left(4,0\right)$ and the $y$ -intercept $B$ is $\left(0,-3\right)$ .

Given:

The equation of the line is $3x-4y=12$ .

Calculation:

The given equation is:

  $3x-4y=12$

The $x$ -intercept will be the point where $y=0$ :

  $\begin{array}{c}3x-4\left(0\right)=12\\ 3x=12\\ x=4\end{array}$

The $y$ -intercept will be the point where $x=0$ :

  $\begin{array}{c}3\left(0\right)-4y=12\\ -4y=12\\ y=-3\end{array}$

Hence, the $x$ -intercept $A$ is $\left(4,0\right)$ and the $y$ -intercept $B$ is $\left(0,-3\right)$ .

b.

Find the coordinates of $P$ .

The coordinates of $P$ are $\left(4.6,-0.8\right)$ and $\left(3.4,0.8\right)$ .

Given:

The equation of the line is $3x-4y=12$ .

  $\overrightarrow{AP}$ is perpendicular to the line $3x-4y=12$ .

  $\left|\overrightarrow{AP}\right|=1$

Concept Used:

The relationship between the two lines with slopes $m_1$ and $m_2$ is:

  $m_1\times m_2=-1$

Calculation:

Let the coordinate $P$ is $\left(x_0,y_0\right)$ .

The coordinate of the point $A$ is $\left(4,0\right)$

The slope of the line $\overrightarrow{AP}$ is:

  $\begin{array}{l}\frac{y_0-0}{x_0-4}\\ \frac{y_0}{x_0-4}\end{array}$

The slope of line $3x-4y=12$ is:

  $\begin{array}{c}3x-4y=12\\ 3x-12=4y\\ y=\frac{3}{4}x-3\end{array}$

The slope of line $3x-4y=12$ is $\frac{3}{4}$ .

The line $\overrightarrow{AP}$ and $3x-4y=12$ are perpendicular. Therefore,

  $\begin{array}{c}\left(\frac{y_0}{x_0-4}\right)\left(\frac{3}{4}\right)=-1\\ y_0=-\frac{4}{3}\left(x_0-4\right)\mathrm{...}\left(i\right)\end{array}$

The modulus of $\overrightarrow{AP}$ given is 1. Therefore,

  $\begin{array}{c}\sqrt{{\left(x_0-4\right)}^2+{\left(y_0-0\right)}^2}=1\\ {\left(x_0-4\right)}^2+{\left(y_0-0\right)}^2=1^2\\ {\left(x_0-4\right)}^2+y_0{}^2=1\end{array}$

Substitute the value of $y_0$ from equation $\left(i\right)$ :

  $\begin{array}{c}{\left(x_0-4\right)}^2+{\left(-\frac{4}{3}\left(x_0-4\right)\right)}^2=1\\ {\left(x_0-4\right)}^2+\frac{16}{9}{\left(x_0-4\right)}^2=1\\ {\left(x_0-4\right)}^2\left(1+\frac{16}{9}\right)=1\\ {\left(x_0-4\right)}^2=\frac{9}{25}\\ x_0-4=\pm \frac{3}{5}\\ x_0=4\pm \frac{3}{5}\\ x_0=4+\frac{3}{5}and{x}_0=4-\frac{3}{5}\\ x_0=4.6and{x}_0=3.4\end{array}$

The value of $y_0$ for $x_0=4.6$ is:

  $\begin{array}{l}{y}_0=-\frac{4}{3}\left(4.6-4\right)\\ y_0=-0.8\end{array}$

The value of $y_0$ for $x_0=3.4$ is:

  $\begin{array}{l}{y}_0=-\frac{4}{3}\left(3.4-4\right)\\ y_0=0.8\end{array}$

Hence, the coordinates of $P$ are $\left(4.6,-0.8\right)$ and $\left(3.4,0.8\right)$ .