Chapter 6.2, Problem 42E

Solution

a.

The $x$ -intercept and $y$ -intercept of the line $x+2y=6$ .

#The $x$ -intercept $A$ is $\left(7,0\right)$ and the $y$ -intercept $B$ is $\left(0,-3\right)$ .

Given:

The equation of the line is $x+2y=6$ .

Calculation:

The given equation is:

  $x+2y=6$

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

  $\begin{array}{c}x+2\left(0\right)=6\\ x=6\end{array}$

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

  $\begin{array}{c}0+2y=6\\ y=3\end{array}$

Hence, the $x$ -intercept $A$ is $\left(6,0\right)$ and the $y$ -intercept $B$ is $\left(0,3\right)$ .

b.

Find the coordinates of $P$ .

The coordinates of $P$ are $\left(6.44,0.88\right)$ and $\left(5.55,-0.9\right)$ .

Given:

The equation of the line is $x+2y=6$ .

  $\overrightarrow{AP}$ is perpendicular to the line $x+2y=6$ .

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

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 $\left(6,0\right)$ .

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

  $\begin{array}{l}\frac{y_0-0}{x_0-6}\\ \frac{y_0}{x_0-6}\end{array}$

The slope of line $x+2y=6$ is:

  $\begin{array}{c}x+2y=6\\ 2y=-x+6\\ y=-\frac{1}{2}x+3\end{array}$

The slope of line $x+2y=6$ is $-\frac{1}{2}$ .

The line $\overrightarrow{AP}$ and $x+2y=6$ are perpendicular. Therefore,

  $\begin{array}{c}\left(\frac{y_0}{x_0-6}\right)\left(-\frac{1}{2}\right)=-1\\ y_0=2\left(x_0-6\right)\mathrm{...}\left(i\right)\end{array}$

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

  $\begin{array}{c}\sqrt{{\left(x_0-6\right)}^2+{\left(y_0-0\right)}^2}=1\\ {\left(x_0-6\right)}^2+{\left(y_0-0\right)}^2=1^2\\ {\left(x_0-6\right)}^2+y_0{}^2=1\end{array}$

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

  $\begin{array}{c}{\left(x_0-6\right)}^2+{\left(2\left(x_0-6\right)\right)}^2=1\\ {\left(x_0-6\right)}^2+4{\left(x_0-6\right)}^2=1\\ {\left(x_0-6\right)}^2\left(1+4\right)=1\\ {\left(x_0-6\right)}^2=\frac{1}{5}\\ x_0-6=\pm \frac{1}{\sqrt{5}}\\ x_0=6\pm \frac{1}{\sqrt{5}}\\ x_0=6+\frac{1}{\sqrt{5}}and{x}_0=6-\frac{1}{\sqrt{5}}\\ x_0\approx 6.44and{x}_0\approx 5.55\end{array}$

The value of $y_0$ for $x_0=6.44$ is:

  $\begin{array}{c}{y}_0=2\left(6.44-6\right)\\ y_0=0.88\end{array}$

The value of $y_0$ for $x_0=5.55$ is:

  $\begin{array}{c}{y}_0=2\left(5.55-6\right)\\ y_0=-0.9\end{array}$

Hence, the coordinates of $P$ are $\left(6.44,0.88\right)$ and $\left(5.55,-0.9\right)$ .