Chapter 6.2, Problem 40E

Solution

a.

The $x$ -intercept and $y$ -intercept of the line $-2x+5y=10$

The $x$ -intercept $A$ is $\left(-5,0\right)$ and the $y$ -intercept $B$ is $\left(0,2\right)$ .

Given:

The equation of the line is $-2x+5y=10$ .

Calculation:

The given equation is:

  $-2x+5y=10$

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

  $\begin{array}{c}-2x+5\left(0\right)=10\\ -2x=10\\ x=-5\end{array}$

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

  $\begin{array}{c}-2\left(0\right)+5y=10\\ 5y=10\\ y=2\end{array}$

Hence, the $x$ -intercept $A$ is $\left(-5,0\right)$ and the $y$ -intercept $B$ is $\left(0,2\right)$ .

b.

Find the coordinates of $P$ .

The coordinates of $P$ are $\left(-4.62,-0.95\right)$ and $\left(-5.37,0.92\right)$ .

Given:

The equation of the line is $-2x+5y=10$ .

  $\overrightarrow{AP}$ is perpendicular to the line $-2x+5y=10$ .

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

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 $\left(-5,0\right)$

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

  $\begin{array}{l}\frac{y_0-0}{x_0+5}\\ \frac{y_0}{x_0+5}\end{array}$

The slope of line $-2x+5y=10$ is:

  $\begin{array}{c}-2x+5y=10\\ 5y=2x+10\\ y=\frac{2}{5}x+2\end{array}$

The slope of line $-2x+5y=10$ is $\frac{2}{5}$ .

The line $\overrightarrow{AP}$ and $-2x+5y=10$ are perpendicular. Therefore,

  $\begin{array}{c}\left(\frac{y_0}{x_0+5}\right)\left(\frac{2}{5}\right)=-1\\ y_0=-\frac{5}{2}\left(x_0+5\right)\mathrm{...}\left(i\right)\end{array}$

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

  $\begin{array}{c}\sqrt{{\left(x_0+5\right)}^2+{\left(y_0-0\right)}^2}=1\\ {\left(x_0+5\right)}^2+{\left(y_0-0\right)}^2=1^2\\ {\left(x_0+5\right)}^2+y_0{}^2=1\end{array}$

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

  $\begin{array}{c}{\left(x_0+5\right)}^2+{\left(-\frac{5}{2}\left(x_0+5\right)\right)}^2=1\\ {\left(x_0+5\right)}^2+\frac{25}{4}{\left(x_0+5\right)}^2=1\\ {\left(x_0+5\right)}^2\left(1+\frac{25}{4}\right)=1\\ {\left(x_0+5\right)}^2=\frac{4}{29}\\ x_0+5=\pm \frac{2}{\sqrt{29}}\\ x_0=-5\pm \frac{2}{\sqrt{29}}\\ x_0=-5+\frac{2}{\sqrt{29}}and{x}_0=-5-\frac{2}{\sqrt{29}}\\ x_0\approx -4.62and{x}_0\approx -5.37\end{array}$

The value of $y_0$ for $x_0=-4.62$ is:

  $\begin{array}{c}{y}_0=-\frac{5}{2}\left(-4.62+5\right)\\ y_0=-0.95\end{array}$

The value of $y_0$ for $x_0=-5.37$ is:

  $\begin{array}{c}{y}_0=-\frac{5}{2}\left(-5.37+5\right)\\ y_0\approx 0.92\end{array}$

Hence, the coordinates of $P$ are $\left(-4.62,-0.95\right)$ and $\left(-5.37,0.92\right)$ .