Chapter 4.2, Problem 59SPR

To determine

Tocalculate:

The distance from $P$ to $l$ .

Answer to Problem 59SPR

The distancebetween the $P$ to $l$ is $\sqrt{26}$ units.

Explanation of Solution

Given information:

Line $l$ contains points $\left(0,-2\right)$ and $\left(1,3\right)$ . Point $P$ has coordinates $\left(-4,4\right)$

Calculation:

First find the equation of the line $l$ .

Begin by finding the slope of the line through points $\left(0,-2\right)$ and $\left(1,3\right)$ .

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{3-\left(-2\right)}{1-0}\\ =\frac{5}{1}\\ =5\end{array}$

Then write the equation of this line using the point $\left(0,-2\right)$ on the line.

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\left(-2\right)=5\left(x-\left(0\right)\right)\\ y=5x-0-2\\ y=5x-2\end{array}$

The equation of line $l$ is $y=5x-2$ .

Now, write an equation of the line $w$ perpendicular to line $l$ through $P\left(-4,4\right)$

Since, the slope of line $l$ is $5$ , the slope of a line $p$ is $-\frac{1}{5}$

Write the equation of line $w$ through $P\left(-4,4\right)$ with slope $-\frac{1}{5}$ .

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-4=-\frac{1}{5}\left(x-\left(-4\right)\right)\\ y=-\frac{1}{5}x-\frac{4}{5}+4\\ y=-\frac{1}{5}x+\frac{16}{5}\end{array}$

The equation of line $w$ is $y=-\frac{1}{5}x+\frac{16}{5}$ .

Solve the system of equations to determine the intersection point.

Set $5x-2$ equal to $-\frac{1}{5}x+\frac{16}{5}$ .

  $\begin{array}{l}5x-2=-\frac{1}{5}x+\frac{16}{5}\\ 5x+\frac{1}{5}x=\frac{16}{5}+2\\ \frac{26}{5}x=\frac{26}{5}\\ x=1\end{array}$

Use the value of $x$ to find the value of $y$

  $\begin{array}{l}y=5x-2\\ =5\left(1\right)-2\\ =3\end{array}$

Hence, the point of intersection is $Q\left(1,3\right)$

Use the distance formula to determine the distance between $Q\left(1,3\right)$ and $P\left(-4,4\right)$

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-4-1\right)}^2+{\left(4-3\right)}^2}\\ =\sqrt{25+1}\\ =\sqrt{26}\end{array}$

Hence, the distance between the point and the line is $\sqrt{26}$ units.