Chapter 4.2, Problem 60SPR

To determine

The distance from P to l.

Answer to Problem 60SPR

The distance between the point and line is 3 units.

Explanation of Solution

Given:

Line l contains points $\left(-3,0\right)\text{ and }\left(3,0\right).$

Coordinates of Point P are $\left(4,3\right).$

Calculation:

The equation of line l will be:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{0-0}{3-\left(-3\right)}\\ m=\frac{0}{6}\\ m=0\end{array}$

Use the point $\left(3,0\right)$ to write the equation of the line:

  $\begin{array}{l}{y}_2-y_1=m\left(x_2-x_1\right)\\ y-0=0\left(x-\left(3\right)\right)\\ y=0\end{array}$

The equation of line l is $y=0$ .

The slope of an equation that is perpendicular to line l is undefined and hence it is a vertical line.

The equation of vertical line that passes through point P is $x=4$ .

The point of intersection will be $Q\left(4,0\right).$

Now, finding the distance between Q and P is:

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d=\sqrt{{\left(4-\left(4\right)\right)}^2+{\left(3-0\right)}^2}\\ d=\sqrt{0+9}\\ d=3\end{array}$

So, the distance between the point and line is 3 units.