Chapter 11.6, Problem 52ES

Formulas (1), (2), (3), (5), and (10), which apply to planes in 3-space, have analogs for lines in 2-space.

(a) Draw an analog of Figure 11.6.3 in 2-space to illustrate that the equation of the line that passes through the point $P\left(x_0,y_0\right)$ and is perpendicular to the vector $\text{n}=〈a,b〉$ can be expressed as $\text{n}\cdot \left(\text{r}-{\text{r}}_0\right)=0$ where $\text{r}=〈x,y〉\text{and}{\text{r}}_0=〈x_0,y_0〉.$

(b) Show that the vector equation in part (a) can be expressed as $a\left(x-x_0\right)+b\left(y-y_0\right)=0$

This is called the point-normal form of a line.

(c) using the prof of Theorem 11.6.1 as a guide, show that is a and b are not both zero, then the graph of the equation $ax+by+c=0$ is a line that has $\text{n}=〈a,b〉$ as a normal.

(d) Using the proof of Theorem 11.6.2 as a guide, show that the distance D between a point $P\left(x_0,y_0\right)$ and the line $ax+by+c=0\text{is}$ $D=\frac{\left|a{x}_0+b{y}_0+c\right|}{\sqrt{a^2+b^2}}$

(e) Use the formula in part (d) to find the distance between the point $P\left(-3,5\right)$ and the line $y=-2x+1.$