Chapter 10.5, Problem 30E

To determine

To find a formula for the distance from the point $(x_0,y_0\text{})$ to the line $ax+by=0$ and verify your formula using a point and a line.

Explanation of Solution

Given:

The given point $(x_0,y_0\text{})$ and the given line is $ax+by=0$

Calculation:

The point $(x_0,y_0\text{})$ to the line $ax+by=0$

  

The product of the slopes of the perpendicular lines is $-1$

  $\begin{array}{l}-\frac{a}{b}m=-1\\ m=\frac{b}{a}\end{array}$

Now, find the y-intercept $c$ using $m=\frac{b}{a}$ in the equation $y=mx+c$

  $y=\frac{b}{a}x+\frac{a\cdot y_0-b\cdot x_0}{a}$

Now, solving the following set of equations-

  $\begin{array}{l}ax+by=0\\ y=\frac{b}{a}x+\frac{a\cdot y_0-b\cdot x_0}{a}\end{array}$

On solving the following is the point of intersection obtained −

  $\left(\frac{b^2\cdot x_0-ab\cdot y_0}{a^2+b^2},\frac{a^2\cdot y_0-ab\cdot x_0}{a^2+b^2}\right)$

Now, calculate the distance between two points $(x_0,y_0)$ and $\left(\frac{b^2\cdot x_0-ab\cdot y_0}{a^2+b^2},\frac{a^2\cdot y_0-ab\cdot x_0}{a^2+b^2}\right)$

The distance is obtained as

  $d=\frac{\left(a\cdot x_0+b\cdot y_0\right)}{\sqrt{a^2+b^2}}$

Verification:-

Calculate the distance of point $(5,6)$ to the line $-2x+3y=0$

The distance is

  $d=\frac{\left(a\cdot x_0+b\cdot y_0\right)}{\sqrt{a^2+b^2}}$

Substituting $x_0=5,y_0=6$ in the line $-2x+3y=0$

  $\begin{array}{l}d=\frac{\left(-2\cdot 5+3\cdot 6\right)}{\sqrt{4+9}}\\ d=\frac{8}{\sqrt{13}}\approx 2.2\end{array}$