Chapter 16, Problem 10RP

To determine

To find: the distance between the two graphs represented by the equation.

Answer to Problem 10RP

  $d=0.5$ .

Explanation of Solution

Given:

Two equation are:

  $\begin{array}{l}3x-4y+10=0.\\ 6x-8y+15=0.\end{array}$

Concept used:

The equation of straight line having one point is:

  $y=mx+c$ .

The distance formula for the two equation of straight line:

  $d=\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}$ .

Calculation:

According to the given:

Two equation are:

  $\begin{array}{l}3x-4y+10=0.\\ 6x-8y+15=0.\end{array}$

This above two equations can be written as:

  $\begin{array}{l}3x-4y=-10.\\ 3x-4y=-\frac{15}{2}.\end{array}$

The general equation of the linear equation is:

  $Ax+By=C$ .

Here $A=3,B=-4,C_1=-10\text{ and }{C}_2=-\frac{15}{2}.$

The distance formula for the two equation of straight line is:

  $\begin{array}{l}d=\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}.\\ d=\frac{\left|-10-\left(-\frac{15}{2}\right)\right|}{\sqrt{3^2+{\left(-4\right)}^2}}=\frac{\left|\frac{-20+15}{2}\right|}{5}.\\ d=\frac{\frac{5}{2}}{5}=\frac{1}{2}.\end{array}$

Hence, the distance between two straight line equation is $d=0.5$ .