Chapter 13.3, Problem 15PSC

To determine

To find: The distance between the parallel lines corresponding equations.

Answer to Problem 15PSC

The distance between the parallel lines corresponding equations is $\frac{4\sqrt{5}}{5}$ .

Explanation of Solution

Given:

The corresponding equations of parallel lines are:

  $\begin{array}{c}y=2x+3\left(1\right)\\ y=2x+7\left(2\right)\end{array}$ .

Calculation:

Consider the points $\text{A}=\left(-1,1\right);\text{B}=\left(-3,1\right);\text{C}=\left(-1,5\right)$ .

Here, $\text{A}$ satisfies equation $\left(1\right)$ and $\text{B}\text{\&}\text{C}$ satisfy equation $\left(2\right)$ .

The coordinates $\text{A,B}\text{\&}\text{C}$ formed a triangle.

  $\text{Distance between lines}=\text{length of altitude from A to }\overline{\text{BC}}$ .

Let $\text{D}$ be point where alt intersects $\overline{\text{BC}}$ .

The length of $\text{AB},\text{BC},\text{CA}$ is given as:

  $\begin{array}{c}\text{CA}=\sqrt{{\left(-1-\left(-1\right)\right)}^2+{\left(5-1\right)}^2}\\ =4\\ \text{CB}=\sqrt{{\left(-1-\left(-3\right)\right)}^2+{\left(5-1\right)}^2}\\ =2\sqrt{5}\\ \text{AB}=\sqrt{{\left(-3-\left(-1\right)\right)}^2+{\left(1-1\right)}^2}\\ =2\end{array}$

As $\Delta \text{ABD}\sim \Delta \text{CBA}$ , so

  $\begin{array}{c}\frac{\text{AD}}{\text{AB}}=\frac{\text{CA}}{\text{CB}}\\ \text{AD}=\frac{\left(\text{CA}\right)\left(\text{AB}\right)}{\text{CB}}\\ =\frac{4\sqrt{5}}{5}\end{array}$

So length of altitude is $\frac{4\sqrt{5}}{5}$ .

Therefore the distance between the parallel lines corresponding equations is $\frac{4\sqrt{5}}{5}$ .