Chapter 5, Problem 29CR

To determine

whether the graphs of the two equations are parallel, perpendicular, or neither.

Answer to Problem 29CR

The lines $y+3=\frac{5}{2}x$ and $2x-5y=0$ are neither perpendicular nor parallel.

Explanation of Solution

Given information: The two equations are $2x-5y=0,y+3=\frac{5}{2}x$

Consider the equation $y+3=\frac{5}{2}x$

This equation can be written as $y=\frac{5}{2}x-3$

This equation is in slope intercept form $y=mx+c$ where $m$ represents slope of the line and $c$ represents the y-intercept.

So,

Slope of the line $y=\frac{5}{2}x-3$ is equal to $\frac{5}{2}$ .

Now consider the equation $2x-5y=0$

First convert this equation in slope intercept form.

  $\begin{array}{l}2x-5y=0\\ \Rightarrow 5y=2x\\ \Rightarrow y=\frac{2}{5}x\end{array}$

So,

  $y=\frac{2}{5}x$

Compare this equation with $y=mx+c$

So, slope of the line $y=\frac{2}{5}x$ is $\frac{2}{5}$

Therefore,

Slope of the line $y+3=\frac{5}{2}x$ is $\frac{5}{2}$

And

Slope of the line $2x-5y=0$ is $\frac{2}{5}$

Use the fact that two lines are said to be parallel if their slopes are equal.

Two lines are perpendicular if product of slopes of both the lines is $-1$

Here,

  $\left(\text{Slope of the line}y+3=\frac{5}{2}x\right)\times \left(\text{Slope of the line}2x-5y=0\right)=\frac{5}{2}\left(\frac{2}{5}\right)=1\ne -1$

So, the lines $y+3=\frac{5}{2}x$ and $2x-5y=0$ are not perpendicular

Also,

  $\left(\text{Slope of the line}y+3=\frac{5}{2}x\right)\ne \left(\text{Slope of the line}2x-5y=0\right)$

So, the lines are not parallel.

Therefore,

The lines $y+3=\frac{5}{2}x$ and $2x-5y=0$ are neither perpendicular nor parallel.