Are y= 4x + 5 and y = \frac { 1} { 4} x - 2 perpendicular?

Jun 12, 2017

See a solution explanation below:

Explanation:

Both equations are in the slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

Therefore the slope of the first equation: $y = \textcolor{red}{4} x + \textcolor{b l u e}{5}$ is

${m}_{1} = 4$

And, the slope of the second equation: $y = \textcolor{red}{\frac{1}{4}} x - \textcolor{b l u e}{2}$ is

${m}_{2} = \frac{1}{4}$

Perpendicular lines have negative inverse slopes.

If me call the slope of the perpendicular line ${m}_{p}$ the slope would be:

${m}_{p} = - \frac{1}{m}$

If we substitute the slope of the first line, the slope of a perpendicular line would be:

${m}_{p} = - \frac{1}{4}$

Because $\frac{1}{4}$ does not equal $- \frac{1}{4}$ these lines are not perpendicular.

Another, faster way to determine this without having the exact slopes would be to see the slopes of both lines are positive.

For perpendicular lines one slope will be positive and the other will be negative. Therefore, lines with slope $+ 4$ and $+ \frac{1}{4}$ cannot be perpendicular.