# How do you write the slope-intercept equation of the line perpendicular to y = 5/2 x - 2, which passes through the point (0, 2)?

May 13, 2017

See a solution process below:

#### Explanation:

The equation in the problem is in 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.

$y = \textcolor{red}{\frac{5}{2}} x - \textcolor{b l u e}{2}$

Therefore the slope is: $\textcolor{red}{\frac{5}{2}}$

Next, let's call the slope perpendicular to this line ${m}_{p}$.

The slope of a perpendicular line is: ${m}_{p} = - \frac{1}{m}$

Substituting gives:

${m}_{p} = - \frac{1}{\frac{5}{2}} = - \frac{2}{5}$

Because the point given in the problem has $0$ for the $x$ value this is the y-intercept for the perpendicular line. Substituting for ${m}_{p} = - \frac{2}{5}$ and $2$ for $b$ into the slope-intercept formula gives:

$y = \textcolor{red}{- \frac{2}{5}} x + \textcolor{b l u e}{2}$