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

##### 1 Answer
Sep 20, 2016

$y = - \frac{2}{5} x + 2$

#### Explanation:

Thinks to remember:
[1]$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{y = \textcolor{b l u e}{m} x + b}$ is the slope-intercept form
$\textcolor{w h i t e}{\text{XXXX}}$with a slope of $\textcolor{b l u e}{m}$ and a y-intercept of $\textcolor{red}{b}$

[2]$\textcolor{w h i t e}{\text{XXX}}$If a line has a slope of $\textcolor{b l u e}{m}$
$\textcolor{w h i t e}{\text{XXXX}}$all lines perpendicular to it have a slope of color(blue)(""(-1/m))

Based on this we know that:
given a line $y = \textcolor{b l u e}{\frac{5}{2}} x - 2$
any line perpendicular to this will have a slope of $\textcolor{b l u e}{- \frac{2}{5}}$

and

any such perpendicular line will have the form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{b l u e}{- \frac{2}{5}} x + \textcolor{red}{b}$
where $\textcolor{red}{b}$ is the y-intercept;
that is $\textcolor{red}{b}$ is the value of $y$ when $\textcolor{\mathrm{da} r k m a \ge n t a}{x = 0}$

and

given $\left(\textcolor{\mathrm{da} r k m a \ge n t a}{x} , \textcolor{red}{y}\right) = \left(\textcolor{\mathrm{da} r k m a \ge n t a}{0} , \textcolor{red}{2}\right)$ is a point on the required perpendicular line
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{red}{b} = \textcolor{red}{2}$

So the equation of the required perpendicular line is:
color(white)("XXXXXXXX")color(green)(y=color(blue)(-2/5)x+color(red)(2)