# What is the equation of the line perpendicular to  y=-(2x)/5-12  at x=-1 ?

Jul 27, 2016

Two perpendicular lines have inverse slopes, both additive and multiplicative.

#### Explanation:

First of all, the equation of the perpendicular line will have this form:

$y = a x + b$

where $a$ is the slope and $b$ is the intercept (this is the value of the $y$ coordinate when $x = 0$).

As we have said, $a$ must be inverse to $\left(- \frac{2}{5}\right)$, given that both slopes must be inverse. This is a property required because the perpendicularity of both lines: their directions vector are the same, except that their coordinates are exchanged and one of them changes its sign.

${v}_{\text{line" = (v_1, v_2) leftrightarrow v_"perpendicular line}} = \left(- {v}_{2} , {v}_{1}\right)$

$\text{slope"_"line" = v_2 / v_1 leftrightarrow "slope"_"perpendicular line} = - {v}_{1} / {v}_{2}$

Thus, $a = - {\left(- \frac{2}{5}\right)}^{- 1} = - \left(- \frac{5}{2}\right) = \frac{5}{2}$.

Now, let us find the intercept. We know that both lines intercept at $x = 1$. So $x = 1$ must be in the statement line:

$y = - \frac{2}{5} \cdot 1 - 12 = - \frac{62}{5}$

So point $\left(1 , - \frac{62}{5}\right)$ is in line $y = - \frac{2 x}{5} - 12$.

Now, this point must be also in perpendicular line, with slope $a = \frac{5}{2}$.

$y = \frac{5}{2} x + b \rightarrow - \frac{62}{5} = \frac{5}{2} \cdot 1 + b \rightarrow b = - \frac{149}{10}$

So we have now completed our perpendicular line equation:

$y = \frac{5}{2} x - \frac{149}{10}$

You can watch the graphics of both lines on this link.