# How do you find a line that passes through (3,5) perpendicular to a line whose slope is -1/9?

Feb 17, 2017

$y = 9 x - 22$

#### Explanation:

The standardised equation of a strait line is $y = m x + c$ where $m$ is the gradient.

Any line perpendicular to this will have the gradient of:

$\text{ } \left(- 1\right) \times \frac{1}{m}$

So in this case as $m = - \frac{1}{9}$ the gradient of the perpendicular is:

$\left(- 1\right) \times - \frac{9}{1} = + 9$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus the equation of the line we are after is $y = 9 x + c$

This line passes through the point $\left(x , y\right) \to \left(3 , 5\right)$

so by substitution we have:

$y = 9 x + c \text{ "->" } 5 = 9 \left(3\right) + c$

$\text{ } 5 = 27 + c$

subtract 27 from both sides

$\text{ } 5 - 27 = 27 - 27 + c$

$\text{ } c = - 22$

Thus we have:

$y = 9 x - 22$