# What is the slope of any line perpendicular to the line passing through (-9,8) and (3,-3)?

Feb 7, 2016

12/11

#### Explanation:

The coordinates given can be used to determine the gradient.

In standard form the equation of a straight line is:

$\text{ } y = m x + c$

Where $m$ is the gradient (slope)

The trick is, that the gradient of the line perpendicular to the first is:

$\text{ } - \frac{1}{m}$
So all you need to do is find the gradient of the first line then the other is just a matter of turning the first one 'upside down' and multiplying it by (-1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let $\left({x}_{1} , {y}_{1}\right) \to \left(- 9 , 8\right)$
Let $\left({x}_{2} , {y}_{2}\right) \to \left(3 , - 3\right)$

$m \text{ "=" } \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

=""(-3-8)/(3-(-9))

=""(-11)/(+12)

$= - \frac{11}{12}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So for the perpendicular line its gradient is:

$\left(- 1\right) \times \frac{1}{m} \text{ " ->" } \left(- 1\right) \times \left(- \frac{12}{11}\right) = + \frac{12}{11}$