# What is the slope of any line perpendicular to the line passing through (-6,17) and (2,18)?

Mar 26, 2016

$- 8$

#### Explanation:

First, we need to find the slope of the line passing through $\left(- 6 , 17\right)$ and $\left(2 , 18\right)$. The slope is;

$\frac{18 - 17}{2 - \left(- 6\right)} = \frac{1}{8}$

If we multiply the slope of any line with $- 1$ and then get its reciprocal, we find the slope of the line which is perpendicular to it.

So;

$\frac{1}{8.} - 1 = - \frac{1}{8}$ its reciprocal $\to$ $- 8$

Mar 26, 2016

slope = -8

#### Explanation:

The first step is to calculate the gradient ( slope) of the line passing through the 2 given points using the $\textcolor{b l u e}{\text{ gradient formula }}$

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

where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$

let $\left({x}_{1} , {y}_{1}\right) = \left(- 6 , 17\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(2 , 18\right)$

$\Rightarrow m = \frac{18 - 17}{2 - \left(- 6\right)} = \frac{1}{8}$

If 2 lines with gradients , say ${m}_{1} \text{ and " m_2" are perpendicular}$

Then ${m}_{1} \times {m}_{2} = - 1$

$\Rightarrow \frac{1}{8} \times {m}_{2} = - 1 \Rightarrow {m}_{2} = - \frac{1}{\frac{1}{8}} = - 8$