# What is the slope of any line perpendicular to the line passing through (2,2) and (9,5)?

Feb 27, 2016

$- \frac{7}{3}$

#### Explanation:

the slope of the line passing through the given pts is $\frac{5 - 2}{9 - 2} = \frac{3}{7}$
negative inverse of this slope will be the slope of the line perpendicular to the line joining the given pts.
Hence the slope is $- \frac{7}{3}$

Feb 27, 2016

The gradient of the perpendicular line is$\text{ } - \frac{7}{3}$

#### Explanation:

The standard form equation for a straight line graph is:

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

Where

$x$ is the independent variable (may take on any value you wish)

$y$ is the dependant variable (its value deponds an what value you give $x$)

$c$ is a constant

$m$ is the gradient (slope)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find the gradient of the given line}}$

Let $\left({x}_{1} , {y}_{1}\right) \to \left(2 , 2\right)$
Let $\left({x}_{2} , {y}_{2}\right) \to \left(9 , 5\right)$

Then it follows that

$m \text{ "=" } \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{5 - 2}{9 - 2} = \frac{3}{7}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the slope of any line perpendicular to this}}$

Given that the first line had gradient $m = \frac{3}{7}$

and that the gradient of the perpendicular line is $\left(- 1\right) \times \frac{1}{m}$

Then we have: $\left(- 1\right) \times \frac{7}{3} = - \frac{7}{3}$