# What is the slope of any line perpendicular to the line passing through (15,-12) and (24,27)?

Jan 18, 2016

$- \frac{3}{13}$

#### Explanation:

Let the slope of the line passing through the given points be $m$.

$m = \frac{27 - \left(- 12\right)}{24 - 15} = \frac{27 + 12}{9} = \frac{39}{9} = \frac{13}{3}$

Let the slope of the line perpendicular to the line passing through the given points be $m '$.

Then $m \cdot m ' = - 1 \implies m ' = - \frac{1}{m} = - \frac{1}{\frac{13}{3}}$

$\implies m ' = - \frac{3}{13}$

Hence, the slope of the required line is $- \frac{3}{13}$.

Jan 18, 2016

The slope of any line perpendicular to the given one is: $- \frac{3}{13}$

#### Explanation:

The trick is to just remember that if the gradient of the first line is $m$ the gradient of the one that is perpendicular to it (normal) has the gradient of $\left(- 1\right) \times \frac{1}{m}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Gradient (slope) of the first line}}$

Let ${m}_{1}$ be the gradient of the first line

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

Given that
$\left({x}_{1} , {y}_{1}\right) \to \left(15 , - 12\right)$
$\left({x}_{2} , {y}_{2}\right) \to \left(24 , 27\right)$

We have:
$\textcolor{b l u e}{{m}_{1} = \frac{27 - \left(- 12\right)}{24 - 15} \textcolor{w h i t e}{\ldots .} \to \textcolor{w h i t e}{\ldots .} \frac{39}{9}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Gradient (slope) of the second line}}$

Let ${m}_{2}$ be the gradient of the second line

Then
${m}_{2} = \left(- 1\right) \times \frac{1}{m} _ 1 \textcolor{w h i t e}{\ldots .} \to \textcolor{w h i t e}{\ldots .} \left(- 1\right) \times \frac{9}{39}$

$\textcolor{b l u e}{{m}_{2} = - \frac{9 \div 3}{39 \div 3} = - \frac{3}{13}}$