# What is the equation of the line that passes through (2,1) and is perpendicular to the line that passes through the following points: (8,-5),(3,9) ?

Jan 10, 2016

Let $A = \left(2 , 1\right) , B = \left(8 , - 5\right) , C = \left(3 , 9\right)$

First of all we have to find out the slope of the line that passes through $B$ and $C$.
After knowing the slope of the line passing through $B$ and $C$ we can find the slope of the line that is perpendicular to $B C$.

Slope of $B C = \frac{9 - \left(- 5\right)}{3 - 8} = \frac{9 + 5}{-} 5 = \frac{12}{-} 5 = - \frac{12}{5} = m$

Let the slope of the line perpendicular to $B C$ be $m '$.
Then $m . m ' = - 1$

$\implies m ' = - \frac{1}{m} = - \frac{1}{- \frac{12}{5}} = \frac{5}{12}$

$\implies m ' = \frac{5}{12}$

Now, we know two things about the required line
(i) Slope of that line $m ' = \frac{5}{12}$

(ii) The point through which the line passes $\left(2 , 1\right)$.

These two things are enough to find the equation.

Using point-slope form

$y - {y}_{1} = m ' \left(x - {x}_{1}\right)$

$\implies y - 1 = \frac{5}{12} \left(x - 2\right)$

$\implies 12 y - 12 = 5 x - 10$
$\implies 5 x - 12 y + 2 = 0$

This is the required equation.