# What is the equation of the line perpendicular to y=4/17x  that passes through  (5,7) ?

##### 1 Answer
Dec 5, 2015

Writing the solution in the same format as the question:

$y = - \frac{17}{4} x + \frac{113}{4}$

#### Explanation:

Let any point on the target line be ${P}_{i}$

Given: $y = \frac{4}{17} x \ldots \ldots \ldots . . \left(1\right)$

Standard form of equation is $y = m x + c$

In your case:
$m = \frac{4}{17}$
$c = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The gradient (slope) of you graph is $m = \frac{4}{17}$

Any line that is perpendicular to this line has the gradient of: $- \frac{1}{m}$

So the gradient of you target line is $- \frac{17}{4}$

So we have: $y = - \frac{17}{4} x + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

We know that there is a point on the line for equation (2) where

${P}_{\text{(x,y)}} \to \left(x , y\right) \to \left(5 , 7\right)$

So by substitution in (2) we have we have:

$7 = - \frac{17}{4} \left(5\right) + c$

$c = 7 + \frac{17}{4} \left(5\right) = 28 \frac{1}{4}$

So equation (2) is:

$y = - \frac{17}{4} x + \frac{113}{4}$