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

1 Answer
Dec 5, 2015

Answer:

Writing the solution in the same format as the question:

#y=-17/4x+(113)/4#

Explanation:

Let any point on the target line be #P_i#

Given: #y=4/17x...........(1)#

Standard form of equation is #y=mx+c#

In your case:
#m=4/17#
#c=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The gradient (slope) of you graph is #m=4/17#

Any line that is perpendicular to this line has the gradient of: #-1/m#

So the gradient of you target line is #-17/4#

So we have: #y=-17/4x+c......................(2)#

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

#P_("(x,y)")-> (x,y) -> (5,7) #

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

#7=-17/4(5)+c#

#c=7+17/4(5) =28 1/4 #

So equation (2) is:

#y=-17/4x+(113)/4#