How do you write an equation of a line passing through (0, -9), perpendicular to #7x+6y=5#?

1 Answer
Sep 5, 2017

Answer:

Three steps: 1. find the gradient of the given line, 2. find the gradient of a line perpendicular to it, 3. find the equation of a line with that second gradient that passes through the given point. Solution:
#y=6/7x-9#

Explanation:

Step 1 - find the gradient of the given line:

We need to rearrange it into point-slope form, #y=mx+c# where #m# is the gradient (slope).

#7x+6y=5#

Subtract #7x# from both sides:

#6y = -7x + 5#

Divide both sides by #6#:

#y=-7/6x+5/6#

So the gradient of the given line is #-7/6#.

Step 2 - find the gradient of a line perpendicular to this line:

Call the new gradient #m_1#, then #m_1=-1/m#

That is, to find the gradient of a line perpendicular to another line, we change the sign and take the inverse.

In this case #-1/m=-1/(-7/6)=6/7#

The gradient of the new line we want is #6/7#.

Step 3 - find the equation of the new line:

We know the gradient, and we know that the line passes through the point #(0,-9)#. We can substitute those values into the point-slope form of a line equation:

#y=mx+c#

#-9=6/7(0)+c#

#c=-9#

Therefore the overall equation of the line we want to find is:

#y=6/7x-9#