# Thread: Finding Equation of a Line Perpendicular to...

1. ## Finding Equation of a Line Perpendicular to...

I'm having trouble remembering how to solve this problem.
I know it has something to do with point-slope form of an equation
(y - y1) = m(x - x1)
Here's how it reads:

A line has the equation 7x + 3y = 12. Find the equation of a line perpendicular to this one, passing though the point (-7, 0 )

Help?

2. Originally Posted by rachelstargirlrox
I'm having trouble remembering how to solve this problem.
I know it has something to do with point-slope form of an equation
(y - y1) = m(x - x1)
Here's how it reads:

A line has the equation 7x + 3y = 12. Find the equation of a line perpendicular to this one, passing though the point (-7, 0 )

Help?
see posts 2 and 3 here

3. Ahh, thank you!!

4. Originally Posted by rachelstargirlrox
I'm having trouble remembering how to solve this problem.
I know it has something to do with point-slope form of an equation
(y - y1) = m(x - x1)
Here's how it reads:

A line has the equation 7x + 3y = 12. Find the equation of a line perpendicular to this one, passing though the point (-7, 0 )

Help?
Ok, so first things first. You have $(y - y1) = m(x - x1)$ and you have the point it passes through, which is (-7,0).

If you substitute so far, you get: $(y-0) = m(x- (- 7))\quad\rightarrow\quad y=m(x+7)$

but you need m.

So what is the slope of a line that is perpendicular to another line? It is the negative inverse of the other lines slope!

So first we have to find the slope of the line we want to pass through. You should be able to do this yourself, coming up with the answer of $\frac{-7}{3}$

Now the negative inverse of $\frac{-7}{3}$ is $\frac{3}{7}$

So now substitute that into your point-slope equation to get: $y=\frac{3}{7}(x+7)\quad\rightarrow\quad \boxed{y=\frac{3}{7}x+3}$