# Find equation of line that is perpendicular to the tangent line to the curve

1. Aug 3, 2010

### Kinetica

1. The problem statement, all variables and given/known data

Find the equation of the line that is perpendicular to the tangent line to the curve, y=(3x+1)/(4x-2) at the point (1,2)

2. Relevant equations

3. The attempt at a solution
I am absolutely confused with this problem. I tried taking a derivative of the equation. And I got y'=-10/(4x-2)2
I couldn't set it equal to 0, it will not work. What have I done wrong?

2. Aug 3, 2010

### Bohrok

There's no need to solve -10/(4x-2)2 = 0 because you're not supposed to find when/if the tangent line to the curve is ever horizontal. The problem wants you to first find the slope of the tangent line at the point (1, 2); you need to find the derivative when x = 1.

3. Aug 4, 2010

### Kinetica

Excuse me, what exactly do you mean by that?

4. Aug 4, 2010

### Dick

y'(x) is the slope of the tangent line. Substitute x=1 into your equation for y'(x) to find the slope of the tangent line. What's y'(1)? Now what's the relation between the slope of the tangent line and a perpendicular to it?

5. Aug 5, 2010

### Kinetica

OK, here we go:

y'(1)=-10/4

Normal line is reciprocal of the tangent, which is 4/10.
The normal line equation is:

y-2=0.4(x-1)
y=0.4x+1.6

6. Aug 5, 2010

### Dick

Uh, NEGATIVE reciprocal of the tangent. Works for me. Seem ok to you?

7. Aug 5, 2010

### Kinetica

That's what I exactly meant
Thank you SO much for everything.