# Math Help - tangent line

1. ## tangent line

For each curve, find the equation of the tangent at the given point.

a) x^2 + y^2 = 13 at (2, -3)

I got the slope, which is - x/y but I have no clue how to put this into an equation.

2. Use point-slope form: $y-y_1 = m(x-x_1)$

3. I have tried that. I get y=2/3 (x) + 5/3 (correct me if I'm wrong), but the answer in the back of my textbook is 2x - 3y - 13 = 0

4. $y+3 = \frac{2}{3}(x-2)$

$y = \frac{2}{3}x - \frac{13}{3}$

So... $3y = 2x - 13$

Or $3y-2x + 13 = 0$

Then $2x-3y-13 = 0$