# Math Help - Finding the equations for the lines that are tangent and normal to the given point.?

1. ## Finding the equations for the lines that are tangent and normal to the given point.?

Finding the equations for the lines that are tangent and normal to the given point.?
I was given a derivative formula in which I had to use to find the slope and I ended up having a slope of 1/-2 and the given point is (1,1)

I was asked to find the tangent line which is the x and normal like which is the y.
I really don’t know what its asking so it would be appreciated to have some help with it.

My professor said something about the tan line’s fomula being (x-b)=m(x-a)
And the normal line formula being (y-b)=-1/m (x-a)
But I didn’t quite get it.

Also just another quick question to confirm my answer.
I was asked to give the equation of the line with two points, one having the center which was (2,-3) and radius of 3 and the other was (-3,-2) with radius of 2.

Was I suppose to use y-y1 = (y2-y1/x2-x1)(x-x1)

Because I did that and I got (y+3) = -1/5(x-2)

Thanks for the all the help

2. The tangent at a point is the line having a slope equal to the derivative at that point. So take the derivative (that is, the slope) and the point, and plug-n-chug to get the line equation, like you learned back in algebra.

The "normal" is the perpendicular. You learned back in algebra how to find perpendicular slopes, so do that. Then plug-n-chug, as before.

3. Originally Posted by stapel
The tangent at a point is the line having a slope equal to the derivative at that point. So take the derivative (that is, the slope) and the point, and plug-n-chug to get the line equation, like you learned back in algebra.

The "normal" is the perpendicular. You learned back in algebra how to find perpendicular slopes, so do that. Then plug-n-chug, as before.
can you please just show me how me to, just like do it please, like tell me the 2 numbers, i can figure it out please

4. Originally Posted by stapel
The tangent at a point is the line having a slope equal to the derivative at that point. So take the derivative (that is, the slope) and the point, and plug-n-chug to get the line equation, like you learned back in algebra.

The "normal" is the perpendicular. You learned back in algebra how to find perpendicular slopes, so do that. Then plug-n-chug, as before.
oh hold on i know what your saying now haah
are you saying i should like plug them in
like you know (x-b)=m(x-a)
and it would be (1-b)=1/-2(1-a)
i got that part but how does that help me find the asnwer
like what do i do with the a and the b

5. To learn how slopes work, try here.

To learn how to find line equations from slopes and points, try here.