which gives the point (4,16) and gradient 11. From this you can use the equation of a line:
where and m is the gradient of 11.
As we know the normal passes through the same point we can use the coordinates (4,16) as above. Remember that because the normal is perpendicular to the tangent the products of their gradients must be -1:
= and so
You can then use the same equation as above putting m = -1/11 instead of 11.