Explanation:
If you have a vector that starts at (a,b) and goes to (c,d), then the vector is <c-a, d-b>.
Here, that gives <-9,-12>, meaning the vector should point down to the left.
Additionally, anytime you have a vector <x,y>, you can write it as xi + yj, where i and j are unit vectors, <1,0> and <0,1> respectively.
Here that would be -9i-12j.
The magnitude of a vector <a,b> is equal to \sqrt{a^2+b^2}.
Here that would give you a length of 15.
As for the angle, if you drew a circle centered at the origin, we call 0 degrees the right side, on the x-axis.
You can shift a vector around on a plane and as long as it is parallel to and has the same magnitude as the original, it is the same. So let's put the tail of the vector at (0,0), so that the vector goes from (0,0) to (-9,-12). Now, form a triangle with those two points and the y-axis at (-9,0).
Now, the sin of the angle next to the origin is equal to 12/15, so the angle is arcsin(12/15) which is about 0.9273 radians, or 53.13 degrees. Let's use degrees.
Now, where is this angle? This angle is underneath the x-axis on the left side of the origin. Going from the x-axis on the right side, over the top, to the x-axis on the left side, is 180 degrees. We add 53.13, and get 233.13 degrees. Note that we did this counterclockwise, which you should probably note in your explanation. It is also 180-53.13 = 126.87 degrees clockwise.
In order to find a unit vector (a vector with magnitude 1) in the same direction as the original, just divide by the magnitude, giving us <-9/15, -12/15> or <-3/5, -4/5>. You can verify that the magnitude is equal to 1.