vectors

A(1,-2)  B(3,-2)
P2-P1=(3,-2)-(1,-2)=<2,0>
component AB= square root of 2^2+0^2= square root of 4=2

The directions tell you to express AB in component form.  We are given two points.  We have point A at (1,-2) and point B at (3,-2).  Now you have to find a vector from the two points you have.  To do this, you use the formula you have to know which is P2-P1.  That means point 2 minus point 1.  So now you do (3,-2)-(1,-2) and you just subtract it like normal subtraction and you get <2,0> as your vector.  Now you find the component of AB.  To do this, you use almost the same exact formula as the pathagorean thereom.  Next you use you vector to find the component of AB.  This formula is the square root of x^2+y^2.  So we do the square root of 2^2+0^2 and we got the square root of 4.  That reduces to 2, and that is your answer.