Vector equation help

I have no idea how to go about answering this question, can somebody help please...

Consider the vector equation x=p+t(q-p), where p and q correspond to distinct points P and Q in R2 or R3.
(a) Show that this equation describes the line segment PQ as t varies from 0 to 1.
(b) For which value of t is x the midpoint of PQ and what is x in this case.
(c) Find the midpoint of PQ when P=(2, -3) and Q=(0,1).
(d) Find the midpoint of PQ when P=(1, 0, 1) and Q=(4, 1, -2).
(e) Find the two points that divide PQ in part (c) into three equal parts.
(f) Find the two points that divide PQ in part (d) into three equal parts.

Really need your help! I have no idea how to go about this at all and I have a midterm this week.

This is a typical line .- The director vector is D= q-p , so if vector x is a position vector , a general vector of the line , is why
x-p= t D
segment pq is when x= q , so q-p= t (q-p) , ie , t=1 and t=0 starting at p so 0
( To the end i will give you a short method for this) , Now ,

If x is position vector of a midpoint of PQ , is why
Ix-PI = Iq-xI
q-x= q- (p+tD)
q-x= (q-p) -tD
If you have a vector A , IAI^2 = A dot A ,so

Ix-pI^2 = tD . tD
Iq-xI^2 = ( (q-p) -tD ) . ( (q-p) -tD )
=(q-p).(q-p) -2(q-p).tD + tD.tD

Ix-PI^2 = Iq-xI ^2
tD . tD =(q-p).(q-p) -2(q-p).tD + tD.tD

(q-p).(q-p) -2(q-p).tD =0 , remember that D=(q-p) and D.D= IDI^2

IDI^2 -2tIDI^2=0
IDI^2 ( 1-2t) =0
IDI<>0 so
1-2t=0
t=1/2

Midpoint , P(1,0,1) and Q(4,1-2) ( The other is the same easier )
Director vector D= Q-P = ( 3,1,-3)
x- (1,0,1) = (1/2) (3,1,-3)

x= (1,0,1) +(3/2,1/2,-3/2) = (5/2,1/2,-1/2) ie
Midpoint Xm= (1/2) ( 5,1,-1)

We will divide PQ in 3 equal parts ( The same method you can use for midpoint , easier )

First point is such as Ix-pI = (1/3) IDI
Ix-pI^2 = (1/9) IDI^2

tD.tD = (1/9) IDI^2
t^2 IDI^2 = (1/9) IDI^2
t=1/3

Plug t=1/3 in X with P and Q

Second point
Ix-pI = (2/3) IDI
t^2 IDI^2 = (4/9) IDI^2
t=2/3 plug t in X with P and Q