Answer
a) $x= p+ t(q-p) = p+tq-tp = (1-t)p+tq$
b) Maped into a line of the length $(1-t)T(p) + tT(q)$ if $0\lt(t)\leq1$ . However if $t=0$ then it will be a point as, $(1-(0))T(p)+(0)T(q)=T(p)$. With $T(p)$ being the transformed particular solution it creates a point.
Work Step by Step
a) Going back to exercise #21 in section 1.5 we can see an illustration of the line M going through points p and q. Line M is parallel to line q-p and travels through point p. If we recall the topic of 1.5 (Solution Sets) we can see that this is similar to the particular solution and the associated homogenous solution. Where line M is the line translated by the particular solution p and line q-p is the associated homogenous solution going through the origin. This means the general solution for line M= p+ t(q-p). [particular solution] + [general solution of homogenous system]. It is then algebraically manipulated into the form $(1-t)p+tq$.
b) The range of values $0\leq(t)\leq1$ will impact the appearance of the transformation differently. If $0\lt(t)\leq1$ then it will remain a line of varying length depending on the value of t as the vector q will not be removed. If $t=0$ then it will form a point as vector q will be null and there will only be p remaining which is the particular solution of the line, thus only generating a point.
