Line segment connecting points in a convex set A subset $A$ of $\mathbf{R}^d$ is called $convex$ if for all $\mathbf{a,b} \in A$, the line segment in $\mathbf{R}^d$ connecting $\mathbf{a}$ and $\mathbf{b}$ lies entirely in $A$. Prove:

For any $\mathbf{a}$, $\mathbf{b}\in \mathbf{R}^d$, the line segment connecting $\mathbf{a}$ and $\mathbf{b}$ is the set of all points

$\mathbf{r}(t)=t\mathbf{a}+(1-t)\mathbf{b},0\le t\le 1$.} 

I'm just looking for a hint to get started on this.
Let $r(t_1)=t_1a + (1-t_1)b$ and $r(t_2)=t_2a+(1-t_2)b$ be two points on the line connecting a and b. I claim $tr(t_1)+(1-t)r(t_2)$ also takes the above form of a point on the line connecting and b. So

$tr(t_1)+(1-t)r(t_2)=t(t_1a+(1-t_1)b)+(1-t)(t_2a+(1-t_2)b)=(tt_1+(1-t)t_2)a+(t(1-t_1)+(1-t)(1-t_2))b$.

Since you asked for a hint i did not complete the computation but you start this way and you must continue until you get the equation into the same form as r(t) given in the question. You must check that the two coefficients also add up to 1 and that each coefficient of $a$ and $b$ is in $[0,1]$.
Assume that vector a and b represents the endpoints of the line segment. Now the line made from this line segment will go through direction b-a and to get all the point multiply it by t which gives t(b-a). This line goes through origin. We can shift it by adding b which gives b + t(b-a). Now change t from 0 to 1 and draw out the vectors on paper and you can see why it forms a line segment with a, b as endpoints.

To learn about line in vectors.
https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/vectors/v/linear-algebra-parametric-representations-of-lines