Intuition / Proof behind Affine Set taking form of a line geometrically

I'm independently studying Boyd & Vandenberghe's Convex Optimization and came across the following statement.

Suppose $$x_1 \ne x_2$$ are two points on $$\mathbb{R}^n$$. Points of the form $$y = \theta x_1 + (1 - \theta)x_2$$ where $$\theta \in \mathbb{R}$$, form the line passing through $$x_1$$ and $$x_2$$.

I understand intuitively how this is a line where $$n = 2$$ (i.e. you can write it in slope-intercept form with $$m = x_1 - x_2$$). But what about for more than two dimensions (i.e. $$n \gt 2$$)? Why is this guaranteed to be a line geometrically for any $$n$$?

• Can you picture it in $\mathbb R^3$? – J. W. Tanner Feb 25 at 20:50
• I'm not sure I understand what you're asking. Can you clarify? – Noah Stebbins Feb 25 at 20:52
• Does this help? – J. W. Tanner Feb 25 at 20:58
• It helps outline the different possible forms. Thank you for the link, @J.W.Tanner. – Noah Stebbins Feb 25 at 23:46

The vector $$v=x_2-x_1$$ is a direction vector of the line.
If, from the origin, we first go to point $$x_1$$, then take a vector of direction $$v$$, which is just $$\lambda v$$ for a scalar $$\lambda$$, then we arrive to a point on the line: $$x\ =\ x_1+\lambda v\ =\ x_1+\lambda(x_2-x_1)\ =\ (1-\lambda)x_1+\lambda x_2$$ and every point on the line arises this way.
(For swapping $$x_1,x_2$$, just take $$\theta:=1-\lambda$$.)