$\det (xA + (1-x)B)$ is nothing but a real polynomial in $x$ of degree at most $n.$ So, $\det (xA + (1-x)B) = 0$ is having at most $n$ roots or you can say, it is having finitely many solutions for $x.$

$xA + (1-x)B$ is nothing but a linear combination of Matrices $A$ and $B.$

$xA + (1-x)B$ is nothing but a linear combination of Matrices $A$ and $B.$