There are 11 points in a plane of which 5 points are collinear. Then the total number of distinct quadrilaterals with vertices at these points is

A quadrilateral requires selecting four points with no three collinear. Given 11 total points including five collinear, the valid selections are determined by excluding invalid cases.

The total ways to choose any four points is $C(11, 4)=330$.

Invalid combinations occur if all four points are collinear, which gives $C(5, 4)=5$ cases, or if three are collinear and one is not, yielding $C(5, 3)\times C(6, 1)=60$ cases.

Subtracting these from the total gives $330-5-60=265$ distinct quadrilaterals.

Answer: $\boxed{265}$