The points (0, 5), (0, -9)

False

Here, $x_{1}=0, x_{2}=0, x_{3}=3$ and $y_{1}=5, y_{2}=-9, y_{3}=6$

$\because \quad$ Area of triangle $\Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore \quad \Delta=\frac{1}{2}[0(-9-6)+0(6-5)+3(5+9)]$

$=\frac{1}{2}(0+0+3 \times 14)=21 \neq 0$

If the area of triangle formed by the points (0, 5), (0 – 9) and (3, 6) is zero, then the points are collinear.

Hence, the points are non-collinear.