The points are not collinear.
Solution:
Let A, B and C be (3, -10), (-2, -7) and (0, -5).
If slopes of any two points are same, then the points are collinear.
Slope formula:
[tex]$m=\frac{y_2-y1}{x_2-x_1}[/tex]
Slope of AB:
[tex]$m_1=\frac{-7-(-10)}{-2-3}[/tex]
[tex]$m_1=\frac{-7+10}{-5}[/tex]
[tex]$m_1=-\frac{3}{5}[/tex]
Slope of BC:
[tex]$m_2=\frac{-5-(-7)}{0-(-2)}[/tex]
[tex]$m_2=\frac{-5+7}{2}[/tex]
[tex]$m_2=\frac{2}{2}[/tex]
[tex]m_2=1[/tex]
Slope of CA:
[tex]$m_3=\frac{-10-(-5)}{3-0}[/tex]
[tex]$m_3=\frac{-10+5}{3}[/tex]
[tex]$m_3=-\frac{5}{3}[/tex]
[tex]m_1\neq m_2 \neq m_3[/tex]
Slope of AB ≠ Slope of BC ≠ Slope of AC
Therefore the points are not collinear.
