Answer :
Answer:
The point P does not belong to the line that passes through points A and B.
Step-by-step explanation:
P is inside line AB only line AP is a multiple of the line AB. That is:
[tex]\vec l_{AP} = \alpha \cdot \vec l_{AB}[/tex]
Vectorially speaking, the line AB is equal to:
[tex]\vec l_{AB} = (0-1,-3-1)[/tex]
[tex]\vec l_{AB} = (-1,-4)[/tex]
The vector form of the line AP is:
[tex]\vec l_{AP} = (2-1,3-1)[/tex]
[tex]\vec l_{AP} = (1, 2)[/tex]
The following property must be fulfilled:
[tex](x_{2},y_{2}) = (\alpha \cdot x_{1},\alpha \cdot y_{1})[/tex]
The coefficients of each component are computed:
[tex]\alpha_{x} = \frac{1}{-1}[/tex]
[tex]\alpha_{x} = -1[/tex]
[tex]\alpha_{y} = \frac{2}{-4}[/tex]
[tex]\alpha_{y} = -\frac{1}{2}[/tex]
Since [tex]\alpha_{x} \neq \alpha_{y}[/tex], the point P does not belong to the line that passes through points A and B.