To determine a straight line in the plane, it is necessary to have two points or a point and a vector. A vector director of a straight line is any vector that has the same direction as the given straight line.
As with two points we can easily obtain the vector that exists between them and keep one of the points, we will suppose from now on that we have a point and a vector.
Given the points $$A = (3, 4)$$ and $$B = (-2, 6)$$, obtain the vector that goes from $$A$$ to $$B$$ and the one that goes from $$B$$ to $$A$$. $$$\overrightarrow{AB} = B - A = (-2, 6) - (3, 4) = (-2-3, 6-4) = (-5, 2) \\ \overrightarrow{BA}= A - B = (3, 4) - (-2, 6) = (3-(-2), 4-6) = (5,-2)$$$
Given a point $$P =(p_1,p_2)$$ and a vector $$\overrightarrow{v}=(v_1,v_2)$$, we can describe the points $$(x, y)$$ of the straight line that crosses point $$P$$ and has the direction of the vector $$\overrightarrow{v}$$ thus: $$$\begin{array}{rcl} (x,y) & = & P+k \cdot \overrightarrow{v} \\ (x,y) &=& (p_1,p_2)+ k \cdot (v_1,v_2)\end{array}$$$ where $$k$$ is a free parameter (that is to say, a variable that when we give it any real values we obtain points of the straight line).
Write the vector equation of the straight line $$r$$ that crosses the points $$A=(3,4)$$ and $$B=(-2,6)$$. $$$\overrightarrow{AB}=(-2-3,6-4)=(-5,2)$$$ Vector Equation: $$$(x, y) = A + k \cdot \overrightarrow {AB} = (3, 4) + k \cdot (-5, 2)$$$