Equation of a line Vector equation There are two useful versions of the vector equation of a line, and the one we choose depends upon what information is given: The vector equation of a line can be obtained by using a fixed point on the line and a vector parallel to the line. The vector equation of a line can be obtained by using the position vectors of two points A and B on the line.     Case 1. Given a fixed point on the line and a vector parallel to the line. Suppose the fixed point on the line is $/fs1P$ whose position vector with respect to the origin is OP and the vector with the same direction of the line is $/fs1/vec{u}$ . Let $/fs1X$ be any point on the line and let the position vector of $/fs1X$ be $/fs1/vec{OX}$ . From the diagram we can see that $/fs1/vec{OX}=/vec{OP}+/vec{PX}$. And we know that $/fs1/vec{PX}$ has the same direction as $/fs1/vec{u}$ , and therefore $/fs1/vec{PX}=/lambda/vec{u}$ for some real number $/fs1/lambda$ . So we have: $/fs1/vec{OX}=/vec{OP}+/lambda/vec{u}$ (the vector equation of a line). Case 2. Given two points on the line. Suppose the two given points are A and B. Then find the vector AB and return to case 1. Vector equation of a line $/fs1/vec{OX}=/vec{OP}+/lambda/vec{u}$ Find the vector equation of the line which is parellel to $/fs1/vec{u}=(-2,1)$and passes through the point $/fs1P=(2,3)$ $/fs1/vec{OX}=/vec{OP}+/lambda/vec{u}/;/Rightarrow/;/;/vec{OX}=(2,3)+/lambda(-2,1)/;/;/Rightarrow/;/;(x,y)=(2,3)+/lambda(-2,1)$ Find the vector equation of the line passing through the points P=(-2,3) and Q=(1,4) Firstly, we find the vector PQ: $/fs1/vec{PQ}=(1-(-2),4-1)=(3,3)$. The vector equation is: $/fs1/vec{OX}=/vec{OP}+/lambda/vec{PQ}/;/Rightarrow/;/;/vec{OX}=(-2,3)+/lambda(3,3)/;/;/Rightarrow/;/;(x,y)=(-2,3)+/lambda(3,3)$ Find the vector equation of the line that passes through the points $/fs1P(-5,-3)$ and $/fs1Q(0,9)$ $/fs2/vec{OX}$ =( , ) $/fs2/;+/;/lambda$( , )