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 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 $\fs1P(3,4)$ and $\fs1Q(-1,8)$ $\fs2\vec{OX}$ =( , ) $\fs2\;+\;\lambda$( , )