Equation of a line Parametric equation The vector equation: $/fs1/vec{OX}=/vec{OP}+/lambda/vec{u}$ is the key to finding the parametric equation of a straight line. If we solve this equation for the vector OX: $/fs1/vec{OX}=/vec{OP}+/lambda/vec{u}/;/Rightarrow/;(x,y)=(p_1,p_2)+/lambda(u_1,u_2)$ $/fs1/Rightarrow/;(x,y)=(p_1/;+/;/lambda/;u_1/;,/;p_2/;+/;/lambda/;u_2)$ and write the above vector equation as two separate scalar equations we have: $/left/{{/;/;x/;=/;p_1/;+/;/lambda/;u_1/;/atop/;y/;=/;p_2/;+/;/lambda/;u_2}$ (parametric equation of a straight line) Parametric equation of a straight line $/left/{{/;/;x/;=/;p_1/;+/;/lambda/;u_1/;/atop/;y/;=/;p_2/;+/;/lambda/;u_2}$ Given the vector equation: $/fs1(x,y)=(2,3)+/lambda(-2,1)$ of a straight line, find the parametric equation: Firstly, $/fs1(x,y)=(2,3)+/lambda(-2,1)/;/Rightarrow/;(x,y)=(2-2/lambda,3+/lambda)$ Then the parametric equation is: $/left/{{/;/;x/;=/;2/;-2/lambda/;/atop/;y/;=/;3/;+/;/lambda}$ Find the parametric equation of the line that passes through the points P(1,-1) and Q(0,-3). The vector PQ is: $/fs1/vec{PQ}=(0-1,-3-(-1))=(-1,-2)$ . Then: $/left/{{x/;=/;1/;-/lambda/;/atop/;y/;=/;-1/;-2/lambda}$ Find the parametric equation of the line that passes through the points $/fs1P(-3,9)$ and $/fs1Q(4,-4)$ x= + y= +