Consider the orange line, l, in the diagram below.

Other features in the diagram are:

• P = (1, 3) is a particular point along the line. In vector form p = (1, 3)^T.
• A direction vector for the line, d = (1, –3)^T, is shown by the green arrow pointing diagonally down and to the right.
• A normal vector for the line, n = (3, 1)^T, is shown by the green arrow pointing diagonally up and to the right. This normal vector is perpendicular to the direction vector.

This line can be represented algebraically in a number of ways:

• The general equation of the line is 3x + y = 6, since it passes through the points (2, 0) and (0, 6).
• The slope-intercept form of the line equation is y = –3x + 6, where the slope is –3 (change in y per unit change in x) and the intercept is 6 (value of y when x = 0).
• The normal form of the line equation is n·x = n·p, which in this case is [latex]\left({3\atop 1}\right)\cdot\left({x\atop y}\right)=\left({3\atop 1}\right)\cdot\left({1\atop 3}\right)[/latex], which is equivalent to 3x + y = 6.
• The vector form of the line equation is x = p +td, where t is a scalar. In this case this form is [latex]\left({x\atop y}\right)=\left({1\atop 3}\right)+t\left({1\atop -3}\right)[/latex].
• Expressing each component of the vector form gives the parametric equations of the line: x = 1 + t and y = 3 – 3t.