### Vector calculus in plane and space: Bases, Coordinates and Equations

### Straight lines in the plane in coordinates

Once we have fixed coordinates, we can calculate in #\mathbb{R}^2#. This allows us to use the coordinates, often indicated by #x# and #y#. In particular, we are able to describe a straight line by means of a linear equation.

Let #\ell# be a straight line in the coordinate plane #\mathbb{R}^2# with position vector #\rv{a, b}# and direction vector #\rv{u, v}#. In addition to the parametric representation of the straight line:

\[\rv{x, y} = \rv{a, b}+ \lambda\cdot\rv{ u, v}\] #\ell# can also be described as the set of solutions #\rv{x,y}# of the equation \[v\cdot x - u\cdot y = v\cdot a - u\cdot b\]

Of course we can simply write: #x = a + \lambda\cdot u# and #y = b + \lambda\cdot v#. By eliminating #\lambda# from these two relations, we find an equation of the straight line: multiply #x = a + \lambda \cdot u# by #v#, and #y = b + \lambda\cdot v# by #u#, and subtract: \[v\cdot x - u\cdot y = v\cdot a - u\cdot b\]

a linear equation with unknown #x# and #y#.

Equations of straight lines are not unique, nor are parametric representations. For example, if you multiply an equation by 2, the result describes the same straight line.

A parametric representation of a straight line *explicitly *describes vectors/points of a straight*:* you can find a vector/point on the line (or its coordinates) for each #\lambda#. An equation of a straight line *implicitly *describes the straight line*:* #\rv{y_1, y_2}# is then and only then on the line if the coordinates satisfy the equation.

The line #A# and #B# has parametric representation #A+\lambda\cdot (B-A)#. If we enter the coordinates, we see \[\rv{0,16}+\lambda\cdot\rv{24-0,96-16}=\rv{0+24\lambda,16+80\lambda}\tiny.\] To see whether it includes a point of the line #l# given by equation #2 x -5 y -8=0#, we enter these coordinates for variables #x# and #y# in the equation \[\begin{array}{rclcl} 2( 0+24\lambda)-5 (16+80\lambda) -8 &=& 0 &\phantom{x}&\color{blue}{\text{coordinates entered}}\\ -88-352\lambda &=& 0 &\phantom{x}&\color{blue}{\text{simplified}}\\ \lambda &=& -{{1}\over{4}} &\phantom{x}&\color{blue}{\text{solved}}\\ \end{array} \] Because the value of #\lambda# not is between #0# and #1#, the intersection #A-{{1}\over{4}} \cdot (B-A)# does not belong to the segment #AB#. Hence, the answer is no. See the figure below.

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