Straight lines in the plane in coordinates

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 $v\cdot x - u\cdot y = v\cdot a - u\cdot b$

Let $m$ be the set of solutions $\rv{x,y}$ for the equation $v\cdot x - u\cdot y = v\cdot a - u\cdot b$ . We must demonstrate that $\ell$ and $m$ coincide.

For each value of $\lambda$ , $\rv{x, y} =\rv{a, b}+ \lambda\cdot\rv{ u, v}$ is a solution of this equation. After all, in this situation $x=a+\lambda\cdot u$ and $y=b+\lambda\cdot v$ apply, so $\begin{array}{rcl}v\cdot x - u\cdot y &=& v\cdot \left(a+\lambda\cdot u \right) - u\cdot \left(b+\lambda\cdot v \right)\\&=&v\cdot a+ v\cdot u\cdot\lambda-u\cdot b-u\cdot v\cdot\lambda\\&=&v\cdot a-u\cdot b\\ \end{array}$ This proves that each point of $\ell$ belongs to $m$ .

We still have to show that each point of $m$ belongs to $\ell$ . In order to do so, we use the fact that $m$ is a line. This is true because $\rv{u,v}$ is a direction vector, so not equal to $\rv{0,0}$ . This means $u\ne0$ or $v\ne0$ . If $u\ne0$ , the equation of $m$ can be written as $y=\frac{v}{u}x-\frac{v\cdot a - u\cdot b}{u}$ and if $v\ne0$ , the equation can be written as $x= \frac{u}{v}\cdot y + \frac{v\cdot a - u\cdot b}{v}$ We can regard the equation as a parametric representation in both cases: $\rv{x,y} =\begin{cases}\rv{0,\frac{ u\cdot b-v\cdot a}{u}}+x\cdot\rv{1,\frac{v}{u}}&\text{ if } u\ne0\\\rv{\frac{v\cdot a - u\cdot b}{v},0}+y\cdot\rv{\frac{u}{v},1}&\text{ if }v\ne0\end{cases}$ The solutions of the equation can be given in the parametric representation of a line, and, therefore, forms a line. Because the line $m$ contains the line $\ell$ , they must be equal. This proves the statement.

Sometimes, vectors are also described in columns. The parametric representation of $\ell$ looks like this:
$\left(\begin{array}{l} x \\ y \end{array}\right) = \left(\begin{array}{l} a \\ b \end{array}\right) + \lambda \left(\begin{array}{l} u \\ v \end{array}\right)$ We will discuss this at a later point.

Often, we write $\rv{x_1,x_2}$ instead of $\rv{x,y}$ . Similarly, $\rv{a_1,a_2}$ instead of $\rv{a,b}$ , and $\rv{v_1,v_2}$ instead of $\rv{u,v}$ . This notation is consistent with the notation for higher dimensions.

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.

Does the segment $AB$ , where $A = \rv{20,19}$ and $B = \rv{140,79}$ , intersect the line $l$ given by the equation $6 x -3 y -9=0$ ?

no

The line $A$ and $B$ has parametric representation $A+\lambda\cdot (B-A)$ . If we enter the coordinates, we see $\rv{20,19}+\lambda\cdot\rv{140-20,79-19}=\rv{20+120\lambda,19+60\lambda}\tiny.$ To see whether it includes a point of the line $l$ given by equation $6 x -3 y -9=0$ , we enter these coordinates for variables $x$ and $y$ in the equation $\begin{array}{rclcl} 6( 20+120\lambda)-3 (19+60\lambda) -9 &=& 0 &\phantom{x}&\color{blue}{\text{coordinates entered}}\\ 54+540\lambda &=& 0 &\phantom{x}&\color{blue}{\text{simplified}}\\ \lambda &=& -{{1}\over{10}} &\phantom{x}&\color{blue}{\text{solved}}\\ \end{array}$ Because the value of $\lambda$ not is between $0$ and $1$ , the intersection $A-{{1}\over{10}} \cdot (B-A)$ does not belong to the segment $AB$ . Hence, the answer is no. See the figure below.

There is actually another way to find out whether the line $l$ intersects the segment $AB$ : calculate the value of $6 x -3 y -9$ for $\rv{x,y} = A$ and of $\rv{x,y}=B$ . If the two values have the not equal sign, then the line does intersect $AB$ , otherwise, it does not. The explanation is that all points $\rv{x,y}$ with $6 x -3 y -9\gt 0$ are on one side of $l$ , and all points with $6 x -3 y -9\lt 0$ are on the other side. The value of $6 x -3 y -9$ for $A$ is $54$ , and the value for $B$ is $594$ .

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