Lines in the Plane

A non-zero vector \(\vec{u}\) is a direction vector of a line \(d\) if there exist two distinct points \(A\) and \(B\) on \(d\) such that \(\Vect{u} = \Vect{AB}\).

A line has an infinite number of direction vectors. If \(\vec{u}\) is a direction vector of line \(d\), then any vector \(k\vec{u}\) (where \(k \neq 0\)) is also a direction vector of \(d\).

The line \(d\) passing through point \(A\) with direction vector \(\vec{u}\) is the set of all points \(M\) such that the vectors \(\Vect{AM}\) and \(\vec{u}\) are collinear.$$\Vect{AM} = k\vec{u} \quad \text{for some } k \in \mathbb{R}$$

Two lines \(d_1\) and \(d_2\) with direction vectors \(\vec{u}_1\) and \(\vec{u}_2\) are:

• parallel if and only if \(\vec{u}_1\) and \(\vec{u}_2\) are collinear (\(\det(\vec{u}_1, \vec{u}_2) = 0\)).
• intersecting (sécantes) if and only if \(\vec{u}_1\) and \(\vec{u}_2\) are not collinear (\(\det(\vec{u}_1, \vec{u}_2) \neq 0\)).

Every line \(d\) in the plane has an equation of the form:$$\textcolor{olive}{a}x + \textcolor{olive}{b}y + \textcolor{olive}{c} = 0$$where \(a\) and \(b\) are not both zero. This is called a Cartesian equation of the line.
A direction vector for this line is \(\Vect{u}=\begin{pmatrix} \textcolor{olive}{-b} \\ \textcolor{olive}{a} \end{pmatrix}\).

Find a direction vector for the line with equation \(2x - 3y + 5 = 0\).

The slope-intercept form of a line's equation is:$$y = mx + c$$where \(m\) is the slope (gradient) and \(c\) is the \(y\)-intercept.

• The slope \(m\) tells us how much \(y\) changes when \(x\) increases by \(1\).
• The \(y\)-intercept \(c\) is the value of \(y\) when \(x = 0\), i.e. the point where the line crosses the \(y\)-axis.

This line has slope \(m=-2\) and \(y\)-intercept \(c=1\).

Every line in the plane admits a unique reduced equation:

• If the line \(d\) is not vertical (not parallel to the \(y\)-axis), its reduced equation is of the form: \(\boldsymbol{y = mx + c}\).
• If the line \(d\) is vertical (parallel to the \(y\)-axis), its reduced equation is of the form: \(\boldsymbol{x = k}\) (where \(k\) is a constant).

The slope of a non-vertical line passing through two distinct points \(A\left(\textcolor{colordef}{x_A}, \textcolor{colorprop}{y_A}\right)\) and \(B\left(\textcolor{colordef}{x_B}, \textcolor{colorprop}{y_B}\right)\) is given by the formula:$$\text{slope} = \frac{\textcolor{colorprop}{y_B}-\textcolor{colorprop}{y_A}}{\textcolor{colordef}{x_B}-\textcolor{colordef}{x_A}}, \quad \text{where } \textcolor{colordef}{x_A} \neq \textcolor{colordef}{x_B}$$The order of the points does not matter, as long as we subtract in the same order in the numerator and the denominator.

Find the slope of the line \(\Line{AB}\) for \(A\left(\textcolor{colordef}{1}, \textcolor{colorprop}{2}\right)\) and \(B\left(\textcolor{colordef}{5}, \textcolor{colorprop}{4}\right)\).

Line Type	Equation	Direction Vector
Non-vertical	\(y = mx + p\)	\(\begin{pmatrix} 1 \\ m \end{pmatrix}\)
Vertical	\(x = c\)	\(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\)
General	\(ax + by + c = 0\)	\(\begin{pmatrix} -b \\ a \end{pmatrix}\)

Finding the intersection of two lines \(d\) and \(d'\) in the plane means looking for a point \(M(x;y)\) that belongs to both lines at the same time. This problem is equivalent to finding the solution(s) of a system of two linear equations with two unknowns.

To determine the intersection point of two non-parallel lines:

1. Write the equations of both lines as a system: \(\begin{cases} ax + by + c = 0 \\ a'x + b'y + c' = 0 \end{cases}\)
2. Solve the system using the substitution method or the elimination method.
3. The calculated values \((x_0; y_0)\) are the coordinates of the intersection point.

Find the intersection point of line \(d\) with equation \(x + y - 3 = 0\) and line \(d'\) with equation \(2x - y = 0\).

Let \(\Vect{u}=\begin{pmatrix} -b \\ a \end{pmatrix}\) and \(\Vect{v}=\begin{pmatrix} -b' \\ a' \end{pmatrix}\) be the direction vectors of lines \(d\) and \(d'\). Their determinant is \(\det(\vec{u}, \vec{v}) = ab' - a'b\).