Parallel, Intersecting, and Coincident Lines

Three possible positions

If both lines are written in slope-intercept form $y = mx + b$, the slopes tell you almost everything.

Try the sliders: change the slope $m$ of the first line and watch what happens to its position relative to $y = 2x - 3$.

Some problems call slope-intercept form the explicit equation, standard form $Ax + By + C = 0$ the general equation, and symmetric form (you'll meet it later on this page) the continuous form. They are just different ways of writing the same line.

Comparing coefficients in $Ax + By + C = 0$

Many problems give lines as $Ax + By + C = 0$. (US standard form is usually written $Ax + By = C$, which is the same thing with $C$ moved to the other side.) Solve for $y$ to see the slope:

Comparing slopes and intercepts of two such lines gives a shortcut that needs no rewriting.

Finding the point of intersection

If two lines are intersecting, the point where they cross lies on both lines. So its coordinates satisfy both equations, which means you solve the system of the two equations (substitution or elimination).

Lines given by two points, parametric equations, or symmetric form

Lines don't always arrive as $y = mx + b$. The strategy is always the same: get the slope of each line. If the slopes differ, the lines intersect. If they match, test one point of the first line in the second.

Three ways to read a slope

• Two points $(x_1, y_1)$ and $(x_2, y_2)$: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.

• Parametric equations $x = x_0 + at,\ y = y_0 + bt$: the line passes through $(x_0, y_0)$ (when $t = 0$) and moves $a$ units across and $b$ up for each step of $t$. So the direction vector is $\langle a, b \rangle$ and the slope is $\dfrac{b}{a}$.

• Symmetric form $\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b}$: the line passes through $(x_0, y_0)$ and again the slope is $\dfrac{b}{a}$. To get $y = mx + b$ form, cross-multiply: $b(x - x_0) = a(y - y_0)$.

Two direction vectors point the same way when their components are proportional.

Parallel and perpendicular lines through a point

Often you must build a new line: through a point and parallel to a given line, or through a point and perpendicular to it.

The shortest path from a point to a line is the perpendicular segment. That is why "the shortest cable from a point to a road" is the perpendicular through that point.

Triangles from three lines

Three lines that are pairwise intersecting form a triangle. Its vertices are the three intersection points, one for each pair of lines.

A line and the two coordinate axes also form a triangle. Its vertices are the origin and the two intercepts, and the axes are perpendicular, so it is a right triangle with the intercepts as legs.

Perpendicular feet, reflections, and rectangles

Combine perpendicular lines with intersections and you can solve harder geometry problems.

Reflecting a point over a line

To reflect point $A$ over line $r$: draw the perpendicular from $A$ to $r$, find where it meets $r$ (the foot $M$), and go the same distance beyond. So $M$ is the midpoint of $A$ and its image $A'$:

A ray bouncing off the x-axis

A mirror on the x-axis reflects the picture across the axis: $y$ becomes $-y$. A ray hitting the mirror at a point and leaving with equal angles keeps that point and flips the sign of the slope.

Rectangles

In a rectangle, opposite sides are parallel and adjacent sides are perpendicular. If two given sides are perpendicular and a vertex $P$ is somewhere else, the other two sides both pass through $P$ and are parallel to the given sides.

Summary

Back to the map: two streets that are "really the same street" are coincident lines. You detect that when every coefficient ratio matches, or when the slopes and intercepts are equal.