Lines Revisited: Slope and General Form

Part 13, Chapter 13: Conic Sections in the Coordinate Plane

Learning objectives

Write the general form ax + by + c = 0 of a line
Find the normal vector to a line
Compute the distance from a point to a line
Convert between slope-intercept and general form

The slope-intercept form $y = mx + b$ is the line you first met. It is fine for sketching, but analytic geometry needs a form that is more symmetric and that does not break down for vertical lines. The general form $ax + by + c = 0$ treats $x$ and $y$ on equal footing and instantly reveals the line's normal vector, which is the key to computing distances from points to lines, a tool you will use again and again in the conic sections to follow.

General form

Every straight line in the plane satisfies an equation of the form

$ax + by + c = 0, \qquad (a, b) \neq (0, 0)$

The pair $(a, b)$ controls the direction; the scalar $c$ slides the line parallel to itself. Vertical lines $x = k$ correspond to $b = 0$ ; horizontal lines $y = k$ correspond to $a = 0$ . Slope-intercept cannot express vertical lines, general form can.

The normal vector is the gradient

The vector $\mathbf{n} = (a, b)$ points perpendicular to the line. Why? Because if $(x_1, y_1)$ and $(x_2, y_2)$ both satisfy $ax + by + c = 0$ , subtracting gives $a(x_1 - x_2) + b(y_1 - y_2) = 0$ , i.e. $\mathbf{n} \cdot (x_1 - x_2, y_1 - y_2) = 0$ . The dot product vanishes, so $\mathbf{n}$ is perpendicular to every direction along the line.

The direction vector along the line is $(-b, a)$ (rotate the normal by $\pi/2$ ). Either choice of sign works.

Where this shows up

Computer Vision: The general-form line $ax + by + c = 0$ is the natural output of the Hough transform; $(a, b)$ is the line's normal vector, and $c$ encodes its signed distance from the origin.

Image Processing: Edge-detection filters output gradient vectors that point along the normal $(a, b)$ of local edges; aligning these normals across pixels is how line-detection algorithms find roads, walls, and text.

Machine Learning: Half-spaces $ax + by + c \geq 0$ are the building blocks of linear-programming feasible regions and of support-vector-machine classifiers; both are general-form lines used as decision boundaries.

(Drag the two points to redraw the line. The widget displays both the general form and the slope-intercept form, so you can see them as two views of the same object.)

Distance from a point to a line

The most-used formula in this chapter is

$d = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}$

where $(x_0, y_0)$ is the external point and $ax + by + c = 0$ is the line. The numerator is the signed value of the line equation at the point; the denominator is the length of the normal vector. Dividing turns "how much does this point miss the equation by" into a true Euclidean distance.

Converting between forms

From slope-intercept: $y = mx + k$ becomes $mx - y + k = 0$ , so $(a, b, c) = (m, -1, k)$ .

From general form: if $b \neq 0$ , the slope is $m = -a/b$ and the $y$ -intercept is $-c/b$ . Two lines are parallel when $a_1 b_2 = a_2 b_1$ ; they are perpendicular when $a_1 a_2 + b_1 b_2 = 0$ (dot product of normals is zero).

Try it

Find the distance from $(3, -1)$ to $2x + 3y - 6 = 0$ . (Plug into the formula.)

Write $y = -\tfrac{3}{4}x + 2$ in general form with integer coefficients. (Multiply through by $4$ .)

Find the angle between the two lines $x - y = 1$ and $x + y = 3$ . (Use the normals: their dot product divided by the product of lengths gives $\cos\theta$ .)

Pause: is $y = mx + b$ really general, or is it missing some lines? It is missing all vertical lines, because they have undefined slope. The general form fixes this gap.

A trap to watch for

The distance formula needs the line in general form. If you have $y = mx + b$ and you plug $(x_0, y_0)$ into $y - mx - b$ , you do not get the distance, you get the vertical gap from the point to the line. To get the perpendicular distance, you must first rewrite as $mx - y + b = 0$ , then divide by $\sqrt{m^2 + 1}$ . Forgetting the denominator overestimates the distance by a factor of $\sqrt{m^2 + 1}$ , which can be huge for steep lines.

What you now know

You can write any line in general form $ax + by + c = 0$ (including verticals), identify the normal vector, compute the perpendicular distance from a point, and translate between slope-intercept and general form. The next section turns these coordinate-method tools loose on curved figures, starting with the parabola.

Quick check

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References

Lang, S. (1971). Basic Mathematics. Springer. Chapter 12, §1, the general-form approach to the line in analytic geometry.

Apostol, T. M. (1969). Calculus, Volume 2. Wiley. Chapter 1: lines and planes in vector form, including the point-to-line distance derivation.

Hartshorne, R. (2000). Geometry: Euclid and Beyond. Springer. Chapter 3: analytic and synthetic treatments of lines side by side.

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