Straight Line — General Equation

A railway track runs dead straight from one town to the next. You stand at the station and look along the track. Two things define that track completely: where it starts and which direction it goes. But there are many different ways to describe the same track — you could give its slope and where it crosses a river, or two points it passes through, or its angle with the horizon and how far it sits from a landmark.

In the previous article on forms of a line, you met several of these descriptions: slope-intercept (y = mx + c), point-slope, two-point, intercept, and normal form. Each is useful in its own situation. But each also has a blind spot. Slope-intercept breaks for vertical lines (slope is undefined). Intercept form fails when the line passes through the origin (both intercepts are zero). Normal form needs you to know the perpendicular distance from the origin.

There is one form that handles every line — vertical, horizontal, through the origin, at any angle — with no exceptions. It is the general equation.

The general equation

That last condition — A and B not both zero — is doing real work. If A = B = 0, the equation becomes C = 0, which is either always true (every point in the plane) or always false (no point). Neither of those is a line.

When B \neq 0, you can solve for y:

This is the slope-intercept form y = mx + c with m = -A/B and c = -C/B. The slope is hiding inside the ratio of the two coefficients.

When B = 0 (and A \neq 0), the equation becomes Ax + C = 0, which gives x = -C/A — a vertical line. The general form handles vertical lines with no special syntax, no "undefined slope" exception. That is the whole point.

Angle of inclination

Before converting between forms, there is a more primitive quantity that controls a line's direction: its angle of inclination.

A horizontal line has \theta = 0°. A line tilted upward to the right at 45° has \theta = 45°. A vertical line has \theta = 90°. A line tilted downward to the right has \theta between 90° and 180°.

The slope m and the inclination \theta are related by

For a vertical line, \theta = 90° and \tan 90° is undefined — which is exactly why slope-intercept form breaks for vertical lines. The angle of inclination is the deeper quantity; slope is its tangent.

Indian railways are a good way to visualise this. A track running due east has inclination 0°. A track heading northeast at 45° rises one unit for every unit it moves to the right — slope 1. A track running due north has inclination 90° — it is a vertical line on the map, and no finite slope can describe it. The inclination captures all of these cases in a single, well-defined angle.

Given the general equation Ax + By + C = 0 (with B \neq 0), the inclination is

The two cases ensure \theta lands in the range [0°, 180°). If B = 0, the line is vertical and \theta = 90°.

Converting between forms

Every named form of the line equation is a rearrangement of Ax + By + C = 0 with specific choices of A, B, C. Here is a reference table.

The conversion is always mechanical: rearrange everything to one side so that the equation reads (\ldots) = 0, and read off the coefficients.

From general to slope-intercept. Given 3x + 4y - 12 = 0, solve for y:

The slope is -3/4 and the y-intercept is 3.

From general to intercept form. The same equation 3x + 4y - 12 = 0 can be rewritten as 3x + 4y = 12. Divide both sides by 12:

The x-intercept is 4 and the y-intercept is 3.

From general to normal form. Start with Ax + By + C = 0, rewrite as Ax + By = -C. Divide both sides by \sqrt{A^2 + B^2} (choosing sign so the right side is non-negative):

The coefficients of x and y are now \cos\alpha and \sin\alpha for some angle \alpha, and the right side is the perpendicular distance p from the origin. Why \cos\alpha and \sin\alpha? Because \left(\frac{A}{\sqrt{A^2+B^2}}\right)^2 + \left(\frac{B}{\sqrt{A^2+B^2}}\right)^2 = 1, so the pair lies on the unit circle and can be written as (\cos\alpha, \sin\alpha).

For the line 3x + 4y - 12 = 0: rewrite as 3x + 4y = 12, divide by \sqrt{9 + 16} = 5:

Here \cos\alpha = 3/5, \sin\alpha = 4/5 (so \alpha \approx 53.13°), and p = 12/5 = 2.4. The perpendicular from the origin to the line has length 2.4 and makes an angle of about 53° with the x-axis.

Why the general form matters

You might ask: if slope-intercept is simpler, why bother with Ax + By + C = 0?

Three reasons. First, universality — it handles vertical lines, lines through the origin, and every other case uniformly. Second, the distance formula — the formula for the distance from a point to a line uses A, B, C directly (you will see this in Distance Formulas). Third, systems of equations — when you solve two simultaneous linear equations, the algebra is cleanest in general form, because you can add and subtract the equations without rearranging.

The parametric form (distance form)

All the forms above describe a line as a condition that a point (x, y) must satisfy. The parametric form takes a different approach: it describes a line as a path — a recipe for generating every point on the line by walking along it from a starting position.

Equivalently, you can write this as two separate equations:

The parameter r runs from -\infty to +\infty. When r > 0, you are walking in the direction of inclination \theta. When r < 0, you are walking in the opposite direction. When r = 0, you are standing at the starting point (x_1, y_1).

Why "distance form"? Because |r| is literally the distance from (x_1, y_1) to (x, y). You can verify this: the distance between those two points is

Why: \cos^2\theta + \sin^2\theta = 1 is the Pythagorean identity, so the square root collapses to |r|.

This form is powerful when a problem says something like "find the point on the line that is a distance d from a given point." In slope-intercept form, you would need to set up a distance equation and solve a quadratic. In parametric form, you just set r = \pm d and read off the coordinates.

Deriving the parametric form

Start with a point P = (x_1, y_1) on the line and another point Q = (x, y) on the same line, at distance |r| from P. The direction from P to Q is along the line, which has inclination \theta.

The horizontal displacement from P to Q is the distance times \cos\theta:

Why: if you walk a distance r at angle \theta to the horizontal, the horizontal component is r\cos\theta. The sign of r handles direction automatically.

The vertical displacement is the distance times \sin\theta:

Why: similarly, the vertical component of walking distance r at angle \theta is r\sin\theta.

Dividing one by the other:

This is the parametric form. The derivation is nothing more than resolving a displacement vector into its horizontal and vertical components — the same triangle you have been drawing since you first met trigonometry.

Worked examples

Common confusions

• "The general equation Ax + By + C = 0 is unique." It is not. Multiplying the entire equation by any non-zero constant k gives kAx + kBy + kC = 0 — a different triple (kA, kB, kC) describing the same line. The equation determines the line, but the coefficients are only determined up to a common scalar multiple.

• "A is the slope." No — the slope is -A/B. Getting the sign wrong here is one of the most common mistakes in coordinate geometry. Always double-check: from Ax + By + C = 0, solving for y gives y = (-A/B)x + (-C/B).

• "Angle of inclination can be negative." By convention, \theta is always in [0°, 180°). If your calculator gives you a negative angle from \arctan, add 180° to bring it into range.

• "The parametric form only works when \theta \neq 0° and \theta \neq 90°." It works for all angles. When \theta = 0°, \cos\theta = 1 and \sin\theta = 0, so the parametric equations become x = x_1 + r, y = y_1 — a horizontal line. When \theta = 90°, \cos\theta = 0 and \sin\theta = 1, so x = x_1, y = y_1 + r — a vertical line. In both cases, the formula in the "fraction" form \frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} must be read carefully, since one denominator is zero. The two-equation form x = x_1 + r\cos\theta, y = y_1 + r\sin\theta has no such issue.

• "Normal form and general form are the same thing." Close, but not quite. Normal form is a specific normalisation of the general form where the coefficients of x and y satisfy (\text{coeff of } x)^2 + (\text{coeff of } y)^2 = 1 and the right-hand side is non-negative. Every normal-form equation is a general equation, but most general equations are not in normal form.

Going deeper

The direction vector and normal vector

The general equation Ax + By + C = 0 hides two important vectors.

The normal vector to the line is \vec{n} = (A, B). This is a vector perpendicular to the line. To see why, take any two points (x_1, y_1) and (x_2, y_2) on the line. Both satisfy the equation, so

Subtracting: A(x_2 - x_1) + B(y_2 - y_1) = 0. This says the dot product of (A, B) and (x_2 - x_1, \, y_2 - y_1) is zero. The second vector points along the line (from one point to another); the first vector is perpendicular to it.

The direction vector of the line is \vec{d} = (-B, A) — or equivalently (B, -A). This is a vector parallel to the line. You can verify: A \cdot (-B) + B \cdot A = 0, so \vec{d} is indeed perpendicular to the normal \vec{n}, which means it lies along the line.

The parametric form is the same thing expressed differently: \vec{d} = (\cos\theta, \sin\theta) is the unit direction vector of the line. The parametric equations x = x_1 + r\cos\theta, y = y_1 + r\sin\theta say "start at (x_1, y_1) and walk r units in the direction \vec{d}."

Families of lines

Since Ax + By + C = 0 is determined only up to a scalar multiple, the "true" data of a line is the ratio A : B : C. This is a point in the projective plane — a concept from projective geometry. In JEE Advanced problems, you sometimes encounter families of lines: for instance, \lambda(2x + 3y - 1) + \mu(x - y + 5) = 0 describes every line passing through the intersection of 2x + 3y - 1 = 0 and x - y + 5 = 0, as \lambda and \mu vary. This is the one-parameter family of lines through a fixed point, and it is easiest to express using the general form.

Historical note

The idea of describing geometric objects by algebraic equations goes back to Descartes (1637) and Fermat (working independently around the same time). But the coordinate approach to geometry was also anticipated by Indian mathematicians — Bhaskara II, in the Lilavati (12th century), worked extensively with problems about distances, areas, and positions that are today solved using coordinate geometry. The general equation Ax + By + C = 0 is the algebraic distillation of a line's geometry, and it is the starting point for the entire study of analytic geometry.

Where this leads next

The general equation is the foundation for everything that follows in straight-line geometry. The next set of articles builds directly on it.

• Angle and Conditions — how to find the angle between two lines, and the conditions for two lines to be parallel, perpendicular, or concurrent.
• Distance Formulas — the formula for the distance from a point to a line (which uses A, B, C directly), and the distance between parallel lines.
• Straight Line — Forms — the individual forms in more detail, if you want to revisit slope-intercept, point-slope, or intercept form.
• Angle Bisectors — the equations of lines that bisect the angle between two given lines — another formula that uses A, B, C directly.
• Locus — using the general equation to describe sets of points satisfying geometric conditions.