Find the perpendicular line through a point given a line equation. Calculate point-to-line distance, intersection point, and perpendicular slope with step-by-step solutions.
Finding a perpendicular line through a given point is one of the most common tasks in analytic geometry. Given a line and an external point, this calculator determines: the equation of the perpendicular line through the point, the foot of the perpendicular (the intersection of the original and perpendicular lines), and the shortest distance from the point to the line.
The key relationship is elegantly simple: if a line has slope m₁, any line perpendicular to it has slope m₂ = −1/m₁. Their slopes are negative reciprocals — and their product is always −1. For vertical lines (undefined slope), the perpendicular is horizontal, and vice versa. The point-to-line distance formula d = |ax₀ + by₀ + c|/√(a² + b²) gives the shortest (perpendicular) distance from a point to a line.
The page accepts line equations in multiple formats — slope-intercept (y = mx + b), standard form (ax + by + c = 0), or via two points on the line — and computes everything with a step-by-step breakdown. Presets for common configurations let you explore quickly. Whether you are solving geometry homework, computing distances in CAD software, or deriving normal lines for reflections, this calculator handles it all.
Finding perpendicular lines, intersection points, and point-to-line distances by hand involves converting between line forms, computing negative reciprocals, and solving linear systems — a multi-step process, especially with non-integer slopes. This calculator handles all three input formats and shows every intermediate step.
It is particularly useful for geometry students working through perpendicular line problems, engineers computing normal distances, and programmers verifying geometric calculations in code.
Perpendicular slope: m₂ = −1/m₁. Distance from point (x₀, y₀) to line ax + by + c = 0: d = |ax₀ + by₀ + c| / √(a² + b²). Perpendicular line through (x₀, y₀): y − y₀ = m₂(x − x₀). Foot of perpendicular: solve the system of original and perpendicular line equations.Result: Perpendicular: y = −0.5x + 1.5, Intersection: (−0.6, 1.8), Distance ≈ 1.789
Line: y = 2x + 3 (or 2x − y + 3 = 0). Perpendicular slope = −1/2. Perp line through (1, 1): y = −0.5x + 1.5. Solving gives intersection (−0.6, 1.8). Distance = |2(1) − 1(1) + 3|/√(4 + 1) = 4/√5 ≈ 1.789.
The concept of perpendicularity bridges algebra and geometry. In the Cartesian plane, two lines are perpendicular if and only if the product of their slopes equals −1 (assuming neither is vertical). This elegant condition, m₁ · m₂ = −1, follows from the fact that rotating a direction vector by 90° swaps its components and negates one of them.
The standard form ax + by + c = 0 reveals perpendicularity even more directly: the vector (a, b) is the normal (perpendicular) vector to the line. Any line perpendicular to it has direction vector (a, b), meaning its equation is bx − ay + k = 0 for some constant k.
The distance formula d = |ax₀ + by₀ + c|/√(a² + b²) appears throughout mathematics and engineering: in linear regression (finding the best-fit line minimizes the sum of squared distances), in computational geometry (point-in-polygon tests use signed distances), in physics (computing the impact parameter of trajectories), and in optimization (constraint satisfaction uses perpendicular distances).
The foot of the perpendicular is also known as the orthogonal projection of the point onto the line. This concept generalizes to higher dimensions: projecting a vector onto a subspace involves finding the closest point on that subspace. In linear algebra, this is expressed as proj_u(v) = (v · u / u · u) u, which is the algebraic equivalent of dropping a perpendicular from the tip of v onto the line spanned by u.
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Two lines are perpendicular if they intersect at a 90° angle. For two non-vertical lines with slopes m₁ and m₂, this means m₁ × m₂ = −1.
Take the negative reciprocal of the original slope: m₂ = −1/m₁. For example, if m₁ = 2, then m₂ = −1/2.
The foot of the perpendicular is the point where the perpendicular line meets the original line. It is the closest point on the original line to the external point.
A vertical line (undefined slope, e.g., x = 5) has a horizontal perpendicular (slope = 0). The perpendicular through (x₀, y₀) is simply y = y₀.
Yes (or zero if the point lies on the line). The absolute value in the formula ensures a non-negative result, representing the shortest distance.
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