Supports integers, decimals, signed fractions, and simple radicals such as √2 or sqrt(2).
Parsed preview: enter a linear equation.
Vertical lines do not have a finite slope. Use equation, standard, or two-point mode.
The distance uses the same unit as the coordinates. Full precision is retained internally; rounding affects display only. Press Enter in an input to calculate.
Enter valid values to see a textual summary of everything plotted.
Blue is the infinite line; red is the perpendicular distance segment. The graph preserves equal x/y scale.
The perpendicular distance is the length of the shortest segment from a point P to an infinite line. Write the line as Ax + By + C = 0 and the point as (x0, y0). Here, x0 and y0 are the point coordinates, A and B are the components of a normal vector, and C sets the line's offset. A and B cannot both be zero because then no line direction is defined.
Ax + By + C = 0 uses the main formula directly.y = mx + b to mx − y + b = 0, giving d = |mx₀ − y₀ + b| / √(m² + 1).(x₁,y₁) and (x₂,y₂), use A=y₁−y₂, B=x₂−x₁, and C=x₁y₂−x₂y₁.Let k=(Ax₀+By₀+C)/(A²+B²). The perpendicular foot is F=(x₀−Ak, y₀−Bk). This calculator reports F and plots it with a right-angle marker.
The vector n=(A,B) is normal to the line. The value Ax₀+By₀+C measures the point's offset along that normal, but it is scaled by the length of n. Dividing by ||n||=√(A²+B²) converts that offset to geometric length; the absolute value removes orientation. Equivalently, form a triangle from the point and any two points on the line: its height is twice its area divided by its base length.
A, B, and C by the same nonzero number does not change the distance; the formula is scale-invariant.Identify: P=(2,−1) and 2x−y−3=0, so A=2, B=−1, C=−3.
Substitute: d=|2(2)+(−1)(−1)−3|/√(2²+(−1)²).
Simplify: d=2/√5≈0.894427.
Interpret: The point is about 0.894 coordinate units from the line; its perpendicular foot is F=(1.2,−0.6).
Identify: P=(3,−1) and y=2x+4.
Convert: 2x−y+4=0, so A=2, B=−1, C=4.
Substitute and simplify: d=|2(3)+(−1)(−1)+4|/√5=11/√5≈4.919350.
Interpret: The projection is F=(−1.4,1.2), 4.919 coordinate units away.
Identify: P=(4,1); the line passes through P₁=(0,0) and P₂=(3,3).
Convert: A=0−3=−3, B=3−0=3, C=0, giving −3x+3y=0.
Substitute and simplify: d=|−3(4)+3(1)|/√18=9/(3√2)=3/√2≈2.121320.
Interpret: The projection is F=(2.5,2.5).
Vertical line: from P=(5,−2) to x=1, d=|5−1|=4. Point on the line: P=(2,3) satisfies x+y−5=0, so the numerator and distance are zero and F=P.
Write the line as Ax + By + C = 0, substitute the point (x0, y0), and calculate |Ax0 + By0 + C| / sqrt(A² + B²). The absolute value makes the distance nonnegative.
Rearrange y = mx + b as mx - y + b = 0. Substituting A = m, B = -1, and C = b into the standard formula gives |mx0 - y0 + b| / sqrt(m² + 1).
For distinct line points (x1, y1) and (x2, y2), use A = y1 - y2, B = x2 - x1, and C = x1y2 - x2y1, then apply the standard distance formula.
A vertical line x = k has standard form x - k = 0, so the distance from (x0, y0) is |x0 - k|. Use equation, standard, or two-point mode.
No. Geometric distance is always zero or positive. Removing the absolute value produces a signed distance that indicates which side of an oriented line contains the point.
The formula numerator is zero, so the distance is zero and the projection point is the original point.
The distance uses the same unit as the point and line coordinates. If coordinates are in metres, the distance is in metres.
Not always. This calculator uses an infinite line. For a segment, the closest location can be an endpoint when the perpendicular projection falls outside the segment.
Let k = (Ax0 + By0 + C) / (A² + B²). The perpendicular foot is F = (x0 - Ak, y0 - Bk).
The implementation converts every supported input to Ax+By+C=0, evaluates the normal-vector distance formula, and uses orthogonal projection for F. These reference cases are included as transparent checks:
| Point and line | Expected distance | Expected projection |
|---|---|---|
(2,−1); 2x−y−3=0 | 2/√5≈0.894427 | (1.2,−0.6) |
(5,−2); x=1 | 4 | (1,−2) |
(2,3); x+y−5=0 | 0 | (2,3) |
Limitations: calculations use browser floating-point arithmetic, so very large or nearly degenerate values can accumulate rounding error. The parser accepts linear equations with numeric coefficients, signed fractions, and simple square roots; it intentionally rejects variables in denominators, powers, functions, and nonlinear terms. Results apply to an infinite 2D line, not a ray or segment.