Last updated: 2026-06-05

What is the distance from a point to a line?

The distance from a point P = (p_x, p_y) to a line is the length of the shortest segment connecting P to any point on the line. That shortest segment is always perpendicular to the line, so the distance is called the perpendicular distance. Given a line in standard form Ax + By + C = 0, the formula is:

$d = \frac{|Ap_x + Bp_y + C|}{\sqrt{A^2 + B^2}}$

The denominator normalises the result by the length of the line's normal vector, converting the raw algebraic expression into a true geometric length.

Derivation

The normal vector to the line Ax + By + C = 0 is n = (A, B). Any point Q = (x₀, y₀) on the line satisfies Ax₀ + By₀ + C = 0.

The signed projection of the vector QP = (p_x − x₀, p_y − y₀) onto the unit normal n̂ = (A, B)/|n| equals:

$\text{signed distance} = \frac{A(p_x - x_0) + B(p_y - y_0)}{|\mathbf{n}|} = \frac{Ap_x + Bp_y - (Ax_0 + By_0)}{|\mathbf{n}|}$

Since Q is on the line, Ax₀ + By₀ = −C, so:

$\text{signed distance} = \frac{Ap_x + Bp_y + C}{\sqrt{A^2 + B^2}}$

Taking the absolute value gives the unsigned geometric distance d. The sign indicates which side of the line the point lies on.

Foot of the perpendicular

The foot F is the point on the line closest to P. It is obtained by moving from P along the inward normal by exactly t = (Ap_x + Bp_y + C) / (A² + B²):

$F_x = p_x - \frac{A(Ap_x + Bp_y + C)}{A^2 + B^2}, \qquad F_y = p_y - \frac{B(Ap_x + Bp_y + C)}{A^2 + B^2}$

One can verify that AF_x + BF_y + C = 0 (F is on the line) and that the segment PF has length d.

Worked example

Problem: Find the distance from P = (1, −2) to the line 3x − 4y + 5 = 0, and locate the foot of the perpendicular.

Step 1 — compute the numerator and denominator: $Ap_x + Bp_y + C = 3(1) + (-4)(-2) + 5 = 3 + 8 + 5 = 16$ $\sqrt{A^2 + B^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 2 — distance: $d = \frac{|16|}{5} = 3.2$

Step 3 — foot coordinates: $t = \frac{16}{25} = 0.64$ $F_x = 1 - 3(0.64) = 1 - 1.92 = -0.92, \qquad F_y = -2 - (-4)(0.64) = -2 + 2.56 = 0.56$

Verification: 3(−0.92) + (−4)(0.56) + 5 = −2.76 − 2.24 + 5 = 0 ✓

The point P is 3.2 units from the line, and the closest point on the line is F = (−0.92, 0.56).

Converting from slope-intercept form

If the line is given as y = mx + b, rewrite it as:

$mx - y + b = 0 \quad \Rightarrow \quad A = m,\; B = -1,\; C = b$

For example, the line y = 2x − 3 becomes 2x − y − 3 = 0, so A = 2, B = −1, C = −3.

Note that multiplying all of A, B, and C by the same non-zero constant does not change the line or the distance — the numerator and the denominator scale by the same factor and cancel.

Signed distance

The calculator reports the unsigned distance d ≥ 0. The signed version (Ap_x + Bp_y + C) / √(A² + B²) is positive when P is on the same side as the normal vector direction and negative on the other side. Signed distance is useful in applications such as half-plane containment checks or support-vector classification.

Special cases

• Point on the line: Ap_x + Bp_y + C = 0, so d = 0 and F = P.
• Horizontal line (A = 0): simplifies to d = |Bp_y + C| / |B|.
• Vertical line (B = 0): simplifies to d = |Ap_x + C| / |A|.
• Degenerate equation (A = B = 0): C = 0 means every point satisfies the equation (not a line); C ≠ 0 means no point satisfies it. Either way the distance formula is undefined.

Relationship to the Pythagorean theorem

The derivation can be visualised as a right triangle: P, the foot F, and any third point Q on the line form a triangle with the right angle at F. The hypotenuse PQ satisfies PQ² = PF² + FQ², confirming that PF (the perpendicular distance) is shorter than any other segment from P to the line.