Distance of a Point from a Plane

The distance from a point to a hyperplane is the single geometric quantity that underpins Support Vector Machines, the margin concept, and the intuition behind every linear classifier. Before reaching SVMs, you need this formula cold — in 2D, 3D, and $p$ dimensions — and you need to understand what the sign of that distance tells you.

Distance from a Point to a Line (2D)

A line in 2D is written in general form as $ax+by+c=0$. The distance from a point $(x_0,y_0)$ to this line is the length of the perpendicular from the point to the line.

Why the perpendicular? Any other path from the point to the line is longer — the perpendicular is the shortest. The perpendicular meets the line at the point where the line's normal vector $\mathbf{n}=(a,b)$ points from the line to $(x_0,y_0)$.

Working through the projection, the signed distance is:

$d=\frac{a{x}_0+b{y}_0+c}{\sqrt{a^2+b^2}}$

The absolute value $|d|$ gives the distance; the sign tells you which side of the line the point is on.

Anchor setup: Predict loan default from income ($x_1$, in \k$)anddebtratio($x_2$).Acandidatedecisionlineis$0.5 x_1 - 3 x_2 - 20 = 0$,so$a = 0.5$,$b = -3$,$c = -20$.Thedenominatoris$\sqrt{0.5^2 + 3^2} = \sqrt{0.25 + 9} = \sqrt{9.25} \approx 3.04$.

Computing distances for three anchor points:

Point $(45,0.35)$ — income $45k, debt ratio 0.35:

$d=\frac{0.5\times 45+(-3)\times 0.35+(-20)}{3.04}=\frac{22.5-1.05-20}{3.04}=\frac{1.45}{3.04}\approx 0.477$

Point $(31,0.62)$ — income $31k, debt ratio 0.62:

$d=\frac{0.5\times 31+(-3)\times 0.62+(-20)}{3.04}=\frac{15.5-1.86-20}{3.04}=\frac{-6.36}{3.04}\approx -2.09$

Point $(95,0.12)$ — income $95k, debt ratio 0.12:

$d=\frac{0.5\times 95+(-3)\times 0.12+(-20)}{3.04}=\frac{47.5-0.36-20}{3.04}=\frac{27.14}{3.04}\approx 8.93$

<line x1="60" y1="280" x2="520" y2="280" stroke="#334155" stroke-width="1.5"/> <line x1="60" y1="20" x2="60" y2="280" stroke="#334155" stroke-width="1.5"/> <text x="290" y="310" text-anchor="middle" font-size="12" fill="#334155">income ($k)</text> <text x="20" y="150" text-anchor="middle" font-size="12" fill="#334155" transform="rotate(-90,20,150)">debt ratio</text> <line x1="100" y1="20" x2="500" y2="200" stroke="#3b82f6" stroke-width="1.8"/> <text x="410" y="196" font-size="10" fill="#3b82f6">0.5x₁ − 3x₂ − 20 = 0</text> <circle cx="200" cy="230" r="6" fill="#dc2626"/> <text x="208" y="228" font-size="10" fill="#dc2626">(45, 0.35) default</text> <line x1="200" y1="230" x2="205" y2="210" stroke="#f59e0b" stroke-width="1.5"/> <text x="208" y="222" font-size="9" fill="#f59e0b">d=0.48</text> <circle cx="130" cy="80" r="6" fill="#dc2626"/> <text x="138" y="78" font-size="10" fill="#dc2626">(31, 0.62) default</text> <line x1="130" y1="80" x2="165" y2="95" stroke="#f59e0b" stroke-width="1.5"/> <text x="148" y="85" font-size="9" fill="#f59e0b">d=2.09</text> <circle cx="460" cy="250" r="6" fill="#22c55e"/> <text x="380" y="268" font-size="10" fill="#22c55e">(95, 0.12) no-default</text> <line x1="460" y1="250" x2="415" y2="228" stroke="#f59e0b" stroke-width="1.5"/> <text x="420" y="245" font-size="9" fill="#f59e0b">d=8.93</text> <circle cx="300" cy="200" r="6" fill="#22c55e"/> <circle cx="380" cy="260" r="6" fill="#22c55e"/> <circle cx="150" cy="140" r="6" fill="#dc2626"/>

Signed Distance — Which Side of the Line?

The sign of $a{x}_0+b{x}_0+c$ directly identifies which side of the decision boundary a point is on:

• Positive ($+1.45$ for point (45, 0.35)): on the side where $0.5x_1-3x_2-20>0$. Classifier predicts class +1.
• Negative ($-6.36$ for point (31, 0.62)): on the opposite side. Classifier predicts class −1.

This signed quantity is exactly what SVMs use. The sign is the prediction; the magnitude is the confidence. A point 8.93 units from the decision boundary is much more confidently classified than one 0.48 units away.

Normalizing the Line Equation

Dividing $a$, $b$, $c$ by $\sqrt{a^2+b^2}$ creates a normalized form where the denominator equals 1, simplifying the distance formula to just the numerator:

$a^{\prime }=\frac{0.5}{3.04}=0.164,b^{\prime }=\frac{-3}{3.04}=-0.987,c^{\prime }=\frac{-20}{3.04}=-6.58$

Check for point $(45,0.35)$:

$|0.164\times 45+(-0.987)\times 0.35-6.58|=|7.38-0.35-6.58|=|0.45|\approx 0.48✓$

The small difference from 0.477 is rounding. Normalizing is why SVM theory often assumes $\Vert \mathbf{w}\Vert =1$ — it makes the distance formula clean.

Extension to 3D: Distance from a Point to a Plane

In 3D, the plane equation is $ax+by+cz+d=0$. The distance formula extends naturally:

$\text{dist}=\frac{|a{x}_0+b{y}_0+c{z}_0+d|}{\sqrt{a^2+b^2+c^2}}$

For the plane $x+2y-3z+6=0$ and point $P=(1,2,3)$:

$\text{Numerator:}|1\times 1+2\times 2+(-3)\times 3+6|=|1+4-9+6|=|2|=2$

$\text{Denominator:}\sqrt{1^2+2^2+3^2}=\sqrt{14}\approx 3.742$

$\text{dist}=\frac{2}{3.742}\approx 0.535$

The geometry is identical to the 2D case: the denominator is the length of the normal vector to the plane, and dividing by it projects the point's displacement onto the unit normal.

Extension to p Dimensions: Distance from a Point to a Hyperplane

In $p$ dimensions the hyperplane is $\mathbf{w}\cdot \mathbf{x}+w_0=0$, and the distance from a point ${\mathbf{x}}_0$ to it is:

$d=\frac{|\mathbf{w}\cdot {\mathbf{x}}_0+w_0|}{\Vert \mathbf{w}\Vert }$

This is the margin formula in SVMs. The margin between two classes is defined as $\frac{2}{\Vert \mathbf{w}\Vert }$ — twice the minimum distance from the closest training point to the decision hyperplane. Maximizing the margin means minimizing $\Vert \mathbf{w}\Vert$, which is the SVM optimization objective.

Distance Formula Reference

Related Concepts and Honest Limitations

This distance formula is the mathematical backbone of SVMs (margin maximization), signed output in logistic regression (the log-odds is $\mathbf{w}\cdot \mathbf{x}$), and the geometric interpretation of regularization (constraining $\Vert \mathbf{w}\Vert$ constrains the margin). The formula requires the hyperplane to be parameterized in the $\mathbf{w}\cdot \mathbf{x}+w_0=0$ form — not all representations make the normal vector explicit.

The limitation here is dimensionality. In high dimensions, points that look geometrically distant can actually cluster near the surface of a hyperphere (the curse of dimensionality). Distance-based reasoning becomes unreliable when $p\gg n$, which is part of why SVMs use kernel tricks to work in feature spaces rather than raw input spaces.

Test Your Understanding

1. For the decision line $0.5x_1-3x_2-20=0$, compute the distance for the point $(67,0.28)$ using the formula. Which class does the sign predict?

2. If you multiply both sides of the line equation by $-1$ (so $-0.5x_1+3x_2+20=0$), does the distance change? Does the signed distance change?

3. In SVM, the margin is $2/\Vert \mathbf{w}\Vert$. If $\mathbf{w}=[0.5,-3]$ as in our anchor, what is the margin? To double the margin, what would you need to do to $\mathbf{w}$?

4. The normalization step changes $a=0.5$ to $a^{\prime }=0.164$. If you use the normalized coefficients in the distance formula, why does the denominator disappear?

5. A data point lies exactly on the decision hyperplane. What is its signed distance? What prediction does an SVM make for it, and why is this a problem in practice?