Distance from a Point to a Line Calculator (Foot of the Perpendicular and Parallel Lines)

Choose how the line is written (general form or slope-intercept form), then enter the coefficients and the coordinates of the point. The equation below is linked to the input fields, so you can also edit the numbers in the equation directly.

Find

Form of the line

x coefficient a

y coefficient b

Constant c

2nd line's constant c₂

Slope m

y-intercept k

2nd line's y-intercept k₂

x-coordinate of P, x₀

y-coordinate of P, y₀

Enter numbers only. Decimals, negative numbers and fractions such as 3/4 are OK. A blank coefficient of x or y counts as 1 (x is the same as 1x), and a blank constant or coordinate counts as 0.

Result and figure

Enter the coefficients of the line and the coordinates of the point in the fields on the left and press "Calculate". The result and a figure will appear here.

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What you can do on this page
What is this calculation used for?
How to Use
Formulas and figures
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Enter the coefficients and coordinates, and you get the distance from a point \(\mathrm{P}(x_0,\ y_0)\) to a line \(ax + by + c = 0\) on the spot (the length of the perpendicular from the point to the line, which is the shortest distance)
The answer is shown both as an exact value with a rationalized denominator, such as \(\dfrac{7\sqrt{5}}{5}\), and as a decimal (12 significant digits)
Besides the distance, you also get the coordinates of the foot of the perpendicular \(\mathrm{H}\) (the point on the line closest to \(\mathrm{P}\)) and the equation of the perpendicular line itself
You can enter the line in general form \(ax + by + c = 0\) or in slope-intercept form \(y = mx + k\) (for slope-intercept form, the steps also show how to rewrite it in general form)
The distance between two parallel lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) can be found on the same screen
You can check the result on a figure of the coordinate plane. A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are also on this page

Coefficients and coordinates can be decimals, negative numbers or fractions such as 3/4. This page works with lines in the plane (2D).

What is this calculation used for?

How GPS and map apps know which road you are on

The location your phone or car GPS receives can be off by several feet to over a hundred feet. If the app simply plotted that point, you might appear inside a building or in the middle of a field. So map apps calculate the distance from the measured point to each nearby road (on the map, a set of line segments) and move the point onto the closest road. This is called map matching, and the distance from a point to a line is at its core.
If the app also finds the foot of the perpendicular, it knows exactly where on that road you are, so it can tell you how far it is to the next turn.

Measuring offsets from a reference line in construction and surveying

Building plans have grid lines, and road and railway designs have a centerline. The positions of columns, walls, curbs and stakes are given as "how far from the reference line". This offset is exactly the distance from a point to a line. On site, surveyors calculate it from coordinates measured with a total station and check it against the design value.
Placing a curb at half the road width from the centerline is the same calculation repeated along the road.

How machine learning picks a line that separates two groups

A classification method called the support vector machine draws a line (a plane in higher dimensions) between two groups of data. It picks the line that is as far as possible from the data points of both groups, and that "how far" is the distance from a point to a line. Choosing the line with the widest gap (margin) is believed to make fewer mistakes on data the model has not seen yet.
It has been a standard method for spam filtering and image classification since before deep learning spread, and it is still a strong choice when there is little training data.

Finding lines in images and point clouds (image processing and self-driving cars)

Lane detection in self-driving cars and straightening a scanned document both need the line that best fits a set of detected points. How well a line fits is measured by adding up the squared distances from each point to the line, and the line with the smallest total is chosen.
RANSAC, a method that fits a line while ignoring outliers, also uses the point-to-line distance: points within a set distance of the line count as fitting it.

Simplifying paths in maps and graphics

When you zoom out on a map app, small bends in coastlines and roads are dropped automatically. This uses the Ramer–Douglas–Peucker algorithm. It finds the point farthest from the straight line joining the two ends of a path. If that distance is smaller than a set limit, all the points in between are dropped; if not, the path is split at that point and the check is repeated.
The shape still looks right with far less data, because the algorithm measures "how far off the line" each point is with a proper distance.

Formulas and figures

Distance from a point to a line

Figure

Standard notation (the usual math form)

\(d\) \(=\) \(\left|a x_0 + b y_0 + c\right|\) \(\div\) \(\sqrt{a^{2} + b^{2}}\)

In words (symbols replaced with words)

③ \(d\): distance from \(\mathrm{P}\) to the line \(=\) ① absolute value of \(ax + by + c\) at the point \(\div\) ② \(\sqrt{a^2 + b^2}\): length of the normal vector

The formula in words

① Take the absolute value of \(ax + by + c\) with the point \((x_0,\ y_0)\) plugged in

② divide it by the length \(\sqrt{a^2 + b^2}\) of the normal vector \((a,\ b)\)

③ and you get the distance \(d\) from \(\mathrm{P}\) to the line

Quick example

The distance from the point \(\mathrm{P}(4,\ 3)\) to the line \(x + 2y - 3 = 0\) is

distance \(d\) \(=\) absolute value at the point (\(|4 + 6 - 3| = 7\)) \(\div\) length of the normal vector (\(\sqrt{1 + 4} = \sqrt{5}\))

\(d = \dfrac{\left|1 \times 4 + 2 \times 3 + (-3)\right|}{\sqrt{1^{2} + 2^{2}}} = \dfrac{7}{\sqrt{5}} = \dfrac{7\sqrt{5}}{5} \approx 3.13\)

Key idea

The distance from a point to a line is the length of the perpendicular dropped from the point to the line. A line has endless points, but only the foot of the perpendicular is closest to \(\mathrm{P}\). Every other point on the line is farther away, because in a right triangle the hypotenuse is longer than either leg. That is why "distance = shortest distance = length of the perpendicular" is a clear definition. Rewriting \(\dfrac{7}{\sqrt{5}}\) as \(\dfrac{7\sqrt{5}}{5}\) is called rationalizing the denominator. With a root in the denominator, it is hard to see how big the value is or to add it to other values, so answers are usually written with a rationalized denominator. If \(a\) and \(b\) are both 0, then \(ax + by + c = 0\) is not a line, and the formula cannot be used.

Why this formula gives the distance (using the normal vector)

Figure

Standard notation (the usual math form)

\(d\) \(=\) \(\left|\vec{n}\cdot\overrightarrow{\mathrm{AP}}\right|\) \(\div\) \(\left|\vec{n}\right|\)

In words (symbols replaced with words)

③ \(d\): distance from \(\mathrm{P}\) to the line \(=\) ① absolute value of the dot product of the normal vector and \(\overrightarrow{\mathrm{AP}}\) \(\div\) ② \(|\vec{n}|\): length of the normal vector \(\vec{n}\)

The formula in words

① Let \(\mathrm{A}\) be any point on the line and \(\vec{n} = (a,\ b)\) the vector perpendicular to the line (the normal vector). Take the absolute value of the dot product of \(\vec{n}\) and \(\overrightarrow{\mathrm{AP}}\)

② divide it by the length \(|\vec{n}|\) of the normal vector \(\vec{n}\)

③ and you get the distance \(d\) from \(\mathrm{P}\) to the line

Quick example

Try it with the point \(\mathrm{A}(3,\ 0)\) on the line \(x + 2y - 3 = 0\) and the point \(\mathrm{P}(4,\ 3)\) (\(\vec{n} = (1,\ 2)\), \(\overrightarrow{\mathrm{AP}} = (1,\ 3)\))

distance \(d\) \(=\) absolute value of the dot product (\(|1 \times 1 + 2 \times 3| = 7\)) \(\div\) length of the normal vector (\(\sqrt{5}\))

\(\vec{n}\cdot\overrightarrow{\mathrm{AP}} = 1 \times 1 + 2 \times 3 = 7\)

\(d = \dfrac{\left|7\right|}{\sqrt{5}} = \dfrac{7\sqrt{5}}{5} \approx 3.13\)

Key idea

Take the \(x\) and \(y\) coefficients of the line \(ax + by + c = 0\) as they are, and you get the vector \(\vec{n} = (a,\ b)\). This vector is perpendicular to the line, and it is called the normal vector. To see why, take two points \((x_1,\ y_1)\) and \((x_2,\ y_2)\) on the line. Both satisfy the equation, so subtracting gives \(a(x_2 - x_1) + b(y_2 - y_1) = 0\). The dot product of \(\vec{n}\) and a vector along the line is 0, so they are perpendicular. Since \(\mathrm{A}\) is on the line, \(ax_1 + by_1 + c = 0\). Expanding \(\vec{n}\cdot\overrightarrow{\mathrm{AP}} = a(x_0 - x_1) + b(y_0 - y_1)\) therefore gives exactly \(ax_0 + by_0 + c\). This is why the numerator of the distance formula is "the point's coordinates plugged into the equation". The dot product \(\vec{n}\cdot\overrightarrow{\mathrm{AP}}\) equals the length of the shadow of \(\overrightarrow{\mathrm{AP}}\) cast straight onto the direction of \(\vec{n}\), times \(|\vec{n}|\). (If the shadow falls in the direction opposite to \(\vec{n}\), the sign is negative, so we take the absolute value to get a length.) Why is that shadow the distance we want? Let \(\mathrm{H}\) be the foot of the perpendicular. Then \(\overrightarrow{\mathrm{AP}} = \overrightarrow{\mathrm{AH}} + \overrightarrow{\mathrm{HP}}\). \(\overrightarrow{\mathrm{AH}}\) runs along the line, so it is perpendicular to \(\vec{n}\) and casts no shadow in that direction. The remaining part \(\overrightarrow{\mathrm{HP}}\) is parallel to \(\vec{n}\), so its shadow is its full length, which is the distance \(d\) from \(\mathrm{P}\) to the line. The answer does not depend on where you put \(\mathrm{A}\) on the line: moving \(\mathrm{A}\) only changes \(\overrightarrow{\mathrm{AH}}\), which never affects the shadow. So dividing the dot product by \(|\vec{n}|\) gives back the distance. The \(\sqrt{a^2 + b^2}\) in the denominator is this \(|\vec{n}|\).

Distance when the line is in slope-intercept form \(y = mx + k\)

Standard notation (the usual math form)

\(d\) \(=\) \(\left|m x_0 - y_0 + k\right|\) \(\div\) \(\sqrt{m^{2} + 1}\)

In words (symbols replaced with words)

③ \(d\): distance from \(\mathrm{P}\) to the line \(=\) ① absolute value of \(mx - y + k\) at the point \(\div\) ② \(\sqrt{m^2 + 1}\): length of the normal vector \((m,\ -1)\)

The formula in words

① Rewrite the line \(y = mx + k\) as \(mx - y + k = 0\). Then take the absolute value of \(mx - y + k\) with the point plugged in

② divide it by the length \(\sqrt{m^2 + 1}\) of the normal vector \((m,\ -1)\)

③ and you get the distance \(d\) from \(\mathrm{P}\) to the line

Quick example

The distance from the point \(\mathrm{P}(4,\ 3)\) to the line \(y = 2x + 1\) (in general form, \(2x - y + 1 = 0\)) is

distance \(d\) \(=\) absolute value at the point (\(|8 - 3 + 1| = 6\)) \(\div\) length of the normal vector (\(\sqrt{4 + 1} = \sqrt{5}\))

\(y = 2x + 1 \;\Longrightarrow\; 2x - y + 1 = 0\)

\(d = \dfrac{\left|2 \times 4 - 3 + 1\right|}{\sqrt{2^{2} + 1}} = \dfrac{6}{\sqrt{5}} = \dfrac{6\sqrt{5}}{5} \approx 2.68\)

Key idea

When the line is given as \(y = mx + k\), move everything to the left side to get \(mx - y + k = 0\). Comparing this with the general form \(ax + by + c = 0\) gives \(a = m\), \(b = -1\) and \(c = k\), so the denominator becomes \(\sqrt{m^2 + (-1)^2} = \sqrt{m^2 + 1}\). So there is no new formula to memorize. Just rewrite the line in general form and use the same formula. When you choose "slope-intercept form" in this calculator, it also rewrites the line in general form first and shows how in the steps. A vertical line (parallel to the \(y\)-axis, such as \(x = 3\)) has no defined slope \(m\), so it cannot be written as \(y = mx + k\). For such a line, use general form and enter \(a = 1\), \(b = 0\), \(c = -3\).

Coordinates of the foot of the perpendicular \(\mathrm{H}\)

Figure

Standard notation (the usual math form)

\(\mathrm{H}\) \(=\) \(\left(x_0,\ y_0\right)\) \(-\) \(t\,(a,\ b)\)

In words (symbols replaced with words)

③ \(\mathrm{H}\): foot of the perpendicular \(=\) ① \((x_0,\ y_0)\): coordinates of \(\mathrm{P}\) \(-\) ② normal vector \((a,\ b)\) times \(t\)

The formula in words

① Set the scale factor \(t = \dfrac{ax_0 + by_0 + c}{a^2 + b^2}\). Start from the coordinates \((x_0,\ y_0)\) of \(\mathrm{P}\)

② subtract the normal vector \((a,\ b)\) times \(t\)

③ and you get the coordinates of the foot of the perpendicular \(\mathrm{H}\)

Quick example

The foot of the perpendicular from \(\mathrm{P}(4,\ 3)\) to the line \(x + 2y - 3 = 0\) (\(t = \dfrac{7}{5}\)) is

foot of the perpendicular \(\mathrm{H}\) \(=\) coordinates of \(\mathrm{P}\) (\((4,\ 3)\)) \(-\) normal vector times \(\dfrac{7}{5}\) (\(\left(\dfrac{7}{5},\ \dfrac{14}{5}\right)\))

\(t = \dfrac{1 \times 4 + 2 \times 3 + (-3)}{1^{2} + 2^{2}} = \dfrac{7}{5}\)

\(\mathrm{H} = \left(4 - \dfrac{7}{5} \times 1,\ \ 3 - \dfrac{7}{5} \times 2\right) = \left(\dfrac{13}{5},\ \dfrac{1}{5}\right)\)

Key idea

If you move straight from \(\mathrm{P}\) parallel to the normal vector \(\vec{n} = (a,\ b)\), you always hit the line. The place where you hit it is the foot of the perpendicular \(\mathrm{H}\). As the subtraction in \(\mathrm{H} = (x_0,\ y_0) - t\,(a,\ b)\) shows, the scale factor \(t\) tells how many copies of \(\vec{n}\) you must move back to reach the line. If \(t\) is positive, you move opposite to \(\vec{n}\); if \(t\) is negative, you move in the same direction as \(\vec{n}\). The numerator of \(t\) is the same as the numerator of the distance formula (before taking the absolute value), and the denominator is \(a^2 + b^2\), which is \(|\vec{n}|^2\). The distance formula divides by \(|\vec{n}|\) only once, but here we divide twice, and there is a reason. You need to move the distance \(d\), and one copy of \(\vec{n}\) has length \(|\vec{n}|\), so the scale factor is \(d \div |\vec{n}|\). Put \(d = \dfrac{\left|ax_0 + by_0 + c\right|}{|\vec{n}|}\) into it, and you divide by \(|\vec{n}|\) twice. Once you know the foot of the perpendicular, the reflection of \(\mathrm{P}\) across the line is simply \(2\mathrm{H} - \mathrm{P}\). In geometry problems, this reflection is often the key to the solution.

Distance between two parallel lines

Figure

Standard notation (the usual math form)

\(d\) \(=\) \(\left|c_1 - c_2\right|\) \(\div\) \(\sqrt{a^{2} + b^{2}}\)

In words (symbols replaced with words)

③ \(d\): distance between the lines \(=\) ① \(|c_1 - c_2|\): absolute difference of the constants \(\div\) ② \(\sqrt{a^2 + b^2}\): length of the normal vector

The formula in words

① For two lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) with the same \(x\) and \(y\) coefficients, take the absolute difference of the constants \(|c_1 - c_2|\)

② divide it by the length of the normal vector \(\sqrt{a^2 + b^2}\)

③ and you get the distance \(d\) between the lines

Quick example

The distance between the lines \(x + 2y - 3 = 0\) and \(x + 2y + 7 = 0\) is

distance between the lines \(d\) \(=\) absolute difference of the constants (\(|-3 - 7| = 10\)) \(\div\) length of the normal vector (\(\sqrt{5}\))

\(d = \dfrac{\left|(-3) - 7\right|}{\sqrt{1^{2} + 2^{2}}} = \dfrac{10}{\sqrt{5}} = 2\sqrt{5} \approx 4.47\)

Key idea

Whether two lines are parallel depends on the ratio of their \(x\) and \(y\) coefficients. For \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), they are parallel when \(a_1b_2 - a_2b_1 = 0\). (This is just "the ratios are equal" written as a multiplication, and it works even when a coefficient is 0.) For example, \(2x + 4y - 6 = 0\) and \(x + 2y + 7 = 0\) meet this condition, so they are parallel. But the numerator of this formula only works when the coefficients match exactly, so first multiply one equation by a number to make the \(x\) and \(y\) coefficients the same. This calculator takes just one pair of \(x\) and \(y\) coefficients and two constants, so you cannot forget this step. The distance between two parallel lines is the same wherever you pick a point on one of them. So pick any point on one line and find its distance to the other line with the point-to-line formula, and this formula follows right away. A point on \(ax + by + c_1 = 0\) satisfies \(ax_0 + by_0 = -c_1\), so \(|ax_0 + by_0 + c_2| = |c_2 - c_1|\). If \(c_1\) and \(c_2\) are equal, the two equations describe the same line and the distance is 0. Two lines that are not parallel always cross at one point, so there is no gap between them and their distance is 0. That is why this formula is only for parallel lines.

To find the distance from a point \(\mathrm{P}(x_0,\ y_0)\) to a line \(ax + by + c = 0\), plug the point into the equation, take the absolute value \(|ax_0 + by_0 + c|\), and divide it by the length \(\sqrt{a^2 + b^2}\) of the normal vector \((a,\ b)\). This distance is the length of the perpendicular from the point to the line, and the foot of the perpendicular is \(\mathrm{H} = (x_0 - ta,\ y_0 - tb)\), where \(t = \dfrac{ax_0 + by_0 + c}{a^2 + b^2}\).

Symbols and terms

Symbols

\(d\)	dee	A letter often used for distance, from the first letter of "distance". On this page it is the distance from the point to the line (the length of the perpendicular).
\(a,\ b,\ c\)	a, b, c	The coefficients of the line \(ax + by + c = 0\). \(a\) is the number in front of \(x\), \(b\) the number in front of \(y\), and \(c\) the number-only term (the constant). Letters from the start of the alphabet are usually used for fixed numbers.
\(x_0,\ y_0\)	x naught, y naught	The coordinates of the point \(\mathrm{P}\) you measure from. The small 0 at the lower right is a subscript. It shows that this is the coordinate of one fixed point, not the variable \(x\). (Also read "x sub zero".)
\(\mathrm{P},\ \mathrm{H},\ \mathrm{A}\)	P, H, A	Names of points. \(\mathrm{P}\) is for "point". \(\mathrm{H}\) is a common label for the foot of the perpendicular (think of the foot of a height). \(\mathrm{A}\) is a helper point taken on the line.
\(\left\|\ \right\|\)	absolute value	The absolute value sign. A positive number stays as it is; a negative number loses its minus sign (\(\|-7\| = 7\)). A distance is never negative, so the formula takes the absolute value of the plugged-in value.
\(\sqrt{\ }\)	square root	The square root sign: the number (0 or greater) that gives the number inside when squared. \(\sqrt{5}\) is the number whose square is 5. The sign is said to come from a stretched letter r, for "radix" (root).
\(\vec{n}\)	vector n	The normal vector: a vector that points perpendicular to the line. The letter n comes from "normal" (perpendicular). For the line \(ax + by + c = 0\), \(\vec{n} = (a,\ b)\).
\(\left\|\vec{n}\right\|\)	magnitude of vector n	The magnitude of a vector (the length of the arrow). For \(\vec{n} = (a,\ b)\), \(\|\vec{n}\| = \sqrt{a^2 + b^2}\). This is the denominator of the distance formula.
\(\vec{n}\cdot\overrightarrow{\mathrm{AP}}\)	n dot AP	The dot product of two vectors: multiply matching components and add (for \((a,\ b)\) and \((p,\ q)\), it is \(ap + bq\)). It shows how much the two directions line up. The raised dot is a different operation from ordinary multiplication.
\(t\)	tee	The scale factor that tells how many copies of the normal vector you move from \(\mathrm{P}\) to reach the line. The letter \(t\) (from "time") is commonly used for a number that can vary like this.
\(m,\ k\)	m, k	The slope and the y-intercept in slope-intercept form \(y = mx + k\). \(m\) is the usual letter for slope. Textbooks usually write the y-intercept as \(b\) (\(y = mx + b\)), but this page uses \(k\) so it is not confused with the \(b\) of the general form.
\(c_1,\ c_2\)	c one, c two	The constants of the two parallel lines. The small numbers at the lower right are subscripts that tell the first line from the second.
\(\ell_1,\ \ell_2\)	ell one, ell two	Names of lines. A script letter \(\ell\), from the first letter of "line", is commonly used for lines.

Terms

distance from a point to a line	The length of the perpendicular from the point to the line. It is shorter than the distance to any other point on the line, so it is also the shortest distance between the point and the line. It is usually taught in high school geometry or precalculus.
perpendicular	A line or segment that meets another line at a right angle. If you drop a perpendicular from a point to a line, its length is the distance from the point to the line.
foot of the perpendicular	The point where the perpendicular from a point meets the original line. It is the spot on the line closest to that point.
general form	Writing the equation of a line as \(ax + by + c = 0\). Vertical lines (such as \(x = 3\)) can be written this way too, so every line fits the same form. (Some textbooks call \(ax + by = c\) the standard form; it is the same line with the constant moved to the right side.)
slope-intercept form	Writing the equation of a line as \(y = mx + k\) (in most textbooks \(y = mx + b\)). You can see the slope and where the line crosses the \(y\)-axis at a glance, but vertical lines cannot be written this way.
normal vector	A vector pointing perpendicular to a line (or a plane). The normal vector of the line \(ax + by + c = 0\) is \((a,\ b)\). The point-to-line distance formula finds the length measured in the direction of this normal vector.
dot product	A calculation that makes one number from two vectors: multiply matching components and add (\((a,\ b)\cdot(p,\ q) = ap + bq\)). It also equals \(\|\vec{u}\|\|\vec{v}\|\cos\theta\). If it is 0, the two vectors are perpendicular.
projection	The shadow of one vector cast straight onto the direction of another vector (also called the orthogonal projection). The distance from a point to a line is the length of the shadow of \(\overrightarrow{\mathrm{AP}}\) in the direction of the normal vector.
absolute value	The size of a number without its sign. It is the distance from 0 on the number line (\(\|-7\| = 7\), \(\|7\| = 7\)). A distance is never negative, so the formula uses the absolute value.
rationalizing the denominator	Removing a root from the denominator by multiplying the numerator and denominator by the same number. For example, multiply both parts of \(\dfrac{7}{\sqrt{5}}\) by \(\sqrt{5}\) to get \(\dfrac{7\sqrt{5}}{5}\). It is taught in Algebra 1 or Algebra 2, and final answers are usually written this way.
parallel	Two lines that never meet, however far you extend them. For \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), the lines are parallel or the same when the ratios of the \(x\) and \(y\) coefficients are equal (as an equation, \(a_1b_2 - a_2b_1 = 0\)). If the constants follow the same ratio too, the two lines lie on top of each other.
reflection	The point you get by flipping a point across a line, like a mirror image. Once you know the foot of the perpendicular \(\mathrm{H}\), the reflection of \(\mathrm{P}\) is \(2\mathrm{H} - \mathrm{P}\).
coordinate plane	A plane with an \(x\)-axis and a \(y\)-axis, where each point is written as a pair \((x,\ y)\). It lets you turn geometry problems into calculations.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back to review these topics is the fastest way forward.

The coordinate plane and coordinates of points (Grades 5–6)	Knowing that a point in the plane is written as a pair of numbers \((x,\ y)\) (for example, the point \((4,\ 3)\) is 4 right and 3 up from the origin) Knowing what the \(x\)-axis, the \(y\)-axis and the origin are
Equations of lines (Grade 8 and Algebra 1)	Knowing that \(y = mx + k\) is a line, where \(m\) is the slope (how much \(y\) changes when you move 1 to the right) and \(k\) is the height where it crosses the \(y\)-axis Knowing that the same line can be written as \(ax + by + c = 0\) (general form), and being able to rewrite \(y = 2x + 1\) as \(2x - y + 1 = 0\) Knowing that "the point is on the line" is the same as "the point's coordinates make the equation true"
Square roots and rationalizing the denominator (Algebra 1 and Algebra 2)	Knowing that \(\sqrt{5}\) is the number whose square is 5, and being able to take square factors out of a root, as in \(\sqrt{20} = 2\sqrt{5}\) Being able to rationalize \(\dfrac{7}{\sqrt{5}}\) into \(\dfrac{7\sqrt{5}}{5}\) by multiplying the top and bottom by \(\sqrt{5}\)
Absolute value (Grades 6–7)	Knowing that the absolute value is the size of a number without its sign, as in \(\|-7\| = 7\) Understanding that the absolute value is the distance from 0 on the number line, so it is never negative
The Pythagorean theorem and the distance formula (Grade 8 and Geometry)	Knowing that \(a^2 + b^2 = c^2\) in a right triangle Knowing that the distance between \((x_1,\ y_1)\) and \((x_2,\ y_2)\) is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Knowing that the hypotenuse is the longest side of a right triangle (this is why the perpendicular is the shortest distance)
Vectors and the dot product (Precalculus)	Knowing that a vector has a direction and a magnitude, and in the plane can be written in components as \((a,\ b)\) Knowing that the magnitude of a vector is \(\sqrt{a^2 + b^2}\) Being able to calculate the dot product \((a,\ b)\cdot(p,\ q) = ap + bq\), and knowing that the vectors are perpendicular when it is 0 (you can use the distance formula even if you have not studied this yet)

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the distance from a point to a line

x coefficient a	1
y coefficient b	2
Constant c	-3
x-coordinate of P	4
y-coordinate of P	3
Numerator \|ax0+by0+c\|	=ABS(B1B4+B2B5+B3)
Denominator √(a^2+b^2)	=SQRT(B1^2+B2^2)
Distance d	=B6/B7

Table to check the normal vector explanation

x coefficient a	1
y coefficient b	2
x-coordinate of point A on the line	3
y-coordinate of point A on the line	0
x-coordinate of P	4
y-coordinate of P	3
Dot product n·AP	=B1(B5-B3)+B2(B6-B4)
Length of the normal vector \|n\|	=SQRT(B1^2+B2^2)
Distance d	=ABS(B7)/B8

Table to find the distance for slope-intercept form

Slope m	2
y-intercept k	1
x-coordinate of P	4
y-coordinate of P	3
Distance d	=ABS(B1*B3-B4+B2)/SQRT(B1^2+1)

Table to find the foot of the perpendicular H

x coefficient a	1
y coefficient b	2
Constant c	-3
x-coordinate of P	4
y-coordinate of P	3
Scale factor t	=(B1B4+B2B5+B3)/(B1^2+B2^2)
x-coordinate of H	=B4-B1*B6
y-coordinate of H	=B5-B2*B6

Table to find the distance between parallel lines

x coefficient a	1
y coefficient b	2
1st line's constant c1	-3
2nd line's constant c2	7
Distance d	=ABS(B3-B4)/SQRT(B1^2+B2^2)

After pasting, the upper rows (coefficients and coordinates) are your inputs and the lower rows are calculated automatically.
"^" is a power, ABS gives the absolute value, and SQRT gives the square root.
The first table is the example of the point (4, 3) and the line x + 2y - 3 = 0. The numerator is 7, the denominator is about 2.2360679775, and the distance is about 3.1304951685.
The second table checks that the same distance comes from "dot product of the normal vector and AP ÷ length of the normal vector". The dot product is 7, and the distance is about 3.1304951685, the same as in the first table.
The third table is the example of the point (4, 3) and the line y = 2x + 1. The distance is about 2.683281573.
The fourth table gives 1.4, 2.6 and 0.2, so the foot of the perpendicular is (2.6, 0.2) = (13/5, 1/5).
The fifth table is the distance between x + 2y - 3 = 0 and x + 2y + 7 = 0. The answer is about 4.472135955.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the distance from a point to a line

x coefficient a	1
y coefficient b	2
Constant c	-3
x-coordinate of P	4
y-coordinate of P	3
Numerator \|ax0+by0+c\|	=ABS(B1B4+B2B5+B3)
Denominator √(a^2+b^2)	=SQRT(B1^2+B2^2)
Distance d	=B6/B7

Table to check the normal vector explanation

x coefficient a	1
y coefficient b	2
x-coordinate of point A on the line	3
y-coordinate of point A on the line	0
x-coordinate of P	4
y-coordinate of P	3
Dot product n·AP	=B1(B5-B3)+B2(B6-B4)
Length of the normal vector \|n\|	=SQRT(B1^2+B2^2)
Distance d	=ABS(B7)/B8

Table to find the distance for slope-intercept form

Slope m	2
y-intercept k	1
x-coordinate of P	4
y-coordinate of P	3
Distance d	=ABS(B1*B3-B4+B2)/SQRT(B1^2+1)

Table to find the foot of the perpendicular H

x coefficient a	1
y coefficient b	2
Constant c	-3
x-coordinate of P	4
y-coordinate of P	3
Scale factor t	=(B1B4+B2B5+B3)/(B1^2+B2^2)
x-coordinate of H	=B4-B1*B6
y-coordinate of H	=B5-B2*B6

Table to find the distance between parallel lines

x coefficient a	1
y coefficient b	2
1st line's constant c1	-3
2nd line's constant c2	7
Distance d	=ABS(B3-B4)/SQRT(B1^2+B2^2)

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the coefficients and coordinates with your own numbers.

How to calculate it in Python

from fractions import Fraction

# Line ax + by + c = 0 (a fraction such as 3/4 can be written as Fraction(3, 4))
a = Fraction(1)
b = Fraction(2)
c = Fraction(-3)

# Coordinates of point P
x0 = Fraction(4)
y0 = Fraction(3)

# The point plugged into the equation (with its sign) and the squared length of the normal vector
substituted = a * x0 + b * y0 + c
square = a * a + b * b

distance = abs(float(substituted)) / float(square) ** 0.5
print(f"Numerator |a*x0+b*y0+c| = {abs(substituted)}")
print(f"Denominator √(a^2+b^2) = √{square}")
print(f"Distance d = {distance}")

# Foot of the perpendicular H (move back t times the normal vector)
t = substituted / square
foot_x = x0 - a * t
foot_y = y0 - b * t
print(f"Scale factor t = {t}")
print(f"Foot of the perpendicular H = ({foot_x}, {foot_y})")

With the fractions module from the standard library, the numerator, the denominator and the foot of the perpendicular are calculated as exact fractions with no rounding error. This example finds the distance from the point (4, 3) to the line x + 2y - 3 = 0. Running it shows the numerator 7, the denominator √5, the distance 3.1304951684997055, the scale factor t = 7/5 and the foot of the perpendicular (13/5, 1/5). Change the coefficients and coordinates and run it again.

How to write it in LaTeX and other math languages (copy and paste)

Distance from a point to a line

d = |a·x₀ + b·y₀ + c| ÷ √(a² + b²)

d = \frac{\left|a x_0 + b y_0 + c\right|}{\sqrt{a^{2} + b^{2}}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>d</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>|</mo>
        <mi>a</mi><msub><mi>x</mi><mn>0</mn></msub>
        <mo>+</mo>
        <mi>b</mi><msub><mi>y</mi><mn>0</mn></msub>
        <mo>+</mo>
        <mi>c</mi>
        <mo>|</mo>
      </mrow>
      <msqrt>
        <mrow>
          <msup><mi>a</mi><mn>2</mn></msup>
          <mo>+</mo>
          <msup><mi>b</mi><mn>2</mn></msup>
        </mrow>
      </msqrt>
    </mfrac>
  </mrow>
</math>

d = abs(a*x_0 + b*y_0 + c)/sqrt(a^2 + b^2)

d = Abs[a x0 + b y0 + c]/Sqrt[a^2 + b^2]

d := abs(a*x0 + b*y0 + c)/sqrt(a^2 + b^2);

d = abs(a*x0 + b*y0 + c)/sqrt(a^2 + b^2);

d = |a x_0+b y_0+c|/√(a^2+b^2)

Why this formula gives the distance (using the normal vector)

d = |n·AP| ÷ |n|

d = \frac{\left|\vec{n}\cdot\overrightarrow{\mathrm{AP}}\right|}{\left|\vec{n}\right|}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>d</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>|</mo>
        <mover><mi>n</mi><mo>&#x2192;</mo></mover>
        <mo>&#x22C5;</mo>
        <mover><mrow><mi>A</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover>
        <mo>|</mo>
      </mrow>
      <mrow>
        <mo>|</mo>
        <mover><mi>n</mi><mo>&#x2192;</mo></mover>
        <mo>|</mo>
      </mrow>
    </mfrac>
  </mrow>
</math>

d = abs(vec n * vec(AP))/abs(vec n)

d = Abs[n . ap]/Norm[n]

d := abs(DotProduct(n, ap))/Norm(n, 2);

d = abs(dot(n, ap))/norm(n);

d = |n\vec ∙AP\vec |/|n\vec |

Distance when the line is in slope-intercept form \(y = mx + k\)

d = |m·x₀ − y₀ + k| ÷ √(m² + 1)

d = \frac{\left|m x_0 - y_0 + k\right|}{\sqrt{m^{2} + 1}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>d</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>|</mo>
        <mi>m</mi><msub><mi>x</mi><mn>0</mn></msub>
        <mo>&#x2212;</mo>
        <msub><mi>y</mi><mn>0</mn></msub>
        <mo>+</mo>
        <mi>k</mi>
        <mo>|</mo>
      </mrow>
      <msqrt>
        <mrow>
          <msup><mi>m</mi><mn>2</mn></msup>
          <mo>+</mo>
          <mn>1</mn>
        </mrow>
      </msqrt>
    </mfrac>
  </mrow>
</math>

d = abs(m*x_0 - y_0 + k)/sqrt(m^2 + 1)

d = Abs[m x0 - y0 + k]/Sqrt[m^2 + 1]

d := abs(m*x0 - y0 + k)/sqrt(m^2 + 1);

d = abs(m*x0 - y0 + k)/sqrt(m^2 + 1);

d = |m x_0-y_0+k|/√(m^2+1)

Coordinates of the foot of the perpendicular \(\mathrm{H}\)

H = (x₀ − t·a, y₀ − t·b)　t = (a·x₀ + b·y₀ + c) ÷ (a² + b²)

\mathrm{H} = \left(x_0 - t a,\ y_0 - t b\right),\quad t = \frac{a x_0 + b y_0 + c}{a^{2} + b^{2}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi mathvariant="normal">H</mi>
    <mo>=</mo>
    <mo>(</mo>
    <mrow><msub><mi>x</mi><mn>0</mn></msub><mo>&#x2212;</mo><mi>t</mi><mi>a</mi></mrow>
    <mo>,</mo>
    <mrow><msub><mi>y</mi><mn>0</mn></msub><mo>&#x2212;</mo><mi>t</mi><mi>b</mi></mrow>
    <mo>)</mo>
    <mo>,</mo>
    <mi>t</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mi>a</mi><msub><mi>x</mi><mn>0</mn></msub>
        <mo>+</mo>
        <mi>b</mi><msub><mi>y</mi><mn>0</mn></msub>
        <mo>+</mo>
        <mi>c</mi>
      </mrow>
      <mrow>
        <msup><mi>a</mi><mn>2</mn></msup>
        <mo>+</mo>
        <msup><mi>b</mi><mn>2</mn></msup>
      </mrow>
    </mfrac>
  </mrow>
</math>

H = (x_0 - t*a, y_0 - t*b), t = (a*x_0 + b*y_0 + c)/(a^2 + b^2)

t = (a x0 + b y0 + c)/(a^2 + b^2); h = {x0 - t a, y0 - t b}

t := (a*x0 + b*y0 + c)/(a^2 + b^2); H := [x0 - t*a, y0 - t*b];

t = (a*x0 + b*y0 + c)/(a^2 + b^2); H = [x0 - t*a, y0 - t*b];

H = (x_0-t a, y_0-t b), t = (a x_0+b y_0+c)/(a^2+b^2)

Distance between two parallel lines

d = |c₁ − c₂| ÷ √(a² + b²)

d = \frac{\left|c_1 - c_2\right|}{\sqrt{a^{2} + b^{2}}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>d</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>|</mo>
        <msub><mi>c</mi><mn>1</mn></msub>
        <mo>&#x2212;</mo>
        <msub><mi>c</mi><mn>2</mn></msub>
        <mo>|</mo>
      </mrow>
      <msqrt>
        <mrow>
          <msup><mi>a</mi><mn>2</mn></msup>
          <mo>+</mo>
          <msup><mi>b</mi><mn>2</mn></msup>
        </mrow>
      </msqrt>
    </mfrac>
  </mrow>
</math>

d = abs(c_1 - c_2)/sqrt(a^2 + b^2)

d = Abs[c1 - c2]/Sqrt[a^2 + b^2]

d := abs(c1 - c2)/sqrt(a^2 + b^2);

d = abs(c1 - c2)/sqrt(a^2 + b^2);

d = |c_1-c_2|/√(a^2+b^2)

How to have ChatGPT do the calculation

You are a math assistant for coordinate geometry. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

For the point (4, 3) and the line x + 2y - 3 = 0, find:
1. The distance from the point to the line (as an exact value with a rationalized denominator, and as a decimal)
2. The coordinates of the foot of the perpendicular from the point to the line (as reduced fractions)
3. The equation of that perpendicular (the line through the point, perpendicular to the given line)

In Python, use the fractions module from the standard library and sympy to calculate exactly. Show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula