Cartesian coordinates and distance in the plane

Intuition [Beginner]

A point in the plane needs two numbers to pin it down: how far left or right from a fixed reference point, and how far up or down. The horizontal number is $x$ and the vertical number is $y$. Together they form a coordinate pair $(x,y)$. The reference point where both numbers are zero is the origin $(0,0)$.

The horizontal line through the origin is the $x$-axis and the vertical line is the $y$-axis. These two axes divide the plane into four regions called quadrants, numbered counterclockwise starting from the upper right. In the first quadrant both $x$ and $y$ are positive; the signs then alternate as you move around.

The distance between two points in the plane reduces to a right triangle. The horizontal gap between $(x_1,y_1)$ and $(x_2,y_2)$ is $|x_2-x_1|$ and the vertical gap is $|y_2-y_1|$. These are the two legs of a right triangle whose hypotenuse is the straight-line distance. That is why the Cartesian coordinate system exists: it turns every geometric question about position and distance into arithmetic with pairs of numbers.

Visual [Beginner]

A flat grid with two perpendicular axes crossing at the origin. Two points $P=(1,2)$ and $Q=(4,6)$ are marked with dots. Dashed horizontal and vertical lines from $P$ and $Q$ meet at the corner $(4,2)$, forming a right triangle with the direct segment $PQ$ as the hypotenuse.

The horizontal leg has length $|4-1|=3$ and the vertical leg has length $|6-2|=4$. The Pythagorean theorem gives the hypotenuse: $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

Worked example [Beginner]

Find the distance between $A=(-1,3)$ and $B=(2,7)$.

Step 1. Compute the horizontal gap: $|2-(-1)|=|3|=3$.

Step 2. Compute the vertical gap: $|7-3|=|4|=4$.

Step 3. Apply the Pythagorean theorem: $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

What this tells us: any two points determine a right triangle whose legs are parallel to the axes, and the distance is always the square root of the sum of the squared gaps.

Check your understanding [Beginner]

Formal definition [Intermediate+]

The Cartesian plane ${\mathbb{R}}^2$ is the set of all ordered pairs $(x,y)$ with $x,y\in \mathbb{R}$. The distance between points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ is

$$d(P_1,P_2)=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$

The midpoint of the segment $P_1{P}_2$ is

$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$$

A line in ${\mathbb{R}}^2$ is the solution set of a linear equation $Ax+By+C=0$ with $(A,B)\ne (0,0)$. When $B\ne 0$ the line can be written in slope-intercept form $y=mx+b$, where $m=-A/B$ is the slope and $b=-C/B$ is the $y$-intercept ^{[Lang — Basic Mathematics Ch. 9]}.

Three equivalent forms for the equation of a line:

• Point-slope. Through $(x_0,y_0)$ with slope $m$: $y-y_0=m(x-x_0)$.
• Two-point. Through $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\ne x_2$: $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$.
• Slope-intercept. With slope $m$ and $y$-intercept $b$: $y=mx+b$.

Parallel and perpendicular conditions. Two non-vertical lines with slopes $m_1$ and $m_2$ are parallel iff $m_1=m_2$, and perpendicular iff $m_1{m}_2=-1$.

Counterexamples to common slips

• The distance formula uses squares inside the square root, not absolute values added together. Using $|x_2-x_1|+|y_2-y_1|$ gives the taxicab distance, which exceeds the straight-line distance in general: between $(0,0)$ and $(3,4)$ the Euclidean distance is $5$ but the taxicab distance is $7$.
• The perpendicular condition $m_1{m}_2=-1$ requires both slopes to be finite. A vertical line has undefined slope and is perpendicular to a horizontal line (slope $0$), but the product formula does not apply directly to that case.
• The midpoint formula averages both coordinates independently. Between $(0,0)$ and $(4,6)$ the midpoint is $(2,3)$, not $(2,0)$.

Key theorem with proof [Intermediate+]

Theorem (distance formula). For any two points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ in ${\mathbb{R}}^2$, the straight-line distance between them is

$$d(P_1,P_2)=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$

Proof. The horizontal displacement from $P_1$ to $P_2$ is $x_2-x_1$ and the vertical displacement is $y_2-y_1$. Construct the right triangle with vertices $P_1$, $P_2$, and the corner point $Q=(x_2,y_1)$. The leg $P_{1}Q$ has length $|x_2-x_1|$ and the leg $Q{P}_2$ has length $|y_2-y_1|$. By the Pythagorean theorem ^{[Lang — Basic Mathematics Ch. 6]}, the hypotenuse satisfies

$$d(P_1,P_2{)}^2=|x_2-x_1{|}^2+|y_2-y_1{|}^2.$$

Since $|a{|}^2=a^2$ for every real $a$, the absolute-value bars may be dropped:

$$d(P_1,P_2{)}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2.$$

Taking the positive square root (distance is non-negative) yields the distance formula. $\square$

Bridge. The distance formula builds toward the line-equation framework of 00.03.01, where slope-intercept and general forms describe the geometry of straight lines, and the same Pythagorean structure appears again in 00.06.01 as the hypotenuse-length computation underlying right-triangle trigonometry. The foundational reason the formula works is that the absolute value on $\mathbb{R}$ from 00.01.02 extends to the Euclidean norm on ${\mathbb{R}}^2$ via Pythagoras, and this is exactly the construction that makes the plane a metric space. Putting these together, the bridge is between one-dimensional distance and two-dimensional distance, and the pattern generalises to ${\mathbb{R}}^n$ for any dimension via the same sum-of-squares structure.

Exercises [Intermediate+]

Advanced results [Master]

The Cartesian plane ${\mathbb{R}}^2$ carries a norm and metric inherited from the Euclidean inner product, and the coordinate-geometry results at intermediate tier specialise a general framework for normed vector spaces.

Theorem 1 (Euclidean norm). The function $\Vert (x,y)\Vert :=\sqrt{x^2+y^2}$ on ${\mathbb{R}}^2$ satisfies positivity ($\Vert v\Vert \ge 0$ with equality iff $v=0$), homogeneity ($\Vert \lambda v\Vert =|\lambda |\cdot \Vert v\Vert$), and the triangle inequality ($\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$). Hence $\Vert \cdot \Vert$ is a norm, and $d(v,w)=\Vert v-w\Vert$ is a metric.

Theorem 2 (general form of a line). Every line in ${\mathbb{R}}^2$ is the solution set of an equation $Ax+By+C=0$ with $(A,B)\ne (0,0)$, and every such equation defines a line. Two equations define the same line iff their coefficient triples are proportional.

Theorem 3 (distance from a point to a line). The distance from $(x_0,y_0)$ to the line $Ax+By+C=0$ is

$$d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}.$$

The formula is invariant under scaling the equation by a nonzero constant.

Theorem 4 (Cauchy-Schwarz in ${\mathbb{R}}^2$). For all $(a,b)$ and $(c,d)$ in ${\mathbb{R}}^2$,

$$|ac+bd|\le \sqrt{a^2+b^2}\cdot \sqrt{c^2+d^2}.$$

Equality holds iff $(a,b)$ and $(c,d)$ are linearly dependent.

Theorem 5 (collinearity criterion). Three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ are collinear iff the area of the triangle they form is zero, which is equivalent to the determinant condition

$$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0.$$

Theorem 6 (change of coordinates). Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be an invertible linear transformation with matrix $M$. The distance between points $P$ and $Q$ in the new coordinates $P^{\prime }=MP$, $Q^{\prime }=MQ$ is $d(P^{\prime },Q^{\prime })=\Vert M(Q-P)\Vert$. When $M$ is orthogonal ($M^{\top }M=I$), distances are preserved: $d(P^{\prime },Q^{\prime })=d(P,Q)$.

Synthesis. The Euclidean distance on ${\mathbb{R}}^2$ is the foundational reason the plane admits a metric-space structure, and this is exactly the content that identifies the geometric plane with the normed vector space ${\mathbb{R}}^2$. The bridge is between the Pythagorean theorem (a geometric fact about right triangles) and the algebraic distance formula (a computation with coordinate pairs). The Cauchy-Schwarz inequality generalises to arbitrary inner-product spaces and the pattern recurs wherever angle and length coexist. The collinearity determinant identifies geometric alignment with an algebraic vanishing condition, and the change-of-coordinates theorem identifies distance-preserving transformations with orthogonal matrices. Putting these together, the Cartesian framework identifies every planar-geometry question with an algebraic question about pairs of real numbers, and the distance formula is the load-bearing equivalence that makes the identification coherent.

Full proof set [Master]

Proposition 1 (Cauchy-Schwarz in ${\mathbb{R}}^2$). For all $(a,b),(c,d)\in {\mathbb{R}}^2$,

$$(ac+bd{)}^2\le (a^2+b^2)(c^2+d^2).$$

Proof. Expand the non-negative quantity $(ad-bc{)}^2\ge 0$:

$$a^2{d}^2-2abcd+b^2{c}^2\ge 0.$$

Add $a^2{c}^2+b^2{d}^2$ to both sides:

$$a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2\ge a^2{c}^2+2abcd+b^2{d}^2.$$

The left side factors as $(a^2+b^2)(c^2+d^2)$ and the right side as $(ac+bd{)}^2$. Taking square roots gives the result. $\square$

Proposition 2 (distance from a point to a line). The distance from $(x_0,y_0)$ to $Ax+By+C=0$ is $|A{x}_0+B{y}_0+C|/\sqrt{A^2+B^2}$.

Proof. The normal vector to the line is $n=(A,B)$. The perpendicular through $(x_0,y_0)$ is parameterised as $(x,y)=(x_0+At,y_0+Bt)$. Substituting into the line equation gives

$$A(x_0+At)+B(y_0+Bt)+C=0⟹t=-\frac{A{x}_0+B{y}_0+C}{A^2+B^2}.$$

The foot of the perpendicular is at this value of $t$, and the distance from $(x_0,y_0)$ to the foot is $|t|\cdot \Vert (A,B)\Vert$, which gives

$$d=\frac{|A{x}_0+B{y}_0+C|}{A^2+B^2}\cdot \sqrt{A^2+B^2}=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}.$$

$\square$

Proposition 3 (orthogonal transformations preserve distance). Let $M$ be a $2\times 2$ real matrix with $M^{\top }M=I$. Then for all $v\in {\mathbb{R}}^2$, $\Vert Mv\Vert =\Vert v\Vert$.

Proof. $\Vert Mv{\Vert }^2=(Mv{)}^{\top }(Mv)=v^{\top }{M}^{\top }Mv=v^{\top }Iv=v^{\top }v=\Vert v{\Vert }^2$. Taking square roots gives $\Vert Mv\Vert =\Vert v\Vert$. $\square$

Connections [Master]

• Absolute value and the triangle inequality 00.01.02. The distance formula in this unit extends the one-dimensional distance $|x_2-x_1|$ from the number line to the plane by adding the second coordinate. The absolute value on $\mathbb{R}$ is the one-dimensional ancestor of the Euclidean norm on ${\mathbb{R}}^2$, and the triangle inequality proved for $|x+y|$ in that unit becomes the triangle inequality for $\Vert v+w\Vert$ in this one.

• Linear equations and the line 00.03.01. The line equations developed here (point-slope, slope-intercept, two-point, general form) are the coordinate-geometry realisation of the algebraic line equation $ax+by=c$ studied in the linear-equations unit. The slope and intercept definitions here give geometric content to the algebraic solution sets, and the parallel-perpendicular conditions translate the algebraic classification of line systems into slope relations.

• Right-triangle trigonometry 00.06.01. The Pythagorean theorem, which provides the distance formula in this unit, is the same result that underlies the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ in right-triangle trigonometry. The right triangle connecting two points in the plane is the geometric bridge between coordinate distance and trigonometric ratios, and the distance formula reappears when computing side lengths in the trigonometric context.

Historical & philosophical context [Master]

Rene Descartes 1637 La Geometrie ^{[Descartes 1637]}, published as an appendix to Discours de la methode, introduced the correspondence between algebraic equations and geometric curves by assigning numerical coordinates to points in the plane. Pierre de Fermat 1636, in the manuscript Ad Locos Planos et Solidos Isagoge (published posthumously in 1679) ^{[Fermat 1636]}, independently developed the same idea, arriving at the coordinate description of conics through the locus approach. The two inventions differ in emphasis: Descartes starts from algebraic equations and reads off geometric curves; Fermat starts from geometric conditions on loci and writes the corresponding equations.

The modern formulation of ${\mathbb{R}}^2$ as a normed vector space with the Euclidean distance crystallised with the development of analytic geometry in the eighteenth and nineteenth centuries. Euler 1748 Introductio in analysin infinitorum gave the first systematic treatment of curves via coordinate equations. The metric-space abstraction, in which the distance formula becomes the canonical example of a metric, is due to Frechet 1906 Sur quelques points du calcul fonctionnel (Rend. Circ. Mat. Palermo 22).

Bibliography [Master]

@book{Descartes1637,
  author = {Descartes, Ren{\'e}},
  title = {La G{\'e}om{\'e}trie},
  note = {Appendix to {\em Discours de la m{\'e}thode}, Leiden. Modern edition: Smith, D. E. and Latham, M. L. (transl.), Dover 1954},
  year = {1637}
}

@article{Fermat1636,
  author = {Fermat, Pierre de},
  title = {Ad Locos Planos et Solidos Isagoge},
  journal = {Varia Opera Mathematica},
  note = {Written 1636; published posthumously by Toulouse 1679},
  year = {1636}
}

@book{Euler1748,
  author = {Euler, Leonhard},
  title = {Introductio in analysin infinitorum},
  publisher = {Bousquet},
  address = {Lausanne},
  year = {1748}
}

@article{Frechet1906,
  author = {Fr{\'e}chet, Maurice},
  title = {Sur quelques points du calcul fonctionnel},
  journal = {Rendiconti del Circolo Matematico di Palermo},
  volume = {22},
  pages = {1--74},
  year = {1906}
}