Grade 10

Coordinate methods in the plane

Describing points, lines and circles in the plane by numerical coordinates and equations.

IntuitionTurning Geometry into Algebra

In a city laid out on a numbered street grid, you can locate any building with just two numbers: how many blocks east and how many blocks north of the town square. Coordinate geometry applies the same idea to the Euclidean plane: by fixing two perpendicular number lines meeting at an origin, every point becomes an ordered pair of numbers, every line becomes a first-degree equation, and every circle becomes a second-degree equation. Geometric questions about intersections, distances, and tangency turn into algebraic calculations.

With

a=0

and

b=0

, the polynomial curve reduces to the line

y=cx+d

(or

cx-y+d=0

); adjust the slope

c

and vertical intercept

d

to see how the equation controls the line in the coordinate plane.

SchoolCoordinates, Lines, and Circles

Definition: Cartesian Coordinates and the Distance Formula

In the Cartesian plane $Oxy$ with unit basis vectors $\vec{i}=(1,0)$ and $\vec{j}=(0,1)$ , each point $M$ has unique coordinates $M(x,y)$ defined by $\overrightarrow{OM}=x\vec{i}+y\vec{j}$ . Given two points $A(x_1,y_1)$ and $B(x_2,y_2)$ , the displacement vector is $\overrightarrow{AB}=(x_2-x_1,y_2-y_1)$ , the midpoint of segment $AB$ is $I\!\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ , and the distance between $A$ and $B$ is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

\Delta:\; ax + by + c = 0 \quad (a^2+b^2 > 0), \qquad d(M_0, \Delta) = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}

Every straight line $\Delta$ in the plane can be written in the general equation $ax+by+c=0$ with $a^2+b^2>0$ . The nonzero vector $\vec{n}=(a,b)$ is a normal vector perpendicular to $\Delta$ , while $\vec{u}=(-b,a)$ is a direction vector parallel to $\Delta$ because $\vec{n}\cdot\vec{u}=a(-b)+ba=0$ . The perpendicular distance from any point $M_0(x_0,y_0)$ to the line $\Delta$ is given by $d(M_0,\Delta)=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ .

(x - a)^2 + (y - b)^2 = R^2 \iff x^2 + y^2 - 2ax - 2by + c = 0 \quad (R = \sqrt{a^2+b^2-c} > 0)

A circle with center $I(a,b)$ and radius $R>0$ consists of all points $M(x,y)$ at distance $R$ from $I$ ; squaring $IM=R$ yields the standard circle equation $(x-a)^2+(y-b)^2=R^2$ . Expanding the squares produces the general form $x^2+y^2-2ax-2by+c=0$ , which represents a real circle whenever $a^2+b^2-c>0$ , with center $I(a,b)$ and radius $R=\sqrt{a^2+b^2-c}$ .

Summary of plane coordinate formulas
Geometric object	Coordinate equation / formula	Key parameters
Segment length	$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$	Endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$
General line	$ax+by+c=0$	Normal vector $\vec{n}=(a,b)$ , direction vector $\vec{u}=(-b,a)$
Point-to-line distance	$d(M_0,\Delta)=\dfrac{\|ax_0+by_0+c\|}{\sqrt{a^2+b^2}}$	Point $M_0(x_0,y_0)$ and line $\Delta: ax+by+c=0$
Circle	$(x-a)^2+(y-b)^2=R^2$	Center $I(a,b)$ and radius $R>0$

UndergraduateKey Theorems

Theorem: Euclidean Distance Formula in the Plane

For any two points $A(x_1,y_1)$ and $B(x_2,y_2)$ in the Cartesian plane, the distance between them is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

Why is it true?

Dropping lines parallel to the axes from $A$ and $B$ forms a right triangle whose legs have lengths $|x_2-x_1|$ and $|y_2-y_1|$ , so the hypotenuse $AB$ follows directly from the Pythagorean theorem.

Proof

Introduce the auxiliary point $C(x_2,y_1)$ , which shares the second coordinate $y_1$ with $A(x_1,y_1)$ and the first coordinate $x_2$ with $B(x_2,y_2)$ .

The segment $AC$ is parallel to the first axis with length $AC=|x_2-x_1|$ , while the segment $CB$ is parallel to the second axis with length $CB=|y_2-y_1|$ . Since the two coordinate axes are perpendicular, $\angle ACB=90^\circ$ .

Applying the Pythagorean theorem to the right triangle $\triangle ACB$ gives $AB^2=AC^2+CB^2=|x_2-x_1|^2+|y_2-y_1|^2=(x_2-x_1)^2+(y_2-y_1)^2$ . Taking the nonnegative square root yields $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (which also holds when $x_1=x_2$ or $y_1=y_2$ ).

Theorem: Point-to-Line Distance Formula

The perpendicular distance from a point $M_0(x_0,y_0)$ to a line $\Delta: ax+by+c=0$ (with $a^2+b^2>0$ ) is $d(M_0,\Delta)=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ .

Why is it true?

The shortest segment from $M_0$ to $\Delta$ runs along the normal direction $\vec{n}=(a,b)$ , so projecting the vector from any point on $\Delta$ to $M_0$ onto the unit normal $\dfrac{\vec{n}}{|\vec{n}|}$ measures the exact perpendicular distance.

Proof

Let $H(x_H,y_H)$ be the orthogonal projection of $M_0(x_0,y_0)$ onto $\Delta$ . Because $\overrightarrow{HM_0}$ is parallel to the normal vector $\vec{n}=(a,b)$ , there exists a scalar $t\in\mathbb{R}$ such that $\overrightarrow{HM_0}=t\vec{n}=(ta,tb)$ , which gives $x_H=x_0-ta$ and $y_H=y_0-tb$ .

Since $H$ lies on $\Delta$ , its coordinates satisfy $ax_H+by_H+c=0$ . Substituting $x_H$ and $y_H$ yields $a(x_0-ta)+b(y_0-tb)+c=0$ , or $ax_0+by_0+c-t(a^2+b^2)=0$ .

Solving for $t$ gives $t=\dfrac{ax_0+by_0+c}{a^2+b^2}$ . Therefore the distance is $d(M_0,\Delta)=|\overrightarrow{HM_0}|=|t|\,|\vec{n}|=\dfrac{|ax_0+by_0+c|}{a^2+b^2}\sqrt{a^2+b^2}=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ .

UndergraduateReal-World Applications and Worked Examples

Coordinate geometry powers geographic information systems (GIS), autonomous navigation, and collision detection in robotics. By representing roads as line equations and broadcast ranges or safety zones as circle equations, software can instantly compute whether a vehicle is drifting off course or entering a restricted radius.

Example: Shortest Access Road from a Station to a Highway

On a coordinate map (in kilometers), a straight highway follows the line $\Delta: 3x+4y-12=0$ and a rescue station is located at $M_0(4,5)$ . Find the length of the shortest straight access road connecting $M_0$ to the highway $\Delta$ .

Solution

The shortest road from $M_0(4,5)$ to the line $\Delta: 3x+4y-12=0$ is the perpendicular segment, whose length is given by the point-to-line distance formula with $a=3$ , $b=4$ , and $c=-12$ .

Substituting $(x_0,y_0)=(4,5)$ gives $d(M_0,\Delta)=\dfrac{|3(4)+4(5)-12|}{\sqrt{3^2+4^2}}=\dfrac{|12+20-12|}{\sqrt{25}}=\dfrac{20}{5}=4$ km.

Example: Radar Coverage Circle and Tangent Flight Path

A radar tower at $I(1,-2)$ tracks aircraft within a radius of $R=5$ km. Write the equation of the coverage boundary circle and verify whether an aircraft flying along the line $\Delta: 3x-4y-20=0$ skims tangentially along the boundary.

Solution

With center $I(1,-2)$ and radius $R=5$ , the coverage boundary circle has equation $(x-1)^2+(y+2)^2=25$ .

A line $\Delta$ is tangent to the circle if and only if the distance from the center $I(1,-2)$ to $\Delta$ equals $R$ . Computing this distance gives $d(I,\Delta)=\dfrac{|3(1)-4(-2)-20|}{\sqrt{3^2+(-4)^2}}=\dfrac{|3+8-20|}{5}=\dfrac{9}{5}=1.8$ km. Since $d(I,\Delta)=1.8<5=R$ , the flight path cuts across the interior of the radar zone rather than skimming tangentially.

What is the distance $AB$ between the points $A(-1,2)$ and $B(2,6)$ ?

Which of the following is a normal vector $\vec{n}$ of the line $\Delta: 2x-5y+7=0$ ?

What is the perpendicular distance from the point $M_0(1,2)$ to the line $\Delta: 3x+4y-6=0$ ?

A Wi-Fi router at $I(3,-4)$ covers a circular area of radius $R=6$ m. Which equation describes the boundary circle of its signal range?

References

René Descartes (trans. David Eugene Smith, Marcia L. Latham) (1954). The Geometry of René Descartes
H. S. M. Coxeter (1969). Introduction to Geometry