Lines in the plane

A geometric point has no number until you choose axes. The question on this page is how to name a point so that “how far” and “which line” become equations you can check — and which extra freeze then fails for a vertical line.

The plane here is Cartesian: the $y$-axis is perpendicular to the $x$-axis. René Descartes (1596–1650) put arithmetic and algebra to work on geometric objects this way; the axes still carry his name. This math handbook measures the inclination of a line in radians unless a degree sign is shown.

Name a point with a pair

A point is an ordered pair $(x_1,y_1)$. Extra assumption for every formula below: those coordinates are read on perpendicular axes. Left of the $y$-axis, $x$ is negative; below the $x$-axis, $y$ is negative. Drag $P$. The dashed marks drop to the axes. The four fixed points are example 1: $(2,2)$, $(2,-2)$, $(-2,-2)$, $(-2,2)$.

Drop the right angle between the axes and the next formula — a hypotenuse built from $\Delta x$ and $\Delta y$ — is the wrong distance.

The hypotenuse is the distance

Once two points have pairs, the legs $|x_2-x_1|$ and $|y_2-y_1|$ make a right triangle because the axes are perpendicular. The distance $d$ is that hypotenuse — the Pythagorean theorem:

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

From $(1,1)$ to $(2,2)$ that is $\sqrt{(2-1{)}^2+(2-1{)}^2}=\sqrt{2}$. Drag $A$ or $B$. The dashed legs are the coordinate differences; the solid segment is $d$. A circle is this same $d$, frozen from one center.

Non-vertical: slope and intercept

Among all named points, which ones lie on a line? Extra freeze for this section: the line is not vertical ($x_1\ne x_2$). Then it is

$y=mx+b,$

where $b$ is the intercept on the $y$-axis and $m$ is the slope. If $\alpha$ is the inclination from the positive $x$-axis,

$\tan \alpha =m.$

(i) One point $(x_1,y_1)$ and a slope $m$ determine $b=y_1-m{x}_1$.

(ii) Two points $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1\ne x_2$ give

$m=\frac{y_2-y_1}{x_2-x_1},b=y_1-m{x}_1.$

The line through $(1,1)$ and $(3,2)$ has $m=\frac{2-1}{3-1}=\frac{1}{2}$ and $b=1-\frac{1}{2}=\frac{1}{2}$, so $y=\frac{1}{2}x+\frac{1}{2}$. Drag $B$. The small triangle has run $1$ and rise $m$; $\alpha$ is the marked angle. Make $A$ and $B$ share an $x$-coordinate and this $m$ is undefined — that is the next freeze, after one more writing of the same non-vertical line.

The same line, two intercepts

Divide $y=mx+b$ by $b$ and set $\frac{1}{a}:=-\frac{m}{b}$. The same line becomes

$\frac{x}{a}+\frac{y}{b}=1.$

When $y=0$, the line meets the $x$-axis at $(a,0)$; when $x=0$, it meets the $y$-axis at $(0,b)$. This is not a new line. It is $y=mx+b$ rewritten so the two intercepts are the named numbers. Extra assumption: $b\ne 0$ (you divided by it).

Starting from $y=-8x+4$ and dividing by $4$ gives $\frac{y}{4}=-2x+1$, or $2x+\frac{y}{4}=1$. Then $y=0$ forces $x=\frac{1}{2}$: the $x$-intercept is $\frac{1}{2}$. Drag $A$ or $B$. Example 4 starts at those intercepts $(\frac{1}{2},0)$ and $(0,4)$.

Infinite slope, then every line

The $y$-axis itself is

$x=0.$

That is not a special case of $y=mx+b$; it is the line with infinite slope. Drop “non-vertical” and slope-intercept has nothing to say. Every line in the plane is the set of points satisfying

$Ax+By+C=0$

for constants $A,B,C$ with $A^2+B^2\ne 0$. Taking $A=1$ and $B=C=0$ recovers $x=0$. The general equation is the freeze that does not need a finite $m$.

The object is a named point $(x,y)$ on perpendicular axes. Distance is the hypotenuse of $\Delta x$ and $\Delta y$. A non-vertical line is $y=mx+b$ (or the intercept form of the same line). The $y$-axis forces $Ax+By+C=0$.

The next freeze — all points at one distance from one center — is circles in the plane.