How do you compute the distance between two points?

With |AB| = √((x_B − x_A)² + (y_B − y_A)²). It is the Pythagorean theorem applied to the triangle whose legs are parallel to the axes and equal to the coordinate differences. The order of subtraction makes no difference, since both differences are squared.

How do you find the equation of a line through two points?

First the slope a = (y_B − y_A) / (x_B − x_A), then the intercept b by substituting one of the points into y = ax + b. For A(1, 2) and B(3, 8) the slope is 6/2 = 3, and 2 = 3 · 1 + b gives b = −1, so the line is y = 3x − 1.

What is the condition for two lines to be perpendicular?

The product of their slopes must be −1: a₁ · a₂ = −1. In practice the slope of a perpendicular line is the reciprocal with the sign flipped — for 2 it is −1/2, for −3/4 it is 4/3. The exception is a vertical and a horizontal line: they are perpendicular too, although one of them has no slope.

How do you check whether a point lies on a line?

Substitute its coordinates into the equation and see whether the statement is true. The point (2, 5) lies on y = 3x − 1, because 3 · 2 − 1 = 5. The same method works for a circle: a point lies on it when it satisfies its equation.

How do you read the centre and radius off the equation of a circle?

In the standard form (x − a)² + (y − b)² = r² the centre is S(a, b) and the radius is the square root of the right-hand side. Mind the signs: (x + 2)² + (y − 5)² = 9 has centre (−2, 5), because x + 2 is x − (−2). When the equation is in the general form x² + y² − 2ax − 2by + c = 0, complete both squares first.

Advanced level

Analytic geometry

A point is a pair of numbers, a line is an equation, a circle is another equation — and geometric questions become arithmetic. The distance between points from the Pythagorean theorem, the slope from two points, the conditions for parallel and perpendicular lines, and the centre and radius read off the equation of a circle.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

Navigating a kilometre grid
Two points on a map have coordinates (12, 5) and (20, 11) in kilometres. The straight-line distance is √(8² + 6²) = 10 km — the number a pilot or a rescue team compares with their range before working out a route by road.
Transmitter range as a circle
A transmitter at (4, −2) has a range of 6 km, so the edge of coverage is the circle (x − 4)² + (y + 2)² = 36. To check whether a receiver at (8, 1) is covered, substitute: 16 + 9 = 25 < 36, so it is — one calculation instead of drawing a map.
CAD and a perpendicular wall
An edge of a part lies on the line y = 0.5x + 3. A perpendicular wall must have slope −2, because 0.5 · (−2) = −1. A CAD program computes exactly that product when it checks whether a drawn segment really is perpendicular rather than merely looking so.
Setting out a road through two points
A stretch of road is to join junctions at (0, 40) and (60, 25), in metres. The slope is (25 − 40) / 60 = −0.25, so the centreline is y = −0.25x + 40, and for any chainage the offset can be read straight off it.
Surveying a parallelism check
Two boundaries of a plot have slopes 0.750 and 0.748. They are not equal, so the boundaries are not parallel — over 200 m they drift apart by about 0.4 m. Comparing two numbers catches what the eye never sees on a drawing.

All formulas

Distance between two points
$|AB| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$
the Pythagorean theorem on coordinate differences
Slope
$a = \frac{y_B - y_A}{x_B - x_A}$
the rise divided by the run
Slope-intercept form
$y = ax + b$
b is where the line crosses the y axis
General form of a line
$Ax + By + C = 0$
covers vertical lines too, with B = 0
Condition for parallel lines
$a_1 = a_2 \iff k \parallel l$
equal slopes
Condition for perpendicular lines
$a_1 \cdot a_2 = -1 \iff k \perp l$
the slope flipped over and negated
Equation of a circle
$(x - a)^2 + (y - b)^2 = r^2$
centre S(a, b), radius r

Analytic geometry turns a drawing into arithmetic: a point becomes a pair of numbers, a line and a circle become equations. Questions like "are these lines perpendicular?" or "is this point within range?" are then settled by calculation rather than by a ruler.

The lesson stands on three earlier ones: the coordinate plane, the Pythagorean theorem and the special products.

The distance between two points

A segment parallel to an axis was measured by subtracting. A slanted segment is the hypotenuse of a triangle whose legs are exactly such segments:

|AB| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}

The order of subtraction is irrelevant, since both differences end up squared.

Find the distance between A(−3, 2) and B(1, 5).

The line: slope-intercept form

A line that is not vertical is described by

y = ax + b

where $a$ is the slope (how much $y$ grows when $x$ grows by one) and $b$ is where the line crosses the y axis. Given two points, the slope is the ratio of the increments:

a = \frac{y_B - y_A}{x_B - x_A}

The line k: y = 1.5x + 2 through A and B, and the perpendicular line l: y = −2/3 x + 2. The product of their slopes is 1.5 · (−2/3) = −1.

Find the equation of the line through A(−2, −1) and B(2, 5).

The general form and vertical lines

The slope-intercept form does not cover vertical lines: for them the increment in $x$ is zero, so the slope does not exist. A vertical line has the equation $x = c$ , and that is why a second, more general form exists:

Ax + By + C = 0

For $B \neq 0$ it can be rearranged into the slope-intercept form; for $B = 0$ it describes exactly a vertical line. The general form is determined only up to multiplication by a number — $2x - y + 3 = 0$ and $4x - 2y + 6 = 0$ describe the same line.

A point lies on a line exactly when its coordinates satisfy the equation. That single substitution replaces inspecting a drawing.

Parallel and perpendicular

Two non-vertical lines are parallel when their slopes are equal, and perpendicular when their slopes multiply to $-1$ :

a_1 = a_2 \iff k \parallel l, \qquad a_1 \cdot a_2 = -1 \iff k \perp l

In practice: the slope of a perpendicular line is the reciprocal with the sign flipped. For $2$ it is $-\tfrac{1}{2}$ , for $-\tfrac{3}{4}$ it is $\tfrac{4}{3}$ .

Find the line perpendicular to y = 2x − 5 and passing through P(4, 1).

The equation of a circle

A circle is the set of points at distance $r$ from the centre. Putting that sentence into the distance formula and squaring both sides gives the equation of the circle with centre $S(a,\ b)$ :

(x - a)^2 + (y - b)^2 = r^2

A circle with centre S(3, 1) and radius 5. Every point of that line is exactly 5 away from S.

Mind the signs: in $(x + 2)^2 + (y - 5)^2 = 9$ the centre is $(-2,\ 5)$ , because $x + 2$ is $x - (-2)$ . The radius is the square root of the right-hand side, so for $r^2 = 20$ it is $\sqrt{20} = 2\sqrt{5}$ , not $20$ .

The equation is also often given in general form. Then we return to the standard form by completing both squares with a special product.

Find the centre and radius of the circle x² + y² − 6x + 4y − 12 = 0.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

distance between the points: A(−1, 4), B(−7, 12)

Common mistakes

Reading the centre with the signs as written — $(x + 2)^2$ means an abscissa of $-2$ , not $2$ .
Taking the right-hand side of a circle's equation as the radius — it is $r^2$ , so a square root is still owed.
Flipping the slope without flipping the sign — the line perpendicular to $y = 2x$ has slope $-\tfrac{1}{2}$ , not $\tfrac{1}{2}$ .
Looking for the slope of a vertical line — it has none; its equation is $x = c$ .
Mixing coordinates across the distance formula — the first square holds abscissas only, the second ordinates only.
Treating two general forms as different lines — multiplying the equation by a non-zero number changes nothing.

Formula card

Topic: Analytic geometry

Distance between two points
$|AB| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$
the Pythagorean theorem on coordinate differences
Slope
$a = \frac{y_B - y_A}{x_B - x_A}$
the rise divided by the run
Slope-intercept form
$y = ax + b$
b is where the line crosses the y axis
General form of a line
$Ax + By + C = 0$
covers vertical lines too, with B = 0
Condition for parallel lines
$a_1 = a_2 \iff k \parallel l$
equal slopes
Condition for perpendicular lines
$a_1 \cdot a_2 = -1 \iff k \perp l$
the slope flipped over and negated
Equation of a circle
$(x - a)^2 + (y - b)^2 = r^2$
centre S(a, b), radius r

The line k through A(−2, −1) and B(2, 5) has slope 6/4 = 1.5, so its equation is y = 1.5x + 2. The perpendicular line l has slope −2/3.

The circle (x − 3)² + (y − 1)² = 25. Its centre is S(3, 1) and its radius is 5.

Frequently asked questions

A luminous right triangle with squares built on its sides drawn as a constellation, beside a circle with its radius, against a cosmic nebula in violet and magenta

Geometry

Why we measure area in squares

Perimeter is counted in metres, area in square metres — and that single exponent is behind most of the mistakes people make in plane geometry. It is why 1 m² is not 100 cm², why two plots with identical area can need wildly different amounts of fencing, and why the Pythagorean theorem is really a statement about the areas of squares rather than the lengths of sides.

11 min