Here is a classical geometry question: the three medians of a triangle (each joining a vertex to the midpoint of the opposite side) always meet at a single point. Why? With ruler-and-compass reasoning this takes a clever argument. With coordinates, it is a few lines of arithmetic: write the vertices as \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), compute the point two-thirds of the way along each median, and notice that all three computations give the same answer, \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\).

The reverse direction is just as powerful. The equation \(x^2 + y^2 = 25\) is a dry piece of algebra until you see that its solutions trace out a circle of radius 5. A picture makes the algebra obvious; algebra makes the picture precise.

The idea, usually credited to René Descartes (and independently to Pierre de Fermat) in the 1630s, is to give every point an address: two numbers that say how far to go across and how far to go up from a fixed starting point. Once points are numbers:

That third bullet is the foundation of every graphics engine, map application and CAD program ever written. A computer cannot see a triangle; it can only store and transform coordinates.

Take two perpendicular number lines that cross at their zeros. The horizontal one is the x-axis, with positive numbers to the right; the vertical one is the y-axis, with positive numbers up. Every point \(P\) in the plane now has a unique address \((x, y)\): drop a perpendicular from \(P\) to each axis and read off the numbers.

The anticlockwise numbering is not arbitrary: it is the direction in which a positive angle turns, which you will see again with the unit circle in Chapter 13.

A delivery drone sits at \((-2, 1)\) on a city map marked in kilometres; the customer is at \((4, 9)\). How far is the straight-line flight? And where should a relay station go if it must be exactly halfway?

Any two points that are not level with each other are the ends of the hypotenuse of a right triangle whose other two sides run parallel to the axes. The horizontal leg is the change in \(x\); the vertical leg is the change in \(y\). Pythagoras (from Chapter 9) does the rest.

Let \(P_1 = (x_1, y_1)\) and \(P_2 = (x_2, y_2)\). The corner point \((x_2, y_1)\) makes a right angle, because one leg is horizontal and the other vertical. The legs have lengths \(|x_2 - x_1|\) and \(|y_2 - y_1|\). By Pythagoras:

Two pleasant consequences fall out. Because each difference is squared, it does not matter which point you call "1": \((x_1 - x_2)^2 = (x_2 - x_1)^2\). And if the points happen to be level (\(y_1 = y_2\)), the formula still works: it collapses to \(\sqrt{(x_2-x_1)^2} = |x_2 - x_1|\), which is the ordinary number-line distance.

The midpoint is even simpler. Halfway between two numbers on a line is their average; halfway between two points is the average in each coordinate separately, because moving halfway along the hypotenuse moves you halfway along each leg (the half-size triangle is similar to the full one, with scale factor \(\tfrac12\)).

Drone at \(P_1 = (-2, 1)\), customer at \(P_2 = (4, 9)\).

Sanity check: the 6–8–10 triangle is the familiar 3–4–5 triangle doubled. And the distance from \((-2, 1)\) to the midpoint \((1, 5)\) is \(\sqrt{3^2 + 4^2} = 5\), exactly half. Good.

A second example with an unfriendly answer: from \((1, -3)\) to \((-4, 2)\).

What if you want the point one quarter of the way from \(P_1\) to \(P_2\)? The same similar-triangles argument says: move a quarter of \(\Delta x\) and a quarter of \(\Delta y\). In general, the point a fraction \(t\) of the way along is

At \(t = 0\) you are at \(P_1\); at \(t = 1\) at \(P_2\); at \(t = \tfrac12\) at the midpoint. This is the parametric form of a segment: the parameter \(t\) acts like time, and \(P(t)\) is where you are at that time. Equivalently, \(P(t) = (1-t)P_1 + tP_2\), a weighted average of the endpoints.

By hand, compute the differences first, then square; it keeps signs straight. In code, the standard library already knows the formula.

math.hypot avoids overflow when \(\Delta x\) is enormous (squaring a \(10^{200}\) overflows a double; hypot rescales internally). math.dist works in any number of dimensions, which you will need in Section 10.10.

You know two points a road passes through, or one point and its gradient. You need a single equation that every point on the road satisfies — so a program can test whether a given point is on the road, find where it crosses another road, or draw it. Which equation, and in which form?

A line is pinned down by two pieces of information: either two points, or one point and a slope. Each standard form of the equation is just one of those descriptions, written as an equation in \(x\) and \(y\).

Point-slope form. Suppose the line has slope \(m\) and passes through \((x_1, y_1)\). A point \((x, y)\) (other than \((x_1, y_1)\) itself) is on the line exactly when the slope from \((x_1, y_1)\) to it equals \(m\):

The second line also holds at \((x_1, y_1)\) (both sides are zero), so it describes the whole line.

Slope-intercept form. Take the special point where the line crosses the y-axis, \((0, b)\). Point-slope gives \(y - b = m(x - 0)\), i.e.

Standard form. Move everything to one side and clear fractions:

Conventionally \(A, B, C\) are integers with no common factor and \(A \ge 0\). Standard form is the only one of the three that can describe a vertical line: \(x = 5\) is \(1\cdot x + 0 \cdot y = 5\). Its slope is \(-A/B\) (when \(B \ne 0\)); its intercepts are easy to read: set \(y = 0\) to get the x-intercept \(C/A\), set \(x = 0\) to get the y-intercept \(C/B\).

Two points. Line through \((1, 2)\) and \((3, 8)\).

Check with \((3, 8)\): \(3(3) - 1 = 8\). Correct.

Point and fractional slope. Through \((-4, 1)\) with slope \(-\tfrac{2}{3}\).

Slope-intercept is \(y = -\tfrac23 x - \tfrac53\). Check: \(2(-4) + 3(1) = -8 + 3 = -5\). Correct.

From standard form to a sketch. \(4x - 3y = 12\). Intercepts: \(y = 0 \Rightarrow x = 3\); \(x = 0 \Rightarrow y = -4\). Plot \((3, 0)\) and \((0, -4)\) and join them. Slope: \(-A/B = -4/(-3) = \tfrac43\).

Rule of thumb: build in point-slope, present in slope-intercept for graphing or in standard form for systems and code.

A city planner wants a new street through a given point running parallel to an existing avenue, and a footpath from a building that meets the avenue at a right angle (the shortest route). Given the avenue’s equation, what are the slopes of these new lines?

Parallel lines never meet, so neither can be steeper than the other: they have equal slopes (and different intercepts — equal slope and equal intercept would be the same line). Vertical lines are all parallel to each other.

Perpendicular lines are subtler. Take the run–rise triangle of a line with slope \(m_1 = \tfrac13\): run 3, rise 1. Now rotate the whole picture by 90° anticlockwise. The run of 3 becomes a vertical rise of 3; the rise of 1 becomes a horizontal step of 1 to the left. The rotated line therefore has slope \(\frac{3}{-1} = -3\). The slope flipped upside down and changed sign.

By rotation. A line with slope \(m_1 = \frac{a}{b}\) has direction "run \(b\), rise \(a\)", i.e. moves from the origin to \((b, a)\). Rotating a point 90° anticlockwise about the origin sends \((x, y)\) to \((-y, x)\) (we justify this rule in Section 10.9). So the perpendicular direction is \((-a, b)\), with slope \(m_2 = \frac{b}{-a} = -\frac{b}{a}\). Hence

By Pythagoras. Shift both lines so they pass through the origin (that changes neither slope nor angle): \(y = m_1 x\) and \(y = m_2 x\). They contain the points \(A = (1, m_1)\) and \(B = (1, m_2)\). The angle at \(O\) is a right angle exactly when \(OA^2 + OB^2 = AB^2\) (Pythagoras and its converse, Chapter 9):

Every step is reversible, so the condition is both necessary and sufficient.

Perpendicular through a point. Find the line through \((4, 1)\) perpendicular to \(y = \tfrac23 x + 1\).

Check: \(-\tfrac32(4) + 7 = 1\); \(\tfrac23 \cdot (-\tfrac32) = -1\). In standard form, \(3x + 2y = 14\).

Perpendicular bisector. The perpendicular bisector of a segment is the line through its midpoint at right angles to it. Every point on it is equally far from both endpoints, which makes it the answer to "where are all the places equally close to A and B?". For \(A(1, 2)\) and \(B(7, 6)\):

Check: \((0, 10)\) is on this line. Its squared distance to \(A\) is \(1 + 64 = 65\); to \(B\), \(49 + 16 = 65\). Equal, as promised.

The shortest path from a point to a line is the perpendicular one. Doing the perpendicular construction in general (find the perpendicular through the point, intersect it with the line, apply the distance formula) gives a compact formula: the distance from \((x_0, y_0)\) to the line \(Ax + By = C\) is

Example: from \((5, 4)\) to \(3x + 4y = 12\), \(d = \frac{|15 + 16 - 12|}{\sqrt{9 + 16}} = \frac{19}{5} = 3.8\). To confirm by the long route: the perpendicular through \((5, 4)\) has direction \((3, 4)\), so its points are \((5 + 3t, 4 + 4t)\). Substituting into the line, \(3(5 + 3t) + 4(4 + 4t) = 12 \Rightarrow 31 + 25t = 12 \Rightarrow t = -0.76\), giving foot \((2.72, 0.96)\). The distance is \(|t| \cdot \sqrt{3^2 + 4^2} = 0.76 \times 5 = 3.8\). The two routes agree. Notice the numerator is just \(|F(x_0, y_0)|\), the side-of-line test from the previous CS sidebar: its sign says which side, and divided by \(\sqrt{A^2+B^2}\) its size says how far.

A phone mast has a range of 5 km and stands at \((3, -2)\) on a map grid. Which grid points get signal? Equally often you are handed the problem the other way round: an equation like \(x^2 + y^2 - 6x + 4y - 12 = 0\) comes out of some calculation, and you need to recognise it as a circle and find its centre and radius.

A circle is the set of all points at a fixed distance (the radius) from a fixed point (the centre). That definition is the distance formula. A point \((x, y)\) lies on the circle with centre \((h, k)\) and radius \(r > 0\) exactly when

So the mast’s coverage boundary is \((x - 3)^2 + (y + 2)^2 = 25\), and the points with signal are those with \((x - 3)^2 + (y + 2)^2 \le 25\). Note the sign flip: centre \(k = -2\) appears as \((y - (-2))^2 = (y + 2)^2\).

Expanding \((x - 3)^2 + (y + 2)^2 = 25\) gives \(x^2 - 6x + 9 + y^2 + 4y + 4 = 25\), i.e. \(x^2 + y^2 - 6x + 4y - 12 = 0\). To go backwards — from the general form to centre and radius — you complete the square in \(x\) and in \(y\) separately, exactly as in Chapter 8.

\(x^2 + y^2 - 6x + 4y - 12 = 0\).

Centre \((3, -2)\), radius \(\sqrt{25} = 5\): the phone mast again.

A version with a leading coefficient: \(2x^2 + 2y^2 - 4x + 12y - 30 = 0\).

The second call is \(x^2 + y^2 + 2x + 2y + 10 = 0\), i.e. \((x+1)^2 + (y+1)^2 = -8\): a sum of squares cannot be negative, so no point satisfies it.

Where does the line \(y = x + 1\) meet the circle \(x^2 + y^2 = 25\)? Substitute the line into the circle — the substitution method of Chapter 6: Try it: Line meets circle, discriminant.

So \(x = 3, y = 4\) or \(x = -4, y = -3\): two crossing points. In general, the resulting quadratic’s discriminant (Chapter 8) tells the story: positive means the line cuts the circle twice (a secant), zero means it touches once (a tangent), negative means it misses.

A tangent to a circle is perpendicular to the radius at the point of contact (Chapter 9). That turns "find the tangent" into a slope problem. Tangent to \(x^2 + y^2 = 25\) at \((3, 4)\): the radius has slope \(\tfrac43\), so the tangent has slope \(-\tfrac34\), and \(y - 4 = -\tfrac34(x - 3)\), which simplifies to \(3x + 4y = 25\). Try it: See the tangent touch.

Why are satellite dishes and car headlights shaped the way they are? Why do planets trace ovals and some comets swing past the Sun once and never return? And what do all of those have to do with the circle you just met?

Take a double cone (two ice-cream cones joined at their tips) and slice it with a plane. Depending on the tilt of the slice you get exactly four kinds of curve, the conic sections: Try it: Slice the cone yourself.

In coordinates, every conic is the graph of a second-degree equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). Lines are degree 1; conics are the next step up.

A parabola is the set of points equally far from a focus and a directrix. Put the focus at \((0, p)\) and the directrix at \(y = -p\). For a point \((x, y)\), the distance to the focus is \(\sqrt{x^2 + (y - p)^2}\) and the distance to the line is \(|y + p|\). Setting them equal:

So \(y = \frac{x^2}{4p}\): the familiar parabola of Chapter 8, now with geometric meaning attached to its coefficient. For \(y = \tfrac14 x^2\), \(4p = 4\), so the focus is \((0, 1)\) and the directrix is \(y = -1\). If \(p < 0\) the parabola opens downward. Try it: Focus and directrix of y = x²/4.

The reflective property explains the dish: every ray arriving parallel to the axis bounces off a parabolic mirror straight through the focus. Put the receiver at the focus and it collects all the signal; put a bulb there and the headlight sends out a parallel beam. Try it: Watch the rays reflect.

An ellipse is the set of points whose distances to two foci \(F_1, F_2\) add up to a constant \(2a\). (Pin a loop of string at two points and trace around with a pencil.) With foci \((\pm c, 0)\), the algebra — the same square-twice routine as the parabola, just longer — gives

\(a\) is the semi-major axis (half the longest width), \(b\) the semi-minor axis. When \(c = 0\) the foci merge, \(a = b\), and you get a circle.

A hyperbola is the set of points whose distances to two foci differ by a constant \(2a\) (in absolute value). With foci \((\pm c, 0)\):

with asymptotes \(y = \pm \frac{b}{a} x\) — lines the branches approach but never reach.

Worked example: identify each conic and its key features.

A game character is drawn as a list of vertex coordinates. The player presses "turn left": every vertex must rotate 90° about the character’s centre. Then the camera zooms, the level scrolls, a mirror-image enemy appears. Each of these is a rule that takes a point’s coordinates and produces new ones. What are the rules?

In Chapter 9 you met congruence (same shape and size) and similarity (same shape, scaled). Transformations are the moves that produce congruent or similar copies. On the coordinate plane each move is a function from points to points, written \((x, y) \mapsto (x', y')\) — read "\((x, y)\) maps to \((x', y')\)". The original is the pre-image; the result is the image, conventionally labelled with primes (\(A \mapsto A'\)).

Translation (slide) by \(a\) across and \(b\) up: \((x, y) \mapsto (x + a, y + b)\). Every point moves by the same vector \((a, b)\).

Reflection (mirror):

Rotation about the origin, anticlockwise:

Dilation (scaling) about the origin with scale factor \(k\): \((x, y) \mapsto (kx, ky)\). \(|k| > 1\) enlarges, \(0 < |k| < 1\) shrinks; a negative \(k\) also rotates by 180°.

Think of \((x, y)\) as "\(x\) steps along the x-direction \((1, 0)\), plus \(y\) steps along the y-direction \((0, 1)\)". A 90° anticlockwise turn sends the direction \((1, 0)\) (east) to \((0, 1)\) (north), and sends \((0, 1)\) (north) to \((-1, 0)\) (west). Rotating the whole picture rotates the step directions and keeps the step counts:

The same argument works for any angle \(\theta\): east goes to \((\cos\theta, \sin\theta)\) and north to \((-\sin\theta, \cos\theta)\), so

You will meet \(\cos\) and \(\sin\) in Chapter 13; check that \(\theta = 90^\circ\) (\(\cos = 0\), \(\sin = 1\)) gives back \((-y, x)\). The idea "a transformation is determined by where it sends the two basic directions" is exactly what a matrix records. The 90° rotation is the matrix \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), and composing transformations becomes multiplying matrices — the whole subject of Chapter 27.

The rotation and dilation rules above use the origin as their fixed centre. To rotate or scale about another point, move that point to the origin, transform, and move back.

A triangle, rotated. Rotate \(A(1, 1)\), \(B(4, 1)\), \(C(1, 3)\) by 90° anticlockwise about the origin with \((x, y) \mapsto (-y, x)\): \(A' = (-1, 1)\), \(B' = (-1, 4)\), \(C' = (-3, 1)\). Check a side: \(AB = 3\), and \(A'B' = \sqrt{0^2 + 3^2} = 3\). Distances preserved, as an isometry must. Try it: Triangle rotated 90°; Rotate P(5,3) about C(2,1).

About another centre. Rotate \(P(5, 3)\) by 90° anticlockwise about \(C(2, 1)\).

Check: \(CP = \sqrt{3^2 + 2^2} = \sqrt{13}\) and \(CP' = \sqrt{(-2)^2 + 3^2} = \sqrt{13}\). Equal.

Doing one transformation and then another is called composition. Sometimes the order is irrelevant; often it is not.

Try it: Order matters: reflect, rotate.

A horizontal slide and a reflection in the x-axis commute (the order does not matter), because one changes only \(x\) and the other only \(y\). But "reflect in the x-axis, then rotate 90°" sends \((x, y) \to (x, -y) \to (y, x)\) — a reflection in \(y = x\) — while the opposite order sends \((x, y) \to (-y, x) \to (-y, -x)\), a reflection in \(y = -x\). Different answers. The floating-point rotate agrees with the exact rotate90 only after rounding: \(\cos(\pi/2)\) in floating point is about \(6 \times 10^{-17}\), not 0.

Add a third axis, the z-axis, perpendicular to both the x- and y-axes. A point in space now has three coordinates \((x, y, z)\). The axes are arranged by the right-hand rule: curl the fingers of your right hand from the positive x-axis towards the positive y-axis, and your thumb points along the positive z-axis. The three coordinate planes (the xy-, yz- and xz-planes) divide space into eight octants.

From \(P_1 = (x_1, y_1, z_1)\) to \(P_2 = (x_2, y_2, z_2)\), first travel in the floor (the xy-directions) to the point \(Q = (x_2, y_2, z_1)\) directly below or above \(P_2\); then go straight up. Pythagoras twice:

The pattern continues to any number of dimensions: square each coordinate difference, add, take the root. In Chapter 26 this becomes the length (norm) of a vector, and it works just as well in 300 dimensions as in 3 — which is how machine-learning embeddings measure similarity.

The midpoint is again the coordinate-wise average, and the sphere with centre \((h, k, l)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). A single linear equation \(ax + by + cz = d\) describes a plane, not a line — one equation removes one degree of freedom from three.

The length of the main diagonal of a box 2 by 3 by 6 (from \((0, 0, 0)\) to \((2, 3, 6)\)):

Is the point \((3, 2, 0)\) inside the sphere of radius 4 centred at \((1, 0, -2)\)? Squared distance \(2^2 + 2^2 + 2^2 = 12 < 16 = 4^2\): inside.

Where you’d use this. Will a 2.4 m curtain pole fit in a car boot measuring 1.0 m by 1.1 m by 0.5 m? The longest straight line inside the box is the space diagonal, \(\sqrt{1.0^2 + 1.1^2 + 0.5^2} = \sqrt{2.46} \approx 1.57\) m, so no. In medical imaging, a CT scan is a 3D grid of voxels whose spacing is often different along each axis, say 0.5 mm, 0.5 mm and 1.0 mm. To measure a tumour from voxel \((10, 20, 5)\) to voxel \((40, 60, 15)\), convert to millimetres first (differences \(30 \times 0.5 = 15\), \(40 \times 0.5 = 20\), \(10 \times 1.0 = 10\)) and then apply the formula: \(\sqrt{225 + 400 + 100} = \sqrt{725} \approx 26.9\) mm. Applying the formula to raw voxel indices would give the wrong answer, a classic bug. GPS receivers, drones, 3D printers (every line of G-code is a \((x, y, z)\) move) and robot arms all work in 3D coordinates.

You have plotted \(y = x^2\) on paper. Now draw it in a browser canvas, an image file, or a terminal. The first attempt comes out upside down. Why?

Displays are scanned like text: left to right, top to bottom. So almost every screen, image format and windowing system puts the origin at the top-left corner, with \(x\) increasing to the right and \(y\) increasing downward. Pixel coordinates are also integers, and they number cells, not points: pixel \((0, 0)\) is the little square in the corner.

To draw a mathematical scene centred on \((c_x, c_y)\) at \(s\) pixels per unit in a canvas \(W\) pixels wide and \(H\) high, combine three transformations from Section 10.9: translate the world centre to the origin, dilate by \(s\) (and reflect in the x-axis to flip \(y\)), then translate the origin to the canvas centre: Try it: Try the world-to-screen map.

The single minus sign in the row formula is the whole difference between "math" and "screen".

The world point \((3, 2)\), up and right of centre, lands above the centre row 300, at row 200, as it should.

A line on paper is infinitely thin and continuous. A screen can only light whole pixels. Given two pixel endpoints, which pixels should be lit so the result looks like a straight line — no gaps, no clumps — and can we do it fast, ideally using only integer addition?

Consider a line whose slope is between 0 and 1 (gently rising, more across than up), from \((x_0, y_0)\) to \((x_1, y_1)\) with integer endpoints. Because it is flat-ish, we want exactly one pixel per column. Moving one column to the right, the line rises by \(m < 1\), so the next pixel is either straight across, \((x+1, y)\), or diagonally up, \((x+1, y+1)\). Choose whichever is closer to the true line. The tie-breaker is the midpoint between the two candidates, \((x+1, y+\tfrac12)\): if the true line passes above it, go up; otherwise go across.

The naive way computes \(y = y_0 + m(x - x_0)\) for each column and rounds. It works, but it uses floating point and a multiplication per pixel — expensive on 1960s plotters (where Jack Bresenham invented the method, at IBM in 1962) and still wasteful today. Bresenham’s insight is to track the sign of an integer error term that changes by a fixed amount each step.

Let \(\Delta x = x_1 - x_0 > 0\) and \(\Delta y = y_1 - y_0\), with \(0 \le \Delta y \le \Delta x\). Write the line in the standard form of Section 10.5, scaled to integers:

\(F = 0\) on the line; \(F > 0\) for points below it (smaller \(y\) makes \(-\Delta x\,(y - y_0)\) larger). Evaluate at the midpoint and double to clear the \(\tfrac12\): define the decision variable \(D = 2F(x + 1, y + \tfrac12)\).

How does \(D\) change from one column to the next? \(F\) is linear, so moving one unit in \(x\) adds \(\Delta y\) to \(F\) (that is, \(2\Delta y\) to \(D\)), and moving one unit up in \(y\) subtracts \(\Delta x\) (\(2\Delta x\) from \(D\)). At the start, \(F(x_0, y_0) = 0\), so

Only integer additions and comparisons. No multiplication inside the loop, no rounding, no floating point.

Draw the segment from \((1, 1)\) to \((8, 4)\). Here \(\Delta x = 7\), \(\Delta y = 3\), so \(2\Delta y = 6\), \(2\Delta x = 14\), and \(D_{\text{start}} = 6 - 7 = -1\). Try it: Step through this example.

Pixels: \((1,1), (2,1), (3,2), (4,2), (5,3), (6,3), (7,4), (8,4)\). Compare each plotted \(y\) with the true \(y\) in the last column: every one is the nearest integer. (At \(x = 2\), for example, the true height is 1.429, closer to 1 than to 2, and the algorithm plotted 1.)

The production version handles every direction with one loop by keeping signed step directions \(s_x, s_y\) and a combined error term: Try it: Run the all-octant version.

Today GPUs rasterise lines and triangles in hardware, but the decision-variable idea — evaluate a linear function’s sign incrementally — is exactly how triangle rasterisers decide which pixels a triangle covers.

Try it: See the tie case.

A game has 500 moving objects. Checking whether every pair of detailed outlines overlaps is far too slow (\(500 \times 499 / 2 \approx 125{,}000\) pairs per frame, each expensive). How can coordinate geometry give cheap tests that rule out most pairs immediately?

Wrap each object in the smallest rectangle with sides parallel to the axes that contains it: its axis-aligned bounding box (AABB), described by \((x_{\min}, y_{\min}, x_{\max}, y_{\max})\) — the minimum and maximum of its vertices' coordinates. Two such boxes overlap exactly when their x-intervals overlap and their y-intervals overlap. Two intervals \(\lbrack a_1, a_2]\) and \(\lbrack b_1, b_2]\) overlap when each starts before the other ends: \(a_1 \le b_2\) and \(b_1 \le a_2\).

If the boxes do not overlap, the objects certainly do not; only if they do is the expensive exact test needed. This "cheap test first" pattern is called a broad phase followed by a narrow phase.

A point \((x, y)\) is inside or on the circle with centre \((h, k)\) and radius \(r\) when \((x - h)^2 + (y - k)^2 \le r^2\) — the circle equation from Section 10.7, turned into an inequality, with no square root. Two circles overlap when the distance between centres is at most the sum of the radii: \((h_1 - h_2)^2 + (k_1 - k_2)^2 \le (r_1 + r_2)^2\).

The ship’s box spans \(x \in [2, 6]\), \(y \in [1, 5]\); the rock’s spans \(x \in [5, 8]\), \(y \in [4, 9]\). Both interval pairs overlap, so a narrow-phase test is needed. The point \((5, 6)\) is at squared distance \(9 + 16 = 25\) from \((2, 2)\): exactly on the radius-5 circle, so it counts as inside. The point \((6, 6)\) is at \(16 + 16 = 32 > 25\): outside.

In Manhattan, a taxi cannot drive diagonally through buildings: it moves along streets in a grid. From \(A(1, 1)\) to \(B(5, 4)\) it must cover 4 blocks east and 3 blocks north, 7 blocks in total, whatever route it takes. The straight-line (Euclidean) distance is \(\sqrt{4^2 + 3^2} = 5\). Try it: Euclidean versus Manhattan distance.

Try it: Drag A and B.

Always \(\text{Chebyshev} \le \text{Euclidean} \le \text{Manhattan}\): here \(4 \le 5 \le 7\).

Choosing a metric. Use Euclidean when movement is free in any direction (physics, drones, k-nearest-neighbours on continuous features). Use Manhattan when movement is restricted to a grid (city routing, 4-directional game maps, circuit-board wiring) or when you want a cheaper, absolute-value-only computation that is less dominated by one large difference. Use Chebyshev when diagonal moves cost the same as straight ones (8-directional grids, chess kings, some warehouse cranes that move both axes at once).

Try it: Watch Manhattan mislead A*.