Analytic geometry in the plane

Review coordinates and choose a scale

In an orthonormal Cartesian coordinate system, the axes are perpendicular and use the same length unit. The word orthonormal matters: standard metric formulas assume this geometry. If a graph uses different visual scales on the axes, it can represent coordinates correctly while apparent lengths and angles are distorted.

Point A = (−2, 4) has x-coordinate −2 and y-coordinate 4. Displacement from A to B = (3, 1) has components 3 − (−2) = 5 horizontally and 1 − 4 = −3 vertically. Components can be negative because they indicate direction; distance between points is a length and cannot be negative.

For location problems, choose origin, axes and unit before writing numbers. A map may use kilometres, a floor plan metres. Choosing a coordinate system changes coordinates, but not geometric distance between two points when the transformation preserves lengths. A good choice simplifies calculations without changing the problem.

Derive distance and midpoint

If A = (x₁, y₁) and B = (x₂, y₂), horizontal and vertical projections form a right triangle with legs of lengths |x₂ − x₁| and |y₂ − y₁|. By Pythagoras, distance is:

The squares ensure that swapping A and B does not change the result: d(A, B) = d(B, A). If distance is zero, both differences must be zero and the points coincide. Subtracting coordinates in the same order in both components avoids confusion, although squares eliminate the sign in the final result.

Midpoint M divides the segment into two equal parts and has coordinates M = ((x₁ + x₂)/2, (y₁ + y₂)/2). Each coordinate is the average of the corresponding coordinates. For A = (−2, 4) and B = (3, 1), M = (0,5; 2,5), using a semicolon to separate coordinates with decimal commas. The difference from A to M is half that from A to B, justifying the formula and equality of the two distances.

Line equations: more than one form

A nonvertical line through A = (x₀, y₀) with slope m satisfies y − y₀ = m(x − x₀). The idea is that all displacements along the line have the same ratio of vertical to horizontal change. With two points of different x-coordinates, m = (y₂ − y₁)/(x₂ − x₁). The denominator shows why a vertical line has no defined slope.

Expanding the previous form gives y = mx + p. To check whether a point lies on the line, substitute both its coordinates into the equation. One point does not determine a line, since infinitely many directions are possible; two distinct points do. If both x-coordinates are equal, the equation is x = x₀.

The general form Ax + By + C = 0, with A and B not both zero, includes vertical lines too. If B ≠ 0, isolating y gives slope −A/B. If B = 0, we have x = −C/A. Multiplying the entire equation by a nonzero constant does not change the line; for example, 2x + y − 3 = 0 and 4x + 2y − 6 = 0 represent the same set of points.

Parallel lines, perpendicular lines and intersections

Two nonvertical lines are parallel when they have the same slope; if their y-intercepts also match, they are the same line. Two distinct vertical lines are parallel. Nonparallel lines cross at one point satisfying both equations simultaneously. Finding that point means solving a linear system.

For two perpendicular nonvertical lines, slopes satisfy m₁m₂ = −1. One way to understand the sign is to rotate direction (1, m) by 90°: it becomes (−m, 1), whose slope for m ≠ 0 is −1/m. A horizontal and vertical line are perpendicular too, but that case cannot use the product of two defined slopes.

Do not conclude perpendicularity from the drawing alone: different scales can make a right angle look oblique. For example, a slope-2 line is perpendicular to one of slope −1/2, not slope −2. Both the reciprocal and negative sign matter. To construct the perpendicular through a point, first find the new slope, then use point–slope form.

The circle as a locus

A locus is a set of points satisfying a condition. The circle with center C = (h, k) and radius r > 0 contains points P whose distance from C is r. Applying the distance formula and squaring gives:

Equality describes the boundary. The inequality ≤ r² describes the closed disk, including interior and boundary; < r² describes only the interior. In (x + 2)² + (y − 3)² = 25, the center is (−2, 3) and radius 5. The number on the right is radius squared, not radius.

We can recover the center from an expanded equation by completing squares. In x² + y² − 6x + 4y − 12 = 0, we have (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0; hence (x − 3)² + (y + 2)² = 25. Center (3, −2), radius 5. If the final right side were negative, there would be no real points; if zero, only the center would remain, a degenerate case.

The ellipse: a distance condition producing an equation

An ellipse is the set of points whose distance sum to two fixed points, the foci, is constant. Start with foci F₁ = (−c,0) and F₂ = (c,0), with c ≥ 0, and distance sum 2a, with a > c. The center is the origin. The major axis measures 2a; we write b² = a² − c², with b > 0. Here c is distance from center to a focus, not a line's constant term.

We can derive the equation. For P = (x,y), call distances to the foci r₁ and r₂. We have r₁ + r₂ = 2a. By the distance formula, r₁² − r₂² = (x + c)² − (x − c)² = 4cx. Factoring the difference of squares, (r₁ − r₂)(r₁ + r₂) = 4cx, so r₁ − r₂ = 2cx/a. Adding the two relationships and dividing by 2 gives r₁ = a + cx/a.

Now r₁² = (x + c)² + y² gives a² + 2cx + c²x²/a² = x² + 2cx + c² + y². Cancel 2cx and rearrange: a² − c² = x²(1 − c²/a²) + y². Using b² = a² − c² and dividing by b², we reach x²/a² + y²/b² = 1. The lengths and condition a > c ensure b² is positive.

The major-axis vertices are at (±a,0), and minor-axis endpoints at (0,±b). If the major axis is vertical, the term with a² is associated with y². For example, x²/9 + y²/25 = 1 has vertical semimajor axis 5, semiminor axis 3 and foci (0,±4). The larger denominator need not lie under x.

For center (x₀,y₀), replace x with x − x₀ and y with y − y₀. The ellipse (x − 2)²/25 + (y + 1)²/9 = 1 has center (2,−1), semiaxes 5 and 3 and c = √(25 − 9) = 4. Its foci are (−2,−1) and (6,−1). To check a point, substitute both coordinates and verify equality rather than judging only the figure's appearance.

A circle is the case where a = b and both foci coincide at the center. An ellipse drawn with different scales on the axes can appear more or less elongated than it really is. Geometric values should come from the equation and scale, not a visual impression.

Sets of points and problem solving

A line and circle can have no common points, touch at one point or cross at two. Substituting the line expression into the circle equation produces a quadratic equation; the number of real solutions expresses these three possibilities. After finding one coordinate, calculate the other and check the pairs in both equations.

Another useful locus is the set of points equidistant from A and B, with A ≠ B. Squaring the condition d(P, A) = d(P, B) cancels x² and y² and produces a line: the perpendicular bisector of segment AB. Geometrically, this line passes through the midpoint and is perpendicular to the segment. Algebra and geometry confirm the same description.

When solving a situation, decide whether you seek a point, distance, line or region. An equation may describe infinitely many points, while a system can select some of them. The answer must respect this difference: giving only the x-coordinate does not identify a point, and giving only a drawing does not verify a numerical condition. Final units depend on the quantity requested.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Subtracting coordinates can give a negative distance.

Directed differences can be negative; distance uses squares and a nonnegative square root.

Two perpendicular lines have opposite slopes.

For defined, nonzero slopes, they are negative reciprocals, not merely opposites.

In a circle equation, the right side is the radius.

It is r²; take the positive root to obtain r.

A circle and disk are the same set.

The circle is the boundary; the disk includes interior points.