6.1 The Coordinate Plane, Distance & Midpoint

Key Takeaways

  • The Cartesian plane is split by y = 0 (x-axis) and x = 0 (y-axis) into Quadrants I (+,+), II (−,+), III (−,−), and IV (+,−); a point on an axis belongs to no quadrant.
  • GMAC's published coordinate content is the plane, quadrants, ordered pairs, slope, intercepts, line equations, and function graphs; distance, midpoint, and coordinate area are supporting algebra, not printed GMAC formulas.
  • Distance between (x1, y1) and (x2, y2) is sqrt((x2 − x1)^2 + (y2 − y1)^2), reducing to |x2 − x1| for horizontal segments and |y2 − y1| for vertical ones; with no calculator, factor a common scale out of Δx and Δy so the 3-4-5 and 5-12-13 families finish the arithmetic mentally.
  • The midpoint is ((x1 + x2)/2, (y1 + y2)/2). A missing endpoint is (2 x_M − x1, 2 y_M − y1), which is the same as repeating the step from the known endpoint to the midpoint.
  • Triangle area from coordinates is algebra: (1/2)|base × height| when a side is axis-parallel, or bounding-box area minus the outer right triangles when sides are slanted.
Last updated: August 2026

Ordered Pairs, Axes, and Quadrants

GMAC removed Euclidean geometry from GMAT Focus Quantitative Reasoning: you will not chase triangle-angle sums, circle theorems, or solids as geometry items. Coordinate work stayed, classified as algebra. The evidence is structural rather than promotional: in the Official Guide Quantitative Review 2026–2027 Math Review, Graphing is sub-topic 3.2.6, sitting inside "3.2 Algebra, Equalities, and Inequalities," and the Reference Sheet in 3.5 carries a "Lines in the Coordinate Plane" entry. Every (x, y) on this exam is a pair of signed numbers you compute with, not a figure you have to "see."

Be precise about what that official entry contains, because it sets the scope of this chapter. GMAC's Graphing sub-topic and Reference Sheet cover the plane and its quadrants, ordered pairs, slope, x- and y-intercepts, equations of lines, two linear equations as intersecting lines, and the graph of a function. They do not print a distance formula, a midpoint formula, or an area-from-vertices rule. Treat the tools in the rest of this section as supporting coordinate algebra — fast, reusable, and built only from the Pythagorean relation, averaging, and subtraction you already own — rather than as published Quant content. They cost minutes to learn and they protect you if a Problem Solving stem describes a segment or a region in coordinates; they are not a claim that GMAC has restored geometry.

The Cartesian plane is two perpendicular number lines. The horizontal x-axis is the line y = 0. The vertical y-axis is the line x = 0. They meet at the origin (0, 0). An ordered pair (x, y) lists horizontal displacement first (positive right, negative left) and vertical displacement second (positive up, negative down).

The axes cut the plane into four quadrants:

QuadrantSign of xSign of ySample point
Ipositivepositive(3, 5)
IInegativepositive(−4, 2)
IIInegativenegative(−6, −1)
IVpositivenegative(7, −3)

A point on an axis is not in any quadrant. (5, 0) sits on the x-axis; (0, −8) sits on the y-axis. If a stem says P(a, b) lies in Quadrant II, you know a < 0, b > 0, and neither coordinate is 0.

Sign products are a fast filter:

  • xy > 0 means Quadrant I or III (same signs).
  • xy < 0 means Quadrant II or IV (opposite signs).
  • x + y = 0 means the point is on the line y = −x.
  • x = y means the point is on the line y = x.

Reflections as Coordinate Algebra

Reflecting a point is a sign change or a swap, not a geometry construction:

  • Across the x-axis: (x, y) → (x, −y)
  • Across the y-axis: (x, y) → (−x, y)
  • Across the origin: (x, y) → (−x, −y)
  • Across the line y = x: (x, y) → (y, x)
  • Across the line y = −x: (x, y) → (−y, −x)

Worked example. Reflect A(−3, 5) across the y-axis: (3, 5). Reflect that image across the x-axis: (3, −5). The two-step path matches reflecting A across the origin in one move: −(−3, 5) = (3, −5).

Distance Without a Calculator

The distance between P(x₁, y₁) and Q(x₂, y₂) is

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

That identity is the Pythagorean relation written in coordinates. On Quant it is algebra, not a geometry proof. The two legs of the right triangle with hypotenuse PQ are the axis-parallel changes:

  • Horizontal leg |Δx| = |x₂ − x₁|
  • Vertical leg |Δy| = |y₂ − y₁|

Two cases skip the radical entirely:

  • Horizontal segment (y₁ = y₂): d = |x₂ − x₁|
  • Vertical segment (x₁ = x₂): d = |y₂ − y₁|

Because there is no calculator, factor a common scale out of Δx and Δy before you square. If |Δx| and |Δy| are a scaled 3-4-5, 5-12-13, 8-15-17, or 7-24-25 pair, the distance is the scaled third number — the same factoring habit as Chapter 1 mental math, not a Euclidean theorem to memorize as geometry.

Δx and Δy (either order)Distance
3 and 45
6 and 810
9 and 1215
5 and 1213
10 and 2426
8 and 1517
7 and 2425
equal legs k and kk√2

Worked example (scaled 3-4-5). Distance from (12, 18) to (36, 50): Δx = 24, Δy = 32. Factor 8: 24 = 8×3 and 32 = 8×4. Then d = 8×5 = 40. You never hand-compute √(576 + 1,024) as a long square root if you recognize 3-4-5 scaled by 8.

Worked example (equal legs). From (2, −1) to (6, 3): Δx = 4, Δy = 4. Then d = √(16 + 16) = √32 = 4√2. Equal horizontal and vertical change always simplifies through √(2k²) = k√2. That is radical arithmetic, not a 45-45-90 geometry lesson.

Worked example (unknown coordinate). The distance from (t, 1) to (−2, 13) is 13. Then √[(−2 − t)² + 12²] = 13, so (−2 − t)² = 25. Thus −2 − t = 5 or −2 − t = −5, giving t = −7 or t = 3. If the stem also says t is negative, keep t = −7 and throw out 3. Squaring both sides of a distance equation almost always creates two algebraic roots; the stem's sign or quadrant restriction picks one.

When the radical is not an integer, leave it simplified. √(9 + 16) is 5, but √(9 + 4) = √13 stays √13. Do not decimal-approximate on Quant.

Midpoint and Missing Endpoints

The midpoint M of segment AB with A(x₁, y₁) and B(x₂, y₂) is the average of the coordinates:

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

x and y average independently. Distance does not appear in the midpoint formula.

Worked example. A(−6, 8) and B(4, −2): x_M = (−6 + 4)/2 = −1 and y_M = (8 − 2)/2 = 3. Midpoint (−1, 3).

Given one endpoint and the midpoint, the other endpoint is linear:

x₂ = 2x_M − x₁, y₂ = 2y_M − y₁

Equivalently, the step from A to M repeats from M to B:

  • Δx = x_M − x₁, then x_B = x_M + Δx
  • Δy = y_M − y₁, then y_B = y_M + Δy

Worked example. Endpoint A(5, −3), midpoint M(2, 4). Horizontal step 2 − 5 = −3, so x_B = 2 + (−3) = −1. Vertical step 4 − (−3) = 7, so y_B = 4 + 7 = 11. Missing endpoint (−1, 11). Check: midpoint of (5, −3) and (−1, 11) is ((5 − 1)/2, (−3 + 11)/2) = (2, 4).

A point a fraction of the way along a segment

If R divides PQ so that the displacement P → R is the fraction t of P → Q (t = 1/2 is the midpoint; t = 1/4 is a quarter of the way from P toward Q), then

R = ( x_P + t(x_Q − x_P) , y_P + t(y_Q − y_P) )

Worked example. P(−4, 7) and midpoint M(2, 1) of PQ. Recover Q first: steps P → M are +6 and −6, so Q = (2 + 6, 1 − 6) = (8, −5). Point R is 1/4 of the way from P to Q. Total displacement P → Q is (12, −12); one-fourth is (3, −3). Then R = (−4 + 3, 7 − 3) = (−1, 4).

Collinear grid points often hide this ratio. If three points have equally spaced x-coordinates, check whether the y-coordinates are equally spaced. Equal spacing in both coordinates is the midpoint (or equal-step) test.

Area from Coordinates (Still Algebra)

A triangle on the coordinate plane is not a Euclidean-geometry item on Focus Quant, and GMAC does not publish an area-from-vertices rule. If a Problem Solving stem does describe a region by its corner coordinates, the area is still reachable with nothing but subtraction and halving. Two algebraic methods cover almost every Problem Solving stem.

1. Base and height parallel to the axes. If one side is horizontal or vertical, that side is a base and the perpendicular coordinate difference is a height. Area = (1/2)|base × height|.

Worked example. Vertices (0, 0), (0, 6), and (5, 2). The side from (0, 0) to (0, 6) is vertical of length 6. The horizontal distance from x = 0 to x = 5 is the height. Area = (1/2)(6)(5) = 15.

2. Bounding-box subtraction. For a triangle whose sides are slanted, draw the axis-aligned rectangle that just contains the three vertices, compute that rectangle's area, and subtract the right triangles (and any corner rectangles) that lie outside the target.

Protocol:

  1. Read x_min, x_max, y_min, y_max from the three vertices. Box area = (x_max − x_min)(y_max − y_min).
  2. Each box corner that is not a triangle vertex is the right-angle corner of an outer right triangle (or a rectangle) whose legs run along the box.
  3. Subtract those outer areas from the box.

Worked example. Vertices A(1, 2), B(7, 4), C(3, 8).

  • Box: x from 1 to 7 (width 6), y from 2 to 8 (height 6). Box area = 36.
  • Outer piece under AB: legs 6 and 2, area (1/2)(6)(2) = 6.
  • Outer piece right of BC: legs 4 and 4, area (1/2)(4)(4) = 8.
  • Outer piece left of AC: legs 2 and 6, area (1/2)(2)(6) = 6.
  • Triangle area = 36 − (6 + 8 + 6) = 16.

Shoelace is the same arithmetic in one line. Listing vertices in order around the triangle,

Area = (1/2) |x₁y₂ + x₂y₃ + x₃y₁ − (y₁x₂ + y₂x₃ + y₃x₁)|

Use axis-parallel base-height when a side is horizontal or vertical. Use the box when nothing is axis-parallel. Perimeter, if asked, is three distances. Add simplified radicals only when they match: 3 + 4 + 5 = 12, but 5 + √2 does not combine.

Formula sheet for this section

TaskFormula
Distance√[(x₂ − x₁)² + (y₂ − y₁)²]
Horizontal length|x₂ − x₁| when y is constant
Vertical length|y₂ − y₁| when x is constant
Midpoint((x₁ + x₂)/2, (y₁ + y₂)/2)
Other endpoint(2x_M − x₁, 2y_M − y₁)
Axis-parallel-leg triangle(1/2)|Δx||Δy|
Box methodbox area minus outer right triangles
Loading diagram...
Choosing a Distance Method on Quant
Test Your Knowledge

In the xy-plane, the distance between P(m, −4) and Q(8, 5) is 15. If m is a negative integer, what is m?

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Test Your Knowledge

Segment AB has midpoint M(1, −2). Endpoint A is (−5, 4), and C is the midpoint of segment MB. What are the coordinates of C?

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Test Your Knowledge

What is the area of the triangle in the xy-plane with vertices (0, 2), (10, 4), and (2, 8)?

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