Coordinate Geometry

GE.H.1Distance & midpoint01/04

Distance and midpoint formulas.

The two points

A (2, 1) and B (8, 9)

Step01of 04

Distance formula — the Pythagorean theorem applied to the change in x and the change in y.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The horizontal leg has length $∣ x_{2} - x_{1} ∣$ , the vertical leg has length $∣ y_{2} - y_{1} ∣$ ; their hypotenuse is the distance between the points.

GE.H.1 — Distance and midpoint formulas.

The two points

A (2, 1) and B (8, 9)

Step 01 of 04

Distance formula — the Pythagorean theorem applied to the change in x and the change in y.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The horizontal leg has length $∣ x_{2} - x_{1} ∣$ , the vertical leg has length $∣ y_{2} - y_{1} ∣$ ; their hypotenuse is the distance between the points.

Step 02 of 04

Apply to A(2, 1) and B(8, 9).

d = (8 - 2)^{2} + (9 - 1)^{2} = 36 + 64 = 100 = 10

Distance AB = 10.

Step 03 of 04

Midpoint formula — the AVERAGE of the coordinates.

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Apply to the same two points:

M = (\frac{2 + 8}{2}, \frac{1 + 9}{2}) = (5, 5)

Step 04 of 04

One coordinate at a time. Distance and midpoint formulas treat x and y independently — change in x stays in x, change in y stays in y.

Reverse-midpoint problems. If the midpoint of $A$ and $B$ is $M$ , and you know $A$ and $M$ , find $B$ by doubling and subtracting: $B = 2 M - A$ for each coordinate.

Key insight

Distance = Pythagorean theorem on the coordinate differences. Midpoint = average of the coordinates. Both apply componentwise — handle x and y separately, never mix them.

GE.H.2 — Slopes of parallel and perpendicular lines.

Step 01 of 04

In the coordinate plane, the relationship between two lines is decided by their slopes alone.

Parallel	→same slope $m_{1} = m_{2}$
Perpendicular	→negative reciprocals $m_{1} \cdot m_{2} = - 1$
Same line	→same slope AND same y-intercept
Intersecting (oblique)	→different slopes, not perpendicular

Step 02 of 04

Worked example — parallel. Find a line through $(4, - 1)$ parallel to $y = 3 x + 2$ .

Slope must be 3. Use point-slope:

y - (- 1) = 3 (x - 4) \Rightarrow y = 3 x - 13

Step 03 of 04

Worked example — perpendicular. Find a line through $(4, - 1)$ perpendicular to $y = 3 x + 2$ .

Slope is $- 1/3$ (flip and negate 3). Point-slope:

y + 1 = - \frac{1}{3} (x - 4) \Rightarrow y = - \frac{1}{3} x + \frac{1}{3}

Verify: $(3) \cdot (- 1/3) = - 1$ $✓$ .

Step 04 of 04

Special cases.

Horizontal line	→ $y = c$ , slope 0
Vertical line	→ $x = c$ , slope undefined
Perp. to horizontal	→vertical (slope undefined)
Perp. to vertical	→horizontal (slope 0)

Key insight

Read the slope, decide what to do with it: keep it (parallel) or flip and negate it (perpendicular). The given point only sets the y-intercept. Same skill as A1.C.3, restated in the geometry context.

GE.H.3 — Polygons on the coordinate plane.

Step 01 of 04

Coordinate geometry lets you PROVE properties of a polygon by computing side lengths, slopes, and midpoints — instead of relying on diagrams. The recipe:

Distance formula	→side lengths
Slope formula	→parallel / perpendicular sides
Midpoint formula	→diagonals bisecting each other, midsegments

Step 02 of 04

Worked example. Show that $A (0, 0), B (5, 0), C (5, 3), D (0, 3)$ is a rectangle.

Strategy: show all four angles are right by computing slopes of adjacent sides.

slope A B = \frac{0 - 0}{5 - 0} = 0 (horizontal)

slope B C = \frac{3 - 0}{5 - 5} = undef. (vertical)

Horizontal ⟂ vertical, so $∠ B = 90°$ . Same logic at every vertex — all four right. Combined with opposite sides equal length: rectangle.

Step 03 of 04

Classify a quadrilateral from coordinates — the decision tree.

All 4 sides equal AND 4 right angles	→square
4 right angles, opposite sides equal	→rectangle
All 4 sides equal	→rhombus
Both pairs of opposite sides parallel	→parallelogram
Exactly one pair parallel	→trapezoid
Two pairs of consecutive sides equal	→kite

Compute slopes (for parallel/perpendicular) and distances (for length equality), then match the property pattern.

Step 04 of 04

Triangle classifications. Compute the three side lengths.

All 3 equal	→equilateral
Exactly 2 equal	→isosceles
All 3 different	→scalene
$a^{2} + b^{2} = c^{2}$ (largest is c)	→right triangle (apply Pythagorean)

A right isosceles triangle is both — the two short sides are equal AND they're the legs of a right angle.

Key insight

Coordinate geometry replaces visual reasoning with arithmetic: distance for length, slope for parallel/perpendicular, midpoint for bisection. Pick the property you want to prove, compute, conclude.

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GE.H.1 – GE.H.3 · CCSS G.GPE.B.7 / G.GPE.B.5·Full curriculum →