10.2 Distance Formula and Pythagorean Theorem

10.2 Distance Formula and Pythagorean Theorem

The second Algebra 1 geometry skill on College Board Table 11 is using the distance formula and the Pythagorean theorem. On a 20-item AAF CAT this still sits inside the same 5–10% / 1–2 item Geometry-for-Algebra-1 bucket as area and transformations, so one well-aimed item can be a right triangle, a coordinate distance, or both at once.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Distance is Pythagorean on the plane

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

d = √[(x₂ − x₁)² + (y₂ − y₁)²].

Draw the right triangle whose legs are the horizontal run |x₂ − x₁| and the vertical rise |y₂ − y₁|. The segment joining the points is the hypotenuse. The distance formula is a² + b² = c² with a = Δx, b = Δy, and c = d. You do not need a picture if you square both differences, add, and take the nonnegative square root.

Order of subtraction does not matter because of the squares: (x₁ − x₂)² = (x₂ − x₁)². Mixing the order across coordinates — using x₂ − x₁ with y₁ − y₂ — is still fine after squaring. What is not fine is adding the unsquared differences, or stopping at the sum of squares and calling that the distance.

The midpoint of the same segment is the average of the coordinates:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2).

Midpoint is an arithmetic mean, not a Pythagorean length. Do not put a square root on a midpoint item.

Worked example: distance from (−1, 2) to (3, 5)

Let P = (−1, 2) and Q = (3, 5).

1. Horizontal change: Δx = 3 − (−1) = 4.
2. Vertical change: Δy = 5 − 2 = 3.
3. Distance: d = √(4² + 3²) = √(16 + 9) = √25 = 5.

The connecting segment is a 3-4-5 right triangle sitting on the grid. The length is 5, not 7 (the taxicab path 4 + 3), not 25 (the sum of squares with the root forgotten), and not 12 (the product of the legs).

Midpoint:

M = ((−1 + 3)/2, (2 + 5)/2) = (2/2, 7/2) = (1, 7/2).

An option (1, 3) averaged the y-coordinates as integers and dropped the 1/2. An option (2, 7) forgot to divide by 2. An option (4, 3) used the differences as if they were the midpoint.

Slope from the same two points (preview of the parallel-lines section)

The slope of line PQ is m = Δy/Δx = 3/4. A line parallel to PQ has the same slope 3/4 and a different intercept. A line through (0, 1) parallel to PQ is y = (3/4)x + 1. Perpendicular slope would be the negative reciprocal −4/3. Stop there: writing parallel and perpendicular equations is the linear-applications skill, not this geometry skill. If an AAF stem asks for the length of PQ, the answer is 5, not 3/4.

A vertical segment such as (2, −1) to (2, 7) has Δx = 0 and distance |7 − (−1)| = 8. The formula still works: √(0 + 64) = 8. Slope is undefined, so there is no parallel-line slope to copy until you switch to a vertical-line equation x = 2. A horizontal segment (−4, 3) to (5, 3) has distance 9 and slope 0.

Pythagorean theorem off the axes

In a right triangle with legs a, b and hypotenuse c (the side opposite the right angle),

a² + b² = c².

Solving for a missing side:

• hypotenuse: c = √(a² + b²)
• a leg: a = √(c² − b²)

The minus sign is the trap when the unknown is a leg. Using √(c² + b²) invents a longer hypotenuse instead of a shorter leg.

Families you should recognize without a calculator

If the legs are 5 and 12, the hypotenuse is 13, not 17 and not √(5 + 12). If the hypotenuse is 26 and one leg is 10, you are in a doubled 5-12-13, so the other leg is 24.

A 6-8-10 triangle is similar to 3-4-5; every length is scaled by 2, and the area is scaled by 2² = 4. That scaling is the same idea as a dilation in Section 10.3.

Worked example: missing side and a coordinate check

A right triangle has legs 8 and 15. Hypotenuse:

c = √(64 + 225) = √289 = 17.

That is the 8-15-17 family. Checking 8² + 15² = 64 + 225 = 289 = 17² is faster than a decimal square root.

A different triangle has hypotenuse 13 and one leg 5. The other leg is √(169 − 25) = √144 = 12, the 5-12-13 family. If you add instead of subtract, √(169 + 25) = √194 is not 12 and is not among the options unless the item is hunting that exact mistake.

Place the same 5-12-13 on the plane with a right angle at the origin: A(0, 0), B(5, 0), C(0, 12). Then AB = 5, AC = 12, and BC = √(5² + 12²) = 13. Move the right angle off the origin — A(1, 1), B(4, 1), C(4, 5) — and you still have legs 3 and 4 with hypotenuse 5. Distance does not care where the triangle sits, only about the differences.

Finding a missing coordinate

Point R is at (0, 0) and point S is at (6, y) with RS = 10. Then

√(6² + y²) = 10 → 36 + y² = 100 → y² = 64 → y = ±8.

Both (6, 8) and (6, −8) are 10 units from the origin (a 6-8-10 triangle). The stem’s quadrant, a graph, or a second condition such as “above the x-axis” picks the sign. Do not discard the negative root unless the context forbids it.

Converse: is this triangle right?

If a² + b² = c² for the longest side c, the triangle is right. If a² + b² > c², it is acute. If a² + b² < c², it is obtuse. On AAF, the usual task is the equality case. Sides 9, 12, 15 satisfy 81 + 144 = 225, so the triangle is right (a scaled 3-4-5). Sides 6, 8, 11 give 36 + 64 = 100 < 121, so it is obtuse — 11 is too long to be a 6-8-10 hypotenuse.

Traps

• Forgetting the square root. 16 + 9 = 25 is not the distance; 5 is.
• Adding the legs. Taxicab length 3 + 4 = 7 is not Euclidean distance.
• Treating a leg as the hypotenuse. The longest side is c.
• Subtracting coordinates without squaring, then taking an absolute value of a single number.
• Putting a radical on the midpoint.
• Returning slope when the stem asked for distance (or the reverse). From (−1, 2) to (3, 5), slope is 3/4 and distance is 5.

College Board’s calculator policy still applies: some items show an on-screen calculator; many distance items are integer triples that should never need it. Handheld calculators are not allowed except with an approved accommodation.

Nonprism volume and circle equations in the coordinate plane continue in Algebra 2 geometry, starting at Volume of Cylinders, Cones, and Spheres. For the parallel-line slope skill, use Parallel and Perpendicular Lines.