Distance & Midpoint
1. Core Idea
Two formulas, and both are easier than they look:
- Distance is Pythagoras. The horizontal gap and the vertical gap are the legs of a right triangle; the distance is the hypotenuse.
- Midpoint is an average. Average the x's, average the y's.
If a student can draw the right triangle, they never need to memorise the distance formula.
2. Must-Know Rules
| Concept | Formula |
|---|---|
| Distance between (x1,y1) and (x2,y2) | sqrt[(x2-x1)^2 + (y2-y1)^2] |
| Midpoint | ((x1+x2)/2, (y1+y2)/2) |
| Distance from the origin | sqrt(x^2 + y^2) |
| Horizontal distance (same y) | |x2 - x1| |
| Vertical distance (same x) | |y2 - y1| |
Finding an endpoint from the midpoint. If M is the midpoint of AB and you know A and M, then B = (2Mx - Ax, 2My - Ay). "Double the midpoint, subtract the known endpoint."
Distance is symmetric — the order of the points does not matter, because the differences are squared.
Watch for triples. If the horizontal and vertical gaps are 3 and 4, or 6 and 8, or 5 and 12, the distance is a whole number. This happens constantly on the GRE.
3. Worked Examples
Example 1 — Distance
Find the distance between (2, 3) and (10, 9).
Horizontal gap: 10 - 2 = 8. Vertical gap: 9 - 3 = 6.
Distance = sqrt(64 + 36) = sqrt(100) = 10. (A 6-8-10 triangle.)
Example 2 — Midpoint
Find the midpoint of (-4, 7) and (10, -1).
x: (-4 + 10)/2 = 3. y: (7 + (-1))/2 = 3.
Midpoint = (3, 3).
Example 3 — Missing endpoint
M(5, -2) is the midpoint of AB. If A is (1, 4), find B.
B_x = 2(5) - 1 = 9. B_y = 2(-2) - 4 = -8.
B = (9, -8).
Check: midpoint of (1,4) and (9,-8) is ((1+9)/2, (4-8)/2) = (5, -2). Correct.
4. GRE Traps
- Subtracting inside the midpoint formula. Midpoint uses a sum divided by 2.
- Adding inside the distance formula. Distance uses differences.
- Forgetting the square root. sqrt(64 + 36) = 10, not 100.
- Sign errors with negative coordinates. 9 - (-1) = 10, not 8.
- sqrt(a^2 + b^2) = a + b. Never.
- Mixing up which point is which — harmless for distance (squares), fatal for the missing-endpoint formula.
5. Speed Tricks
- Draw the right triangle. Horizontal leg, vertical leg, hypotenuse. Then Pythagoras.
- Look for a Pythagorean triple in the two gaps before reaching for a calculator.
- Midpoint = average. Two averages, done.
- "Double the midpoint minus the known endpoint" for the missing-endpoint type.
- For a circle's centre given two endpoints of a diameter, the centre is the midpoint — and the radius is half the distance.
- Distance from the origin is just sqrt(x^2 + y^2).
6. Self-Check
Q1. Find the distance between (-1, 2) and (4, 14).
Q2. Find the midpoint of (3, -6) and (-9, 2).
Q3. The endpoints of a circle's diameter are (2, 1) and (8, 9). Find the centre and the radius.
Answers
A1. Gaps 5 and 12 -> distance = 13.
A2. ((3-9)/2, (-6+2)/2) = (-3, -2).
A3. Centre = midpoint = (5, 5). Diameter = sqrt(36 + 64) = 10, so radius = 5.
7. One-Line Summary for the Board
Distance is Pythagoras on the gaps. Midpoint is the average of the coordinates.