SAT coordinate geometry: find distance and midpoint
Turn two points into a right triangle for distance, or average their coordinates for the point halfway between them.
SPSAT Practice Editorial5 min read
The short version
What you'll take away
- Distance comes from the horizontal and vertical changes: √((x₂ − x₁)² + (y₂ − y₁)²).
- The midpoint averages x-coordinates and y-coordinates separately.
- Check a distance result against a simple horizontal or vertical estimate.
In this article
Build a right triangle between two points
From A(1, 2) to B(5, 5), move 4 units right and 3 units up. These moves form perpendicular legs, so the segment AB is the hypotenuse. The Pythagorean theorem gives \(\displaystyle AB=\sqrt{4^2+3^2}=5\).
For any two points, use \(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). Squaring makes the direction of each move irrelevant. The formula works even when a coordinate is negative.
The study lab
Two points 5 units apart
A(1, 2) and B(5, 5) are 4 units apart horizontally and 3 units apart vertically.
- Segment AB lies on y = ¾x + 1¼
Average the coordinates for the midpoint
The midpoint is halfway along the segment. Average the x-values and the y-values separately: \(\displaystyle M=((x_1+x_2)/2,(y_1+y_2)/2)\). For A(1, 2) and B(5, 5), the midpoint is \(\displaystyle (3,3.5)\). It is 2 units right and 1.5 units up from A, exactly half of each change to B.
If one endpoint and the midpoint are known, reverse the averages. If A is (1, 2) and M is (3, 3.5), then the other x-coordinate is \(\displaystyle 2(3)-1=5\), and the other y-coordinate is \(\displaystyle 2(3.5)-2=5\).
Try the distance formula
Subtract coordinates in the same order and square both differences.
Your turn
Quick practice
What is the distance between (−2, 1) and (4, 9)?
Hint: Find the horizontal and vertical legs first.
Sources & further reading
One step from here
Put the idea to work.
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