The Midpoint Formula
The midpoint is just the average of the two endpoints — average the $x$'s and average the $y$'s.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. It's an average, so you add and divide — you don't subtract.
How to solve one, step by step
Example: midpoint of $(2, 4)$ and $(6, 10)$.
- Average the $x$'s: $\frac{2 + 6}{2} = 4$.
- Average the $y$'s: $\frac{4 + 10}{2} = 7$. Midpoint $(4, 7)$.
The mistakes that cost points
- Subtracting instead of averaging. The midpoint averages the coordinates; subtraction gives distance components, not the midpoint.
- Averaging only one coordinate. Do both $x$ and $y$.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The midpoint of segment $AB$ is $(5, k)$. Point $A$ is at $(2, 3)$ and point $B$ is at $(8, 9)$. What is the value of $k$?
- A$3$
- B$\sqrt{(8-2)^2+(9-3)^2}$
- C$12$
- D$6$
Show solution
Answer: D, $6$. Midpoint $= \left(\frac{2+8}{2}, \frac{3+9}{2}\right) = (5, 6)$. So $k = 6$.
Medium
The midpoint of segment $AB$ is $(k, 2k)$. Point $A$ is at $(2, 4)$ and point $B$ is at $(6, b)$., what is the value of $k$?
- A$2$
- B$\sqrt{16+something}$
- C$8$
- D$4$
Show solution
Answer: D, $4$. $x$-midpoint: $\frac{2+6}{2} = 4 = k$. So $k = 4$.
Hard
Point $M$ is the midpoint of $AB$. $A$ is at $(3k, k)$ and $B$ is at $(k, 3k)$. The midpoint $M$ lies on the line $y = x + 2$. What is the value of $k$?
- A$-2$
- B$4$
- C$\sqrt{8}$
- D$2$
Show solution
Answer: D, $2$. Midpoint $= (2k, 3k)$. On $y = x + 2$: $3k = 2k + 2$, giving $k = 2$.
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