The Distance Formula
The distance between two points is the Pythagorean theorem applied to their coordinate differences.
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
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Square each difference separately, add, then take the square root.
How to solve one, step by step
Example: distance from $(1, 2)$ to $(4, 6)$.
- Differences: $3$ and $4$, so $\sqrt{3^2 + 4^2}$.
- $\sqrt{9 + 16} = \sqrt{25} = 5$.
The mistakes that cost points
- Adding the differences instead of squaring. Square each difference before adding, then root the sum.
- Forgetting to square each difference individually. Both the $x$ and $y$ differences get squared.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
Points $P(k, 0)$ and $Q(0, k)$ are in the coordinate plane. If the distance $PQ = 6\sqrt{2}$, what is the value of $k$? (Assume $k > 0$.)
- A$12$
- B$3\sqrt{2}$
- C$6\sqrt{2}$
- D$6$
Show solution
Answer: D, $6$. $PQ = \sqrt{(0-k)^2 + (k-0)^2} = \sqrt{k^2 + k^2} = \sqrt{2k^2} = k\sqrt{2} = 6\sqrt{2}$. So $k = 6$.
Medium
Point $A$ is at $(k, 2)$ and point $B$ is at $(3, k+4)$. If the distance $AB = \sqrt{45}$, what is the value of $k$? (Take the positive solution.)
- A$-2$
- B$4$
- C$6$
- D$8$
Show solution
Answer: C, $6$. $AB = \sqrt{(2-k)^2 + (k-2)^2} = \sqrt{2(k-2)^2} = |k-2|\sqrt{2} = 4\sqrt{2}$. So $|k-2| = 4$, giving $k = 6$ or $k = -2$. Taking the positive solution, $k = 6$.
Hard
Points $A(0, 0)$, $B(k, 0)$, and $C(0, k)$ form a right triangle. The hypotenuse $BC$ has length $k\sqrt{2}$. If the perimeter of the triangle is $24 + 8\sqrt{2}$, what is the value of $k$?
- A$3\sqrt{2}$
- B$6$
- C$12$
- D$6\sqrt{2}$
Show solution
Answer: B, $6$. Perimeter $= k + k + k\sqrt{2} = k(2 + \sqrt{2}) = 12 + 6\sqrt{2} = 6(2+\sqrt{2})$. So $k = 6$.
Find your exact gaps
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