ACT Math: Distances
Distances is a Coordinate Geometry topic on the ACT that builds directly on the Pythagorean theorem, and it makes up a meaningful share of the coordinate geometry questions you'll see. This free, complete lesson covers the distance formula, the midpoint formula, finding a missing endpoint when you know the midpoint, and applications like finding a triangle's perimeter or checking whether it's a right triangle on the coordinate plane. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. The Distance Formula
The distance formula finds the straight-line distance between two points, and it comes directly from the Pythagorean theorem, the horizontal and vertical differences between the points form the two legs of a right triangle, and the distance is the hypotenuse.
d = √((x₂ − x₁)² + (y₂ − y₁)²)
The distance is the hypotenuse of a right triangle
Worked Example: Find the distance between (2, 3) and (6, 7).
d = √((6 − 2)² + (7 − 3)²) = √(16 + 16) = √32 = 4√2
1. Find the distance between (1, 2) and (4, 6).
Explanation: d = √((4 − 1)² + (6 − 2)²) = √(9 + 16) = √25 = 5.
2. Find the distance between (−3, 5) and (2, −7).
Explanation: d = √((2 − (−3))² + (−7 − 5)²) = √(25 + 144) = √169 = 13.
3. Find the distance between (0, 0) and (8, 15).
Explanation: d = √(8² + 15²) = √(64 + 225) = √289 = 17.
4. Find the distance between (4, −2) and (4, 7).
Explanation: Both points share x = 4, so the distance is simply the difference in y-values: |7 − (−2)| = 9.
5. Find the distance between (−1, −1) and (2, 3).
Explanation: d = √((2 − (−1))² + (3 − (−1))²) = √(9 + 16) = √25 = 5.
2. The Midpoint Formula
The midpoint formula finds the point exactly halfway between two endpoints, by averaging the x-coordinates and averaging the y-coordinates.
M = (x₁ + x₂2, y₁ + y₂2)
Worked Example: Find the midpoint of the segment connecting (2, 3) and (8, 11).
M = (2 + 82, 3 + 112) = (5, 7)
1. Find the midpoint of the segment connecting (4, 6) and (10, 2).
Explanation: M = (4 + 102, 6 + 22) = (7, 4).
2. Find the midpoint of the segment connecting (−5, 3) and (7, −9).
Explanation: M = (−5 + 72, 3 + (−9)2) = (1, −3).
3. Find the midpoint of the segment connecting (0, 0) and (10, 14).
Explanation: M = (102, 142) = (5, 7).
4. On a city grid, Janice's office is at (33, 7) and Mark's office is at (15, 5). What is the midpoint where they could meet?
Explanation: M = (33 + 152, 7 + 52) = (24, 6).
5. A line segment has midpoint (4, −2). If one endpoint is (1, 3), what is the y-coordinate of the other endpoint?
Explanation: 3 + y2 = −2, so 3 + y = −4, giving y = −7.
3. Finding a Missing Endpoint
If you know the midpoint and one endpoint, you can work backward through the midpoint formula to find the other endpoint.
Worked Example: A segment has midpoint (5, 3) and one endpoint (2, 1). Find the other endpoint.
2 + x2 = 5, so x = 8 1 + y2 = 3, so y = 5
The other endpoint is (8, 5).
1. A segment has midpoint (6, 4) and one endpoint (2, 7). Find the other endpoint.
Explanation: 2 + x2 = 6, so x = 10. 7 + y2 = 4, so y = 1. Other endpoint: (10, 1).
2. A segment has midpoint (−2, 5) and one endpoint (3, −1). Find the other endpoint.
Explanation: 3 + x2 = −2, so x = −7. −1 + y2 = 5, so y = 11. Other endpoint: (−7, 11).
3. A segment has midpoint (0, 0) and one endpoint (6, −8). Find the other endpoint.
Explanation: When the midpoint is the origin, the other endpoint is simply the negative of the given point: (−6, 8).
4. A segment has midpoint (7, 2) and one endpoint (10, 6). Find the other endpoint.
Explanation: 10 + x2 = 7, so x = 4. 6 + y2 = 2, so y = −2. Other endpoint: (4, −2).
4. Applications
Combine the distance formula with the Pythagorean theorem's converse to classify triangles, or find perimeters, directly from coordinates.
Worked Example: A triangle has vertices A(0, 0), B(6, 0), C(0, 8). Find its perimeter.
AB = 6 (along the x-axis). AC = 8 (along the y-axis).
BC = √((6 − 0)² + (0 − 8)²) = √(36 + 64) = √100 = 10
Perimeter = 6 + 8 + 10 = 24.
1. A triangle has vertices A(0, 0), B(3, 0), C(0, 4). What is its perimeter?
Explanation: AB = 3, AC = 4, and BC = √(9 + 16) = 5. Perimeter: 3 + 4 + 5 = 12.
2. Is the triangle with vertices A(1, 1), B(4, 1), C(1, 5) a right triangle?
Explanation: AB = 3, AC = 4, and BC = √((4−1)²+(1−5)²) = √(9+16) = 5. Since 3² + 4² = 9 + 16 = 25 = 5², the converse of the Pythagorean theorem confirms this is a right triangle.
3. Two locations on a map are at (0, 0) and (9, 12), where each unit represents 1 mile. How far apart are the two locations?
Explanation: d = √(9² + 12²) = √(81 + 144) = √225 = 15 miles.
4. A circle has its center at (3, 4) and passes through the point (7, 1). What is the radius of the circle?
Explanation: The radius is the distance from the center to any point on the circle: √((7−3)²+(1−4)²) = √(16+9) = √25 = 5.
Common Mistakes to Avoid
- Forgetting to square the differences in the distance formula. Without squaring, a negative difference could cancel out a positive one, producing an incorrect result. Squaring also removes any concern about which point you subtract first.
- Confusing the distance formula with the midpoint formula. The distance formula produces a single number (a length); the midpoint formula produces a point (a pair of coordinates). Mixing them up is a common but easy-to-catch error, just check what kind of answer the question expects.
- Solving a missing-endpoint problem by simply averaging the wrong pair of points. Set up "known endpoint + unknown endpoint, divided by 2, equals midpoint" for each coordinate separately, then solve for the unknown.
- Assuming two segments meeting at a corner are perpendicular just because they look that way in a sketch. Confirm with actual slopes or the Pythagorean theorem before concluding a right angle exists.
- Leaving a distance answer in unsimplified radical form when the answer choices are simplified (or vice versa). Double-check whether the given values form a perfect square under the radical, like √25 = 5, before assuming the answer must include a radical.
Frequently Asked Questions
Do I need to memorize the distance and midpoint formulas separately, or can I derive one from the other?
They're related but distinct: the distance formula comes from the Pythagorean theorem, while the midpoint formula is just an average of coordinates. Since the ACT provides no formula sheet, it's worth memorizing both directly rather than trying to re-derive them under time pressure.
Is there a shortcut for recognizing distances without doing the full calculation?
Recognizing common Pythagorean triples, like 3-4-5, 5-12-13, and 8-15-17 (and their multiples, like 6-8-10), lets you spot the distance instantly once you see the horizontal and vertical differences, without working through the square root.
How does the distance formula connect to checking whether a triangle is a right triangle?
Once you know all three side lengths (using the distance formula for any sides that aren't purely horizontal or vertical), the converse of the Pythagorean theorem says that if the square of the longest side equals the sum of the squares of the other two, the triangle must be a right triangle.
Where can I practice more problems like these?
The Distances quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.