Distance and Midpoint
📖 Learn
Coordinate geometry connects algebra and geometry by placing geometric figures on a coordinate plane. Two fundamental tools in coordinate geometry are the distance formula and the midpoint formula.
The Distance Formula
The distance d between two points (x₁, y₁) and (x₂, y₂) is:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
This formula is derived from the Pythagorean theorem. The horizontal distance |x₂ - x₁| and vertical distance |y₂ - y₁| form the legs of a right triangle, and the distance between the points is the hypotenuse.
- It doesn't matter which point you call (x₁, y₁) and which you call (x₂, y₂)
- The differences are squared, so negative values become positive
- The result is always positive (distance is never negative)
- When points share a coordinate (same x or same y), the formula simplifies to the absolute difference of the other coordinate
The Midpoint Formula
The midpoint M of a segment with endpoints (x₁, y₁) and (x₂, y₂) is:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
The midpoint is simply the average of the x-coordinates and the average of the y-coordinates.
Key Applications
| Application | Formula Used | Purpose |
|---|---|---|
| Finding segment length | Distance | Determine the length of a line segment |
| Checking if triangle is isosceles | Distance | Compare side lengths |
| Finding the center of a segment | Midpoint | Locate the exact middle point |
| Verifying parallelograms | Both | Diagonals bisect each other (same midpoint) |
| Finding an endpoint | Midpoint (reversed) | Given midpoint and one endpoint, find the other |
If M = (mx, my) is the midpoint of segment AB, and A = (ax, ay) is known, then B = (bx, by) can be found:
- bx = 2mx - ax (solve: (ax + bx)/2 = mx)
- by = 2my - ay (solve: (ay + by)/2 = my)
💡 Examples
Example 1: Finding Distance Between Two Points
Problem: Find the distance between A(2, 3) and B(8, 11).
Solution:
Step 1: Identify coordinates: (x₁, y₁) = (2, 3) and (x₂, y₂) = (8, 11)
Step 2: Apply the distance formula
d = √[(8 - 2)² + (11 - 3)²]
d = √[(6)² + (8)²]
d = √[36 + 64]
d = √100
Answer: d = 10 units
Example 2: Finding the Midpoint
Problem: Find the midpoint of the segment with endpoints P(-4, 7) and Q(6, -3).
Solution:
Step 1: Identify coordinates: (x₁, y₁) = (-4, 7) and (x₂, y₂) = (6, -3)
Step 2: Apply the midpoint formula
M = ((-4 + 6)/2, (7 + (-3))/2)
M = (2/2, 4/2)
Answer: M = (1, 2)
Example 3: Finding an Unknown Endpoint
Problem: The midpoint of segment AB is M(5, -2). If A = (3, 4), find the coordinates of B.
Solution:
Step 1: Set up equations using the midpoint formula
(3 + bx)/2 = 5 and (4 + by)/2 = -2
Step 2: Solve for bx
3 + bx = 10
bx = 7
Step 3: Solve for by
4 + by = -4
by = -8
Answer: B = (7, -8)
Example 4: Classifying a Triangle
Problem: Triangle ABC has vertices A(0, 0), B(4, 0), and C(2, 3). Is it isosceles?
Solution:
Step 1: Find AB
AB = √[(4-0)² + (0-0)²] = √16 = 4
Step 2: Find BC
BC = √[(2-4)² + (3-0)²] = √[4 + 9] = √13
Step 3: Find AC
AC = √[(2-0)² + (3-0)²] = √[4 + 9] = √13
Step 4: Compare side lengths
BC = AC = √13, so two sides are equal
Answer: Yes, triangle ABC is isosceles (BC = AC)
Example 5: Verifying a Rectangle
Problem: Show that the diagonals of rectangle ABCD with A(0,0), B(6,0), C(6,4), D(0,4) bisect each other.
Solution:
Step 1: Find the midpoint of diagonal AC
M₁ = ((0+6)/2, (0+4)/2) = (3, 2)
Step 2: Find the midpoint of diagonal BD
M₂ = ((6+0)/2, (0+4)/2) = (3, 2)
Step 3: Compare midpoints
M₁ = M₂ = (3, 2)
Answer: The diagonals have the same midpoint (3, 2), so they bisect each other.
✏️ Practice
Try these problems on your own. Choose the best answer for each question.
1. What is the distance between (1, 2) and (4, 6)?
A) 5
B) 7
C) √25
D) Both A and C
2. Find the midpoint of the segment from (-2, 5) to (8, -3).
A) (3, 1)
B) (6, 2)
C) (5, 4)
D) (3, 4)
3. The midpoint of AB is (4, -1). If A = (2, 3), what are the coordinates of B?
A) (6, -5)
B) (3, 1)
C) (6, 5)
D) (0, -5)
4. What is the distance from the origin to (5, 12)?
A) 17
B) 13
C) √17
D) 169
5. Points A(-3, 2) and B(1, 2) are connected. What is the length of AB?
A) 0
B) 2
C) 4
D) √20
6. Find the distance between (-1, -4) and (2, 0).
A) 5
B) √7
C) 7
D) 25
7. If M(0, 0) is the midpoint of segment PQ and P = (-5, 8), what is Q?
A) (5, -8)
B) (-5, 8)
C) (5, 8)
D) (-2.5, 4)
8. Which point is equidistant from A(0, 0) and B(6, 0)?
A) (3, 0)
B) (3, 5)
C) (3, -2)
D) All of the above
9. The distance from (a, 3) to (2, 7) is 5. If a is positive, what is a?
A) 5
B) 2
C) -1
D) 7
10. Triangle PQR has P(1, 1), Q(5, 1), R(3, 4). What is the perimeter?
A) 4 + 2√13
B) 4 + √13
C) 8 + √13
D) 4 + 2√5
Click to reveal answers
- D) Both A and C — d = √[(4-1)² + (6-2)²] = √[9 + 16] = √25 = 5
- A) (3, 1) — M = ((-2+8)/2, (5+(-3))/2) = (6/2, 2/2) = (3, 1)
- A) (6, -5) — bx = 2(4) - 2 = 6; by = 2(-1) - 3 = -5
- B) 13 — d = √[(5-0)² + (12-0)²] = √[25 + 144] = √169 = 13 (5-12-13 Pythagorean triple)
- C) 4 — Points have same y-coordinate, so distance = |1 - (-3)| = 4
- A) 5 — d = √[(2-(-1))² + (0-(-4))²] = √[9 + 16] = √25 = 5 (3-4-5 triple)
- A) (5, -8) — qx = 2(0) - (-5) = 5; qy = 2(0) - 8 = -8
- D) All of the above — Any point with x = 3 is equidistant from A and B (lies on perpendicular bisector)
- A) 5 — 5 = √[(2-a)² + 16], so 25 = (2-a)² + 16, (2-a)² = 9, 2-a = ±3. If a is positive: a = 5 or a = -1, so a = 5
- A) 4 + 2√13 — PQ = 4; PR = √[(3-1)² + (4-1)²] = √13; QR = √[(3-5)² + (4-1)²] = √13
✅ Check Your Understanding
Question 1: How is the distance formula related to the Pythagorean theorem?
Reveal Answer
The distance formula is a direct application of the Pythagorean theorem. When you connect two points on a coordinate plane, you can form a right triangle where: the horizontal leg has length |x₂ - x₁|, the vertical leg has length |y₂ - y₁|, and the hypotenuse is the distance between the points. By the Pythagorean theorem, c² = a² + b², which gives us d² = (x₂ - x₁)² + (y₂ - y₁)², and taking the square root yields the distance formula.
Question 2: Why does the midpoint formula use addition and averaging instead of subtraction?
Reveal Answer
The midpoint represents the center point between two locations, which is found by averaging. Think of it like finding the average of two test scores - you add them and divide by 2. The x-coordinate of the midpoint is the average of the two x-coordinates, and the y-coordinate of the midpoint is the average of the two y-coordinates. Subtraction is used in the distance formula because we're finding how far apart values are, while addition/averaging is used in the midpoint formula because we're finding where the middle lies.
Question 3: If two points have the same y-coordinate, how does this simplify the distance formula?
Reveal Answer
When two points have the same y-coordinate (like (2, 5) and (7, 5)), the points lie on a horizontal line. The distance formula simplifies because (y₂ - y₁)² = 0² = 0. So d = √[(x₂ - x₁)² + 0] = √[(x₂ - x₁)²] = |x₂ - x₁|. The distance is simply the absolute value of the difference in x-coordinates. Similarly, if two points have the same x-coordinate, the distance equals |y₂ - y₁|.
Question 4: How can you check your answer when finding an unknown endpoint using the midpoint?
Reveal Answer
After finding the unknown endpoint, verify by applying the midpoint formula to both endpoints. For example, if you found B = (7, -8) given A = (3, 4) and M = (5, -2), check: ((3 + 7)/2, (4 + (-8))/2) = (10/2, -4/2) = (5, -2). This should equal the given midpoint. If it does, your answer is correct. This verification step catches arithmetic errors and confirms your work.
🚀 Next Steps
- Review any concepts that felt challenging
- Move on to the next lesson when ready
- Return to practice problems periodically for review