Distance and Midpoint Formulas

Working with coordinates in the plane

Distance and Midpoint Formulas

Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ :

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Derivation: This comes from the Pythagorean Theorem!

Midpoint Formula

The midpoint between two points $(x_1, y_1)$ and $(x_2, y_2)$ :

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Memory aid: Average the x-coordinates, average the y-coordinates.

Applications

Perimeter: Add distances between consecutive vertices

Proving shapes:

Square: All 4 sides equal, diagonals equal
Rectangle: Opposite sides equal, diagonals equal
Rhombus: All 4 sides equal
Isosceles triangle: Two sides equal

Segment Partitioning

To find a point that divides a segment in ratio $m:n$ :

Use weighted average based on the ratio

Coordinate Proof Strategy

Place figure on coordinate plane strategically
Use distance/midpoint formulas
Show required properties

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the distance between points $(3, 4)$ and $(7, 1)$ .

💡 Show Solution

Use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(7 - 3)^2 + (1 - 4)^2}$

$d = \sqrt{4^2 + (-3)^2}$

$d = \sqrt{16 + 9}$

$d = \sqrt{25} = 5$

Answer: 5 units

2Problem 2medium

❓ Question:

Find the midpoint of the segment connecting $(-2, 5)$ and $(6, -3)$ .

💡 Show Solution

Use the midpoint formula: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

$M = \left(\frac{-2 + 6}{2}, \frac{5 + (-3)}{2}\right)$

$M = \left(\frac{4}{2}, \frac{2}{2}\right)$

$M = (2, 1)$

Answer: $(2, 1)$

3Problem 3hard

❓ Question:

Prove that the triangle with vertices $A(0, 0)$ , $B(5, 0)$ , and $C(5, 12)$ is a right triangle.

💡 Show Solution

Strategy: Show that the sides satisfy the Pythagorean Theorem.

Find all three side lengths:

$AB$ : from $(0,0)$ to $(5,0)$ $AB = \sqrt{(5-0)^2 + (0-0)^2} = 5$

$BC$ : from $(5,0)$ to $(5,12)$ $BC = \sqrt{(5-5)^2 + (12-0)^2} = 12$

$AC$ : from $(0,0)$ to $(5,12)$ $AC = \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Check Pythagorean Theorem: $AB^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169$ $AC^2 = 13^2 = 169$

Since $AB^2 + BC^2 = AC^2$ , the triangle is a right triangle.

Answer: Yes, it's a right triangle (in fact, a 5-12-13 right triangle)

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