Finding Distance Between Two Points

Find the distance between $(2, 3)$ and $(6, 6)$.

Identify coordinates: $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (6, 6)$ = Points identified

Find horizontal distance: $x_2 - x_1 = 6 - 2 = 4$ = Horizontal: $4$

Find vertical distance: $y_2 - y_1 = 6 - 3 = 3$ = Vertical: $3$

Apply the formula: $d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25}$ = $d = 5$

Answer: The distance is $5$ units.

Distance with Negative Coordinates

Find the distance between $(-3, 2)$ and $(1, -1)$.

Identify coordinates: $(x_1, y_1) = (-3, 2)$ and $(x_2, y_2) = (1, -1)$ = Points identified

Calculate $x_2 - x_1$: $1 - (-3) = 1 + 3 = 4$ = Horizontal: $4$

Calculate $y_2 - y_1$: $-1 - 2 = -3$ = Vertical: $-3$

Apply the formula: $d = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25}$ = $d = 5$

Answer: The distance is $5$ units. Note: squaring makes negative values positive.

Distance with Irrational Result

Find the distance between $(0, 0)$ and $(3, 5)$.

Identify coordinates: $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (3, 5)$ = Origin to point

Calculate differences: $x_2 - x_1 = 3$, $y_2 - y_1 = 5$ = Horizontal: $3$, Vertical: $5$

Apply the formula: $d = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$ = $d = \sqrt{34}$

Approximate if needed: $\sqrt{34} \approx 5.83$ = $d \approx 5.83$ units

Answer: The exact distance is $\sqrt{34}$ units, or approximately $5.83$ units.

Mistake: Forgetting to square the differences before adding

Why: Students sometimes write $\sqrt{(x_2-x_1) + (y_2-y_1)}$ instead of $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Correct: Always square each difference: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Mistake: Taking the square root of each term separately

Why: $\sqrt{a^2 + b^2} \neq \sqrt{a^2} + \sqrt{b^2}$. The square root applies to the entire sum.

Correct: Add the squared terms first, then take one square root of the total.

Mistake: Getting confused with negative coordinates

Why: Subtracting a negative number means adding: $3 - (-2) = 3 + 2 = 5$

Correct: Be careful with signs. Squaring will always give a positive result anyway.

GPS Navigation

GPS devices use coordinate systems to calculate distances between locations. While Earth's coordinates are more complex (latitude/longitude on a sphere), the fundamental principle is the same.

A GPS calculates you are at coordinates $(3, 7)$ km and your destination is at $(9, 15)$ km. The straight-line distance is $\sqrt{(9-3)^2 + (15-7)^2} = \sqrt{36 + 64} = 10$ km.

Video Game Design

Game developers use the distance formula constantly to calculate collision detection, enemy AI range, and movement physics.

In a game, your character is at $(50, 30)$ pixels and an enemy is at $(80, 70)$. The distance is $\sqrt{(80-50)^2 + (70-30)^2} = \sqrt{900 + 1600} = 50$ pixels.

Q: Does it matter which point is $(x_1, y_1)$ and which is $(x_2, y_2)$?

A: No! Since we square the differences, the order does not matter. $(5-2)^2 = 9$ and $(2-5)^2 = 9$ give the same result.

Q: Why do we square and then take a square root?

A: Squaring converts negative differences to positive values and sets up the Pythagorean theorem calculation. The square root then gives us the actual distance (the hypotenuse length).

Q: What if my answer has a square root I cannot simplify?

A: That is perfectly fine! Exact answers like $\sqrt{34}$ are often preferred. You can also give a decimal approximation if needed.