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 never gives a negative result, so a sign slip in the difference only matters if the number itself is wrong.

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.