3. Distance Formula: How Far Apart Are Two Points?

The distance between two points measures the straight-line length connecting them. Use the distance formula:

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

This formula comes from the Pythagorean theorem: the two points form a right triangle, and the distance is the hypotenuse.

Worked Example 1: Find the distance between \((1, 2)\) and \((4, 6)\).

\[ d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Worked Example 2: Find the distance between \((-1, 3)\) and \((2, -1)\).

\[ d = \sqrt{(2-(-1))^2 + (-1-3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Notice: negative numbers become positive when squared, so the signs do not matter.

Real-world context: A surveyor marks two property corners at \((10, 20)\) and \((34, 56)\) on a grid (in feet). The distance is:
\[ d = \sqrt{(34-10)^2 + (56-20)^2} = \sqrt{24^2 + 36^2} = \sqrt{576 + 1296} = \sqrt{1872} \approx 43.3 \text{ feet} \]

Common mistake: forgetting to square root. The value under the radical is not the distance by itself.

💡 Tip: If the two points form a 3-4-5 or 5-12-13 right triangle, the distance is that Pythagorean triple — no calculator needed.

Deriving the Distance Formula

The distance formula is not a separate fact from geometry. It is the Pythagorean theorem written in coordinate language. Given \(A(x_1,y_1)\) and \(B(x_2,y_2)\), draw a horizontal segment from A and a vertical segment to B. These segments form the legs of a right triangle. Their lengths are \(|x_2-x_1|\) and \(|y_2-y_1|\). If the straight-line distance \(AB\) is \(d\), then
\[ d^2=|x_2-x_1|^2+|y_2-y_1|^2. \]
Squaring makes the absolute-value bars unnecessary because \(|a|^2=a^2\). Taking the nonnegative square root gives
\[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \]
The formula therefore records three geometric moves: find horizontal change, find vertical change, and combine perpendicular lengths with the Pythagorean theorem.

A Structured Calculation Method

Use a table before substituting:
\[ \Delta x=x_2-x_1,\qquad \Delta y=y_2-y_1, \qquad d=\sqrt{(\Delta x)^2+(\Delta y)^2}. \]
This separates sign work from squaring and reduces errors.

Worked example: Find the distance between \(A(-4,7)\) and \(B(8,-2)\).
\[ \Delta x=8-(-4)=12,\qquad \Delta y=-2-7=-9. \]
Then
\[ d=\sqrt{12^2+(-9)^2}=\sqrt{144+81}=\sqrt{225}=15. \]
A sketch confirms that the horizontal and vertical magnitudes are 12 and 9. The straight segment must be longer than either leg but shorter than their sum: \(12<15<21\). This bound is a fast reasonableness check.

Point order does not change distance. Reversing the endpoints changes \((12,-9)\) to \((-12,9)\), but the squares are unchanged. However, use one order consistently within a calculation. Mixing \(x_2-x_1\) with \(y_1-y_2\) happens to give the same distance after squaring, but that inconsistent habit causes slope errors later.

Exact and Approximate Forms

Some distances are whole numbers; others are irrational. For \((1,2)\) and \((6,6)\), the changes are 5 and 4, so
\[ d=\sqrt{5^2+4^2}=\sqrt{41}. \]
Because 41 has no perfect-square factor greater than 1, \(\sqrt{41}\) is the exact answer. A calculator gives approximately \(6.4\). Keep the radical during the work and round only once at the end. Premature rounding can distort later perimeter or area calculations.

To simplify \(\sqrt{72}\), factor out the largest perfect square:
\[ \sqrt{72}=\sqrt{36\cdot2}=6\sqrt2. \]
On a GED® question, the requested form and answer choices determine whether exact radical form or a decimal approximation is most useful.

Pythagorean Triples and Scale

Recognize common right-triangle patterns:
\[ (3,4,5),\quad(5,12,13),\quad(8,15,17),\quad(7,24,25). \]
Multiples also work. Changes of 6 and 8 produce distance 10; changes of 9 and 12 produce distance 15. Recognition is not a substitute for setup, but it speeds calculation and provides a check.

If a diagram has scale, first calculate in coordinate units and then convert. Suppose two map locations differ by 6 horizontal units and 8 vertical units, with 2.5 kilometers per grid unit. Their grid distance is 10 units, so their actual straight-line distance is \(10(2.5)=25\) kilometers. Apply the linear scale once to the final length.

Comparing Route Distance and Straight-Line Distance

The distance formula measures the direct segment between points, sometimes called Euclidean distance. A route constrained to horizontal and vertical streets may be longer. For changes of 6 and 8, the direct distance is 10, while the grid-route distance is \(|6|+|8|=14\). Read the context carefully: “straight-line,” “direct,” or “as the crow flies” indicates the distance formula; movement along rectangular blocks indicates addition of horizontal and vertical distances.

The triangle inequality explains why the direct route cannot exceed the two-leg route:
\[ \sqrt{(\Delta x)^2+(\Delta y)^2}\le |\Delta x|+|\Delta y|. \]
You need not prove this for the GED®, but it is a useful check.

Using Distance to Test Geometry

Distance can verify equal sides. For points \(A(0,0)\), \(B(4,3)\), and \(C(8,0)\),
\[ AB=\sqrt{4^2+3^2}=5, \qquad BC=\sqrt{4^2+(-3)^2}=5. \]
Thus triangle ABC is isosceles. Also, \(AC=8\). Comparing squared lengths can avoid radicals: \(AB^2=25\), \(BC^2=25\), and \(AC^2=64\). Equal squared lengths imply equal lengths.

Squared distances also test right triangles. If the largest squared side equals the sum of the other two squared sides, the angle opposite the largest side is right. This connects coordinate geometry back to the converse of the Pythagorean theorem.

Error Analysis

• Adding before squaring: \((\Delta x+\Delta y)^2\) is not the formula. The legs are squared separately and then added.
• Dropping a negative coordinate: \(3-(-5)=8\), not \(-2\). Use parentheses.
• Forgetting the square root: The sum of squares is \(d^2\), not \(d\).
• Adding absolute changes for a direct distance: That gives a grid route, not the hypotenuse.
• Rounding intermediate values: Preserve exact squares and radicals until the final step.
• Trusting visual scale: Derive changes from coordinates even if the segment looks longer or shorter.

Mastery Check

Find the distance between \((-5,-1)\) and \((7,4)\). The changes are 12 and 5, so
\[ d=\sqrt{12^2+5^2}=\sqrt{169}=13. \]
Then explain why 17 is a tempting wrong answer: \(12+5=17\) is the horizontal-plus-vertical route. The direct distance is the hypotenuse, 13.