Question

graph a right triangle with the two points forming the hypotenuse. using the sides, find the distance between the two points in simplest radical form.
(8, 6) and (5, 2)
click twice to draw a line. click a segment to erase it.

Explanation:

Step1: Find horizontal and vertical distances

For points \((8,6)\) and \((5,2)\), horizontal distance (Δx) is \(|8 - 5| = 3\), vertical distance (Δy) is \(|6 - 2| = 4\).

Step2: Apply Pythagorean theorem

Let distance be \(d\). Then \(d^2 = 3^2 + 4^2 = 9 + 16 = 25\)? Wait, no, \(3^2 + 4^2 = 9 + 16 = 25\)? Wait, no, wait: \(3^2=9\), \(4^2 = 16\), sum is \(25\), so \(d=\sqrt{25}\)? Wait, no, wait, 8-5 is 3, 6-2 is 4. Then \(d = \sqrt{(8 - 5)^2 + (6 - 2)^2}=\sqrt{3^2 + 4^2}=\sqrt{9 + 16}=\sqrt{25}\)? Wait, no, \(\sqrt{25}=5\)? Wait, but 3-4-5 triangle. Wait, yes. Wait, but let's recalculate: \((8 - 5)=3\), \((6 - 2)=4\). So \(d=\sqrt{3^2 + 4^2}=\sqrt{9 + 16}=\sqrt{25}=5\)? Wait, but that's a perfect square. Wait, maybe I made a mistake? Wait, 8-5 is 3, 6-2 is 4. So the horizontal leg is 3 units, vertical leg is 4 units. Then hypotenuse is 5. So distance is 5. Wait, but the problem says "simplest radical form". But 5 is an integer, which is a radical form (since \(\sqrt{25}=5\)). So the distance is 5.

Answer:

\(5\)