Question

pythagorean distance
what is the distance c?
points: ( 8, -6 ) ( -5, 3 )
a = δx =13
b = δy = 9
c = ?
pythagorean formula c² = a² + b²

Explanation:

Step1: Recall Pythagorean Formula

The formula for the distance (hypotenuse \( c \)) in a right triangle is \( c^{2}=a^{2}+b^{2} \), where \( a = 13 \) and \( b = 9 \).

Step2: Calculate \( a^{2} \) and \( b^{2} \)

First, find \( a^{2} \): \( a^{2}=13^{2}=169 \). Then, find \( b^{2} \): \( b^{2}=9^{2} = 81 \).

Step3: Sum \( a^{2} \) and \( b^{2} \)

Add the two results: \( a^{2}+b^{2}=169 + 81=250 \).

Step4: Find \( c \)

Take the square root of the sum: \( c=\sqrt{250}\approx15.8 \). But wait, let's check the calculation again. Wait, \( \Delta x \) is \( 8-(-5)=13 \), correct. \( \Delta y \) is \( 3 - (-6)=9 \), correct. Then \( c^{2}=13^{2}+9^{2}=169 + 81 = 250 \), so \( c=\sqrt{250}\approx15.8 \). But looking at the options, \( \sqrt{250}\approx15.8 \) is one of the options (the last one in the bottom right: \( \sqrt{250}\approx15.8 \)). Wait, but maybe I made a mistake? Wait, no, let's recalculate \( \Delta x \): \( 8 - (-5)=13 \), \( \Delta y \): \( 3-(-6)=9 \). Then \( c=\sqrt{(13)^{2}+(9)^{2}}=\sqrt{169 + 81}=\sqrt{250}\approx15.8 \).

Wait, but the options have \( \sqrt{250}\approx15.8 \) as the last option in the bottom row (rightmost). So that's the correct one.

Answer:

\( \sqrt{250}\approx15.8 \) (the option with \( \sqrt{250}\approx15.8 \))