A man walks 5 miles North, 3 miles West, 8 miles South, and 7 miles East. What is his distance from his starting point?

sri

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- Mar 28th 2008, 09:23 AMsri340Help me this problem
A man walks 5 miles North, 3 miles West, 8 miles South, and 7 miles East. What is his distance from his starting point?

sri - Mar 28th 2008, 09:35 AMiknowone
Mark a point on a grid representing where he starts $\displaystyle (0,0)$. Draw a line 5 units up (north), 3 units left (west), 8 units down (south), 7 units right (east).

Now you have a starting point and an ending point. The distance between two points $\displaystyle (a,b)$ and $\displaystyle (c,d)$ is given by the formula $\displaystyle \sqrt{(c-a)^2+(d-b)^2}$. Can you solve this now?

You can also do it in your head if you want. 5-8 = -3, so he walked a net distance of 3 miles south. 7-3 = 4, so he walked a net distance of 4 miles east.

If he started at (0,0) he ended at (4,-3) is that what you arrived at? - Mar 28th 2008, 09:35 AMTheEmptySet
immagine this on a coordinate system

North is positive y south i negative y west is negative x and east is positive x.

is he starts at the origin (0,0) then

north 5 is (0,0)+(0,5)=(0,5)

west 3 is (-3,0)+(0,5)=(-3,5)

south 8 is (0,-8)+(-3,5)=(-3,-3)

east 7 is (7,0)+(-3,-3)=(4,-3)

so he is 5 miles from the origin becuase he is on a 3,4,5 right triangle

or we could use the distance formula

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

using the ending (4,-3) and starting (0,0) points we get...

$\displaystyle d=\sqrt{(4-0)^2+(-3-0)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

I hope this helps.