# Find Shortest Distance

• September 27th 2008, 09:57 PM
magentarita
Find Shortest Distance
To get from his high school to his home, Jamal travels 5.0 miles eastand then 4.0 miles north. When Sheila goes to her home from the same high school, she travels 8.0 miles east and 2.0 miles south. What is the measure of the shortest distance, to the nearest tenth of a mile, between Jamal’s home and Sheila’s home?

I sketched 2 right triangles. I sketched a map with east, west, north and south and was not able to find the answer.
• September 27th 2008, 10:15 PM
11rdc11
$\sqrt{45}$
• September 27th 2008, 10:56 PM
Chop Suey
Think of this in the coordinate plane. Assume that school is on (0,0). Then Jamal travels east from (0,0) to (5,0), and then from (5,0) to (5,4). P(5, 4) is his home. Now find the point where Sheila lives, and use the distance formula to find the distance between the two points:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
• September 28th 2008, 06:47 AM
magentarita
Can you.....
Quote:

Originally Posted by Chop Suey
Think of this in the coordinate plane. Assume that school is on (0,0). Then Jamal travels east from (0,0) to (5,0), and then from (5,0) to (5,4). P(5, 4) is his home. Now find the point where Sheila lives, and use the distance formula to find the distance between the two points:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Which coordinates go into the formula?

Can you place the right coordinates into the given formula and I'll take it from there?

Thanks
• September 28th 2008, 06:48 AM
magentarita
Yes...
Quote:

Originally Posted by 11rdc11
$\sqrt{45}$

Yes, the answer is sqrt{45} but how did you get it?
• September 28th 2008, 07:23 AM
Chop Suey
Quote:

Originally Posted by magentarita
Can you place the right coordinates into the given formula and I'll take it from there?

No. You should know that a point in a 2D coordinate plane is in the form of (x, y)

Here's a different approach.
• September 28th 2008, 08:30 AM
11rdc11
Quote:

Originally Posted by magentarita
Yes, the answer is sqrt{45} but how did you get it?

I solved it the way chop suey told you.
• September 28th 2008, 11:59 AM
Soroban
Hello, magentarita!

Quote:

To get from his high school to his home, Jamal travels 5 miles east and 4 miles north.
When Sheila goes to her home from the same high school,
she travels 8 miles east and 2 miles south.
What is the measure of the shortest distance, to the nearest tenth of a mile,
between Jamal’s home and Sheila’s home?

Code:

                J                 *                 |                 | 4       : -  5  - |     S *---------*-------*       : - - - 8 - - - - |                         | 2                         *                         S
The school is at $S.$

Jamal travels 5 miles east and 4 miles north at arrives at his home at $J.$

Sheila travels 8 miles east and 2 miles south and arrives at her home at $S.$

We want the distance from $J$ to $S$ (the length of line segment $JS$).

Re-draw the diagram.
Code:

                J                 *                 |               4 | *                 |     S * - - - - *  *                 |               2 |    *                 |                 * - - - *                     3    S

See that right triangle?

We have: . $JS^2 \:=\:3^2 + 6^2 \:=\:45$

Therefore: . $JS \;=\;\sqrt{45} \;\approx\;6.7$ miles.

• September 28th 2008, 09:24 PM
magentarita
Thanks
Quote:

Originally Posted by Soroban
Hello, magentarita!

Code:

                J                 *                 |                 | 4       : -  5  - |     S *---------*-------*       : - - - 8 - - - - |                         | 2                         *                         S
The school is at $S.$

Jamal travels 5 miles east and 4 miles north at arrives at his home at $J.$

Sheila travels 8 miles east and 2 miles south and arrives at her home at $S.$

We want the distance from $J$ to $S$ (the length of line segment $JS$).

Re-draw the diagram.
Code:

                J                 *                 |               4 | *                 |     S * - - - - *  *                 |               2 |    *                 |                 * - - - *                     3    S
See that right triangle?

We have: . $JS^2 \:=\:3^2 + 6^2 \:=\:45$

Therefore: . $JS \;=\;\sqrt{45} \;\approx\;6.7$ miles.

Thank you, Soroban. I would like you to reply more often to my questions.