Question

A snail is left at the point $(2,2)$ on the coordinate plane. The snail moves $5$ units toward the east and then $6$ units toward the south. Find the shortest distance between the initial and the final position of the snail on the coordinate plane.

• A. $\sqrt{61}$ units
• B. $\sqrt{41}$ units
• C. $\sqrt{71}$ units
• D. $\sqrt{65}$ units

Answer 1

The correct option is A $\sqrt{61}$ units

The snail starts from the point $P(2,2),$ then moves $5$ units east, and then finally $6$ units south.

The movement of the snail is shown in the figure given below



So, we can observe that the snail starts at the point $(2,2)$ and then finally crawls to the point $(7,-4).$

The shortest distance between the initial and the final positions of the snail $=$ Distance between the points $(2,2)$ and $(7,-4)$.

To find the distance between $(2,2)$ and $(7,-4):$
Plot the points in a coordinate plane. Then, draw a right triangle with a hypotenuse that represents the distance between the points.

Use the Pythagorean theorem to find the length of the hypotenuse.

$a^2+b^2=c^2$ (Write the Pythagorean theorem)
$5^2+6^2=c^2$ (Substitute $5$ for a and $6$ for $b)$
$25+36=c^2$ (Evaluate the powers)
$61=c^2$ (Take the positive square root of each side)
$\Rightarrow$ The shortest distance between the initial and final positions of the snail is $\sqrt{61}$ units.

$\Rightarrow$ Option C is correct.