# Find the perimeter of the triangle with the vertices at (2, 2), (-2, 2), and (-2, -3). I keep getting 18 but my online homework website says it is wrong. Find the point (0,b) on the y-axis that is equidistant from the points (5,5) and (4,−3)?

Sep 1, 2015

The perimeter is $9 + \sqrt{41}$, or approximately 15.4.

#### Explanation:

The perimeter of the triangle is equal to the sum of the lengths of the 3 sides.

The 3 sides are the segments from:
$\left(2 , 2\right) \to \left(- 2 , 2\right)$
$\left(- 2 , 2\right) \to \left(- 2 , - 3\right)$
$\left(- 2 , - 3\right) \to \left(2 , 2\right)$

The side from (2, 2) to (-2, 2) is going to be a horizontal line from -2 to 2 on the x-axis. In other words, this side has a length of 4 $\left(2 - \left(- 2\right)\right)$.

The side from (-2, 2) to (-2, -3) is going to be a vertical line (because both x-coordinates are the same) from -3 to 2 on the y-axis. In other words, this side has a length of 5 $\left(2 - \left(- 3\right)\right)$.

The side from (-2, -3) to (2, 2) is not horizontal or vertical. We're going to have to use the pythagorean theorem for this side. The x-value changes by 4 (-2 to 2), and the y-value changes by 5 (-3 to 2). So, this side's length is equal to $\sqrt{{4}^{2} + {5}^{2}}$

$\sqrt{{4}^{2} + {5}^{2}} = \sqrt{16 + 25} = \sqrt{41}$ (this is a prime number so this is as simple as you can make it)

Therefore, the perimeter of the triangle is 4 + 5 + $\sqrt{41}$, or $9 + \sqrt{41}$