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)?

Madison Costa

Madison Costa

Answered question

2022-11-02

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)?

Answer & Explanation

sliceu4i

sliceu4i

Beginner2022-11-03Added 16 answers

The perimeter of the triangle is equal to the sum of the lengths of the 3 sides.
The 3 sides are the segments from:
( 2 , 2 ) ( - 2 , 2 )
( - 2 , 2 ) ( - 2 , - 3 )
( - 2 , - 3 ) ( 2 , 2 )
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 (2−(−2)).
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 (2−(−3)).
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 4 2 + 5 2
4 2 + 5 2 = 16 + 25 = 41 (this is a prime number so this is as simple as you can make it)
Therefore, the perimeter of the triangle is 9 + 41

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