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Klio2033 [76]
3 years ago
11

Find the distance between the points (2,4) and (5,0). (PLEASE HELP!)

Mathematics
2 answers:
Inessa05 [86]3 years ago
5 0

Answer:

5

Step-by-step explanation:

Calculate the distance d using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (2, 4) and (x₂, y₂ ) = (5, 0)

d = \sqrt{(5-2)^2+(0-4)^2}

  = \sqrt{3^2+(-4)^2}

  = \sqrt{9+16} = \sqrt{25} = 5

frozen [14]3 years ago
3 0

The answer is:

5

Hoped I helped

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Assume that p is a relation that contains the points (3, -4). What other point must be included in the relation if p is:1. symme
AlladinOne [14]

The point needed for the relation is P(x,y) = (3, 4) if P and P' are symmetric about the x-axis.

The point needed for the relation is P(x,y) = (-3, -4) if P and P' are symmetric about the y-axis.

In this exercise we are supposed to determine the coordinates of a point P under an assumption of rigid transformation. Now, we must use the following <em>symmetry</em> transformations:

Reflection about the x-axis

S'(x,y) = S(x,y) - 2\cdot (0, s_{y}) (1)

Reflection about the y-axis

S'(x,y) = S(x,y) - 2\cdot (s_{x}, 0) (2)

Where:

  • S(x,y) - Original point.
  • S' (x,y) - Reflected point.
  • s_{x}, s_{y} - Coordinates of point S.

If we know that P'(x,y) = (3, -4), the coordinates for each reflection are, respectively:

Reflection about the x-axis

P(x,y) = (3,-4) - 2\cdot (0, -4)

P(x,y) = (3, 4)

P(x,y) = (3, 4) if P and P' are symmetric about the x-axis.

Reflection about the y-axis

P(x,y) = (3, -4) - 2\cdot (3, 0)

P(x,y) = (-3, -4)

P(x,y) = (-3, -4) if P and P' are symmetric about the y-axis.

We kindly invite to see this question on rigid transformations: brainly.com/question/18613109

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