Best solutions to - "In the upper-plane plane model for hyperbolic geometry,..."

sjeikdom0

Answered question

2021-02-10

In the upper-plane plane model for hyperbolic geometry, calculate the distance between the points

A (0, 4) and B (3, 5) .

Give your answer accurate to three decimals. Hint: Recall the definition of distance in the upper-half plane model.

Liyana Mansell

Skilled2021-02-11Added 97 answers

Step 1
We have the two points

A (0, 4) and B (3, 5)

let

A = (0, 4) \to (x_{1}, x_{2})

∴ x_{1} = 0, x_{2} = 4

B = (3, 5) \to (y_{1}, y_{2})

∴ y_{1} = 3, y_{2} = 5

We have distance,

d i s (\begin{matrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{matrix})

= 2 \ln (\frac{\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} + \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}}{2 \sqrt{y_{1} y_{2}}})

Substitute the values

= 2 \ln (\frac{\sqrt{{(4 - 0)}^{2} + {(5 - 3)}^{2}} + \sqrt{{(4 - 0)}^{2} + {(5 - 3)}^{2}}}{2 \sqrt{3 \times 5}}

Step 2

= 2 \ln (\frac{\sqrt{4^{2} + 3^{2}} + \sqrt{4^{2} + 8^{2}}}{2 \sqrt{15}})

= 2 \ln (\frac{\sqrt{16 + 9} + \sqrt{16 + 64}}{2 \sqrt{15}})

= 2 \ln (\frac{\sqrt{25} + \sqrt{80}}{2 \sqrt{15}})

= 2 \ln (\frac{5 + \sqrt{80}}{2 \sqrt{15}})

\sqrt{80} = \sqrt{16 \times 5}

= \sqrt{16} \times \sqrt{5}

= 4 \sqrt{5}

= 2 \ln (\frac{5 + 4 \sqrt{5}}{2 \sqrt{15}})

= 2 \ln (\frac{13.94407191}{7.7459666924})

= 1.18355406

= 1.184

So the answer is

1.184

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