sjeikdom0

2021-02-10

In the upper-plane plane model for hyperbolic geometry, calculate the distance between the points $A(0,\text{}4)\text{}\text{and}\text{}B(3,\text{}5).$ 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$A(0,\text{}4)\text{}\text{and}\text{}B(3,\text{}5)$ let

$A=(0,4)\to ({x}_{1},{x}_{2})$

$\therefore {x}_{1}=0,{x}_{2}=4$

$B=(3,5)\to ({y}_{1},{y}_{2})$

$\therefore {y}_{1}=3,{y}_{2}=5$

We have distance,

$dis\left(\begin{array}{cc}{x}_{1}& {y}_{1}\\ {x}_{2}& {y}_{2}\end{array}\right)$

$=2\mathrm{ln}\left(\frac{\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}+\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}}{2\sqrt{{y}_{1}{y}_{2}}}\right)$

Substitute the values

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

Step 2

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

$=2\mathrm{ln}\left(\frac{\sqrt{16+9}+\sqrt{16+64}}{2\sqrt{15}}\right)$

$=2\mathrm{ln}\left(\frac{\sqrt{25}+\sqrt{80}}{2\sqrt{15}}\right)$

$=2\mathrm{ln}\left(\frac{5+\sqrt{80}}{2\sqrt{15}}\right)$

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

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

$=4\sqrt{5}$

$=2\mathrm{ln}\left(\frac{5+4\sqrt{5}}{2\sqrt{15}}\right)$

$=2\mathrm{ln}\left(\frac{13.94407191}{7.7459666924}\right)$

$=1.18355406$

$=1.184$

So the answer is$1.184$

We have the two points

We have distance,

Substitute the values

Step 2

So the answer is

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