# In the upper-plane plane model for hyperbolic geometry, calculate the distance between the points A(0, 4) text{and} B(3, 5). Give your answer accurate to three decimals. Hint: Recall the definition of distance in the upper-half plane model.

In the upper-plane plane model for hyperbolic geometry, calculate the distance between the points Give your answer accurate to three decimals. Hint: Recall the definition of distance in the upper-half plane model.
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Step 1
We have the two points let
$A=\left(0,4\right)\to \left({x}_{1},{x}_{2}\right)$
$\therefore {x}_{1}=0,{x}_{2}=4$
$B=\left(3,5\right)\to \left({y}_{1},{y}_{2}\right)$
$\therefore {y}_{1}=3,{y}_{2}=5$
We have distance,
$dis\left(\begin{array}{cc}{x}_{1}& {y}_{1}\\ {x}_{2}& {y}_{2}\end{array}\right)$
$=2\mathrm{ln}\left(\frac{\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}+\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}}{2\sqrt{{y}_{1}{y}_{2}}}\right)$
Substitute the values
$=2\mathrm{ln}\left(\frac{\sqrt{{\left(4-0\right)}^{2}+{\left(5-3\right)}^{2}}+\sqrt{{\left(4-0\right)}^{2}+{\left(5-3\right)}^{2}}}{2\sqrt{3×5}}$
Step 2
$=2\mathrm{ln}\left(\frac{\sqrt{{4}^{2}+{3}^{2}}+\sqrt{{4}^{2}+{8}^{2}}}{2\sqrt{15}}\right)$
$=2\mathrm{ln}\left(\frac{\sqrt{16+9}+\sqrt{16+64}}{2\sqrt{15}}\right)$
$=2\mathrm{ln}\left(\frac{\sqrt{25}+\sqrt{80}}{2\sqrt{15}}\right)$
$=2\mathrm{ln}\left(\frac{5+\sqrt{80}}{2\sqrt{15}}\right)$
$\sqrt{80}=\sqrt{16×5}$
$=\sqrt{16}×\sqrt{5}$
$=4\sqrt{5}$
$=2\mathrm{ln}\left(\frac{5+4\sqrt{5}}{2\sqrt{15}}\right)$
$=2\mathrm{ln}\left(\frac{13.94407191}{7.7459666924}\right)$
$=1.18355406$
$=1.184$
So the answer is $1.184$