Find the distance between the points (¼ , ½) and (5, 6) ???

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Find the distance between the points (¼ , ½) and (5, 6) ???

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Answer 1 · 2018-06-25T17:37:07

distance $d = \sqrt{{(5 - \frac{1}{4})}^{2} + {(6 - \frac{1}{2})}^{2}} = \frac{13 \sqrt{5}}{4}$ $d = \sqrt{(5-1/4)^2+(6-1/2)^2} = {13 sqrt{5}}/4$

Explanation:

When we have a Cartesian plane, there's an easy right triangle between any two points $(a, b)$ $(a,b)$ and $(c, d)$ $(c,d)$ . One leg goes parallel to the $x$ $x$ axis, length $| c - a |$ $|c-a|$ and the other parallel to the $y$ $y$ axis, length $| d - b | .$ $|d-b|.$ So the distance $d$ $d$ between the points satisfies

$d^{2} = {(c - a)}^{2} + {(b - d)}^{2}$ $d^2 = (c-a)^2 + (b-d)^2$

We apply that here and get

$d^{2} = {(5 - \frac{1}{4})}^{2} + {(6 - \frac{1}{2})}^{2} = {(\frac{19}{4})}^{2} + {(\frac{11}{2})}^{2} = \frac{845}{16}$ $d^2 = (5-1/4)^2+(6-1/2)^2 = (19/4)^2+(11/2)^2=845/16$

$d = \frac{\sqrt{845}}{4} = \frac{13 \sqrt{5}}{4}$ $d = sqrt{845}/4 = {13 sqrt{5}}/4$

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Find the distance between the points (¼ , ½) and (5, 6) ???

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