New Notes Blog: `(1,2) , (7,10)` Find the distance between the two points using integration.

Saturday, 19 October 2013

$\displaystyle{\left({1},{2}\right)},{\left({7},{10}\right)}$ Find the distance between the two points using integration.

Given the equation of a line $\displaystyle{y}={m}{x}+{b},$

=> slope = $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={m}$ . Thus, the distance is:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ where $\displaystyle{a}\le{x}\le{b}$

we know the two points $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({1},{2}\right)}$

$\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({7},{10}\right)}$

$\displaystyle{m}=\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}=\frac{{{10}-{2}}}{{{7}-{1}}}=\frac{{8}}{{6}}=\frac{{4}}{{3}}$

so now the length is $\displaystyle{L}={\int_{{1}}^{{7}}}\sqrt{{{1}+{{\left(\frac{{4}}{{3}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{1}}^{{7}}}\sqrt{{{1}+{\left(\frac{{16}}{{9}}\right)}}}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{1}}^{{7}}}\sqrt{{\frac{{25}}{{9}}}}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{1}}^{{7}}}{\left(\frac{{5}}{{3}}\right)}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{5}}{{3}}\right)}{\int_{{1}}^{{7}}}{1}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{5}}{{3}}\right)}{{\mid}_{{1}}^{{7}}}{x}$

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