Notes: `3x - 5y = 3, 3x + 5y = 12` Find the angle theta (in radians and degrees) between the lines

Saturday, June 27, 2009

$\displaystyle{3}{x}-{5}{y}={3},{3}{x}+{5}{y}={12}$ Find the angle theta (in radians and degrees) between the lines

Given:

$\displaystyle{3}{x}-{5}{y}={3}$

$\displaystyle{3}{x}+{5}{y}={12}$

The formula to find the angle between two lines is

$\displaystyle{\tan{{\left(\theta\right)}}}={\left|\frac{{{m}_{{2}}-{m}_{{1}}}}{{{1}+{m}_{{1}}_{2}}}\right|}$

Find the slope of line 1.

$\displaystyle{3}{x}-{5}{y}={3}$

$\displaystyle-{5}{y}=-{3}{x}+{3}$

$\displaystyle{y}=\frac{{3}}{{5}}{x}-\frac{{3}}{{5}}$

The slope of line 1 is $\displaystyle{m}_{{1}}=\frac{{3}}{{5}}$ .

Find the slope of line 2.

$\displaystyle{3}{x}+{5}{y}={12}$

$\displaystyle{5}{y}=-{3}{x}+{12}$

$\displaystyle{y}=-\frac{{3}}{{5}}{x}+\frac{{12}}{{5}}$

The slope of line 2 is $\displaystyle{m}_{{2}}=-\frac{{3}}{{5}}.$

Plug in the slopes into the formula

$\displaystyle{\tan{{\left(\theta\right)}}}={\left|{\left(\frac{{{m}_{{2}}-{m}_{{1}}}}{{{1}+{m}_{{1}}{m}_{{2}}}}\right)}\right|}$

$\displaystyle{\tan{{\left(\theta\right)}}}={\left|\frac{{{\left(-\frac{{3}}{{5}}\right)}-{\left(\frac{{3}}{{5}}\right)}}}{{{1}+{\left(\frac{{3}}{{5}}\right)}{\left(-\frac{{3}}{{5}}\right)}}}\right|}=\frac{{15}}{{8}}$

$\displaystyle\theta={\arctan{{\left(\frac{{15}}{{8}}\right)}}}={{61.9}}^{\circ}={1.0808}$ radians.

The angle between the two lines is 61.9 degrees or 1.0808 radians.

Notes

Saturday, June 27, 2009

$\displaystyle{3}{x}-{5}{y}={3},{3}{x}+{5}{y}={12}$ Find the angle theta (in radians and degrees) between the lines

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