A line L perpendicular to the line 5x - 12y + | vedclass

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(C) The given line is $5x - 12y + 6 = 0$. The slope of this line is $m = \frac{5}{12}$.Since line $L$ is perpendicular to this line,its slope $m_L$ mu

Vedclass · Accepted Answer

$\frac{169}{60}$

Vedclass · Answer

(C) The given line is $5x - 12y + 6 = 0$. The slope of this line is $m = \frac{5}{12}$.Since line $L$ is perpendicular to this line,its slope $m_L$ must satisfy $m_L 	imes \frac{5}{12} = -1$,so $m_L = -\frac{12}{5}$.The equation of line $L$ in normal form is $x \cos 	heta + y \sin 	heta = p$,where $p = 2$ is the distance from the origin.The slope of this line is $-\cot 	heta = -\frac{12}{5}$,which implies $\cot 	heta = \frac{12}{5}$.Thus,$	an 	heta = \frac{5}{12}$.Since the line makes a positive intercept on the $Y$-axis,we check the intercept form: $y = -\frac{12}{5}x + \frac{2}{\sin 	heta}$.Using $\cot 	heta = \frac{12}{5}$,we have $\sin 	heta = \frac{5}{13}$ and $\cos 	heta = \frac{12}{13}$.The intercept is $\frac{2}{\sin 	heta} = \frac{2}{5/13} = \frac{26}{5} > 0$,which is positive.Finally,$	an 	heta + \cot 	heta = \frac{5}{12} + \frac{12}{5} = \frac{25 + 144}{60} = \frac{169}{60}$.

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