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(C) The equation of the line is $12x - 5y + 13 = 0$,which can be rewritten as $12x - 5y = -13$. Multiplying by $-1$,we get $-12x + 5y = 13$. Dividing

Vedclass · Accepted Answer

$\pi - \operatorname{Tan}^{-1} \frac{5}{12}$

Vedclass · Answer

(C) The equation of the line is $12x - 5y + 13 = 0$,which can be rewritten as $12x - 5y = -13$. Multiplying by $-1$,we get $-12x + 5y = 13$. Dividing by $\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = 13$,we get the normal form of the line: $-\frac{12}{13}x + \frac{5}{13}y = 1$. The normal form is $x \cos \alpha + y \sin \alpha = p$,where $\cos \alpha = -\frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$. Since $\cos \alpha  0$,the angle $\alpha$ lies in the second quadrant. We know $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{-12/13} = -\frac{5}{12}$. Thus,$\alpha = \pi - \operatorname{Tan}^{-1} \frac{5}{12}$.

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $12x - 5y + 13 = 0$ with the positive $X$-axis in the anti-clockwise direction,then $\alpha =$

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