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(C) The given equation of the line is $\frac{x-x_1}{\cos 	heta}=\frac{y-y_1}{\sin 	heta}=\gamma$. This represents a line passing through $(x_1, y_1)

Vedclass · Accepted Answer

$\cot 	heta$

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(C) The given equation of the line is $\frac{x-x_1}{\cos 	heta}=\frac{y-y_1}{\sin 	heta}=\gamma$. This represents a line passing through $(x_1, y_1)$ with an inclination $	heta$. The slope of this line is $m_1 = 	an 	heta$. Let the slope of the line perpendicular to this line be $m_2$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. Therefore,$m_2 = -\frac{1}{	an 	heta} = -\cot 	heta$. The equation of the required line is given as $\frac{x}{a}+\frac{y}{b}=1$,which can be rewritten as $y = -\frac{b}{a}x + b$. The slope of this line is $m_2 = -\frac{b}{a}$. Equating the slopes,we get $-\frac{b}{a} = -\cot 	heta$. Thus,$\frac{b}{a} = \cot 	heta$.

The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$,then $\frac{b}{a}$ is equal to

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