The distance of a point (2, 5) from the line | vedclass

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(D) Let the given line be $L: 3x + y + 4 = 0$. The slope of $L$ is $m = -3$.Let the slopes of lines $L_1$ and $L_2$ be $m_1 = \frac{3}{4}$ and $m_2$ r

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(D) Let the given line be $L: 3x + y + 4 = 0$. The slope of $L$ is $m = -3$.Let the slopes of lines $L_1$ and $L_2$ be $m_1 = \frac{3}{4}$ and $m_2$ respectively.Since the distances measured along $L_1$ and $L_2$ from point $A(2, 5)$ to the line $L$ are equal,the angles $	heta_1$ and $	heta_2$ that $L_1$ and $L_2$ make with the line $L$ must satisfy $	an 	heta_1 = 	an 	heta_2$ or $	an 	heta_1 = -	an 	heta_2$.Using the formula for the angle between two lines with slopes $m$ and $m'$,$	an 	heta = \left| \frac{m - m'}{1 + mm'} ight|$.For $L_1$: $	an 	heta = \left| \frac{-3 - 3/4}{1 + (-3)(3/4)} ight| = \left| \frac{-15/4}{1 - 9/4} ight| = \left| \frac{-15/4}{-5/4} ight| = 3$.For $L_2$: $\left| \frac{-3 - m_2}{1 - 3m_2} ight| = 3$.Case $1$: $\frac{-3 - m_2}{1 - 3m_2} = 3$ $\Rightarrow -3 - m_2 = 3 - 9m_2$ $\Rightarrow 8m_2 = 6$ $\Rightarrow m_2 = \frac{3}{4}$ (This is $L_1$ itself).Case $2$: $\frac{-3 - m_2}{1 - 3m_2} = -3$ $\Rightarrow -3 - m_2 = -3 + 9m_2$ $\Rightarrow 10m_2 = 0$ $\Rightarrow m_2 = 0$.Thus,the slope of $L_2$ is $0$.

The distance of a point $(2, 5)$ from the line $3x + y + 4 = 0$ measured along the lines $L_1$ and $L_2$ are the same. If the slope of line $L_1$ is $\frac{3}{4}$,then the slope of the line $L_2$ is

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