The slope of a line perpendicular to xcos60°+ytan30°=1 is

xcos60°+ytan30°=1x12+y13=1⇒2x+3y=1⇒2x+3y−1=0⋯(1)Slope of (1) is m1=−23Slope of a line perpendicular to (1) is m2=32

The slope of a line perpendicular to xcos60°+ytan30°=1 is

xcos60°+ytan30°=1x12+y13=1⇒2x+3y=1⇒2x+3y−1=0⋯(1)Slope of (1) is m1=−23Slope of a line perpendicular to (1) is m2=32

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The slope of a line perpendicular to $\frac{x}{\cos 60 °} + \frac{y}{\tan 30 °} = 1$ is

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$\frac{\sqrt{3}}{2}$

$\frac{2}{\sqrt{3}}$

$- 2$

$\sqrt{5}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{x}{\cos 60 °} + \frac{y}{\tan 30 °} = 1 \\ \frac{x}{\frac{1}{2}} + \frac{y}{\frac{1}{\sqrt{3}}} = 1 \Rightarrow 2 x + \sqrt{3} y = 1 \\ \Rightarrow 2 x + \sqrt{3} y - 1 = 0 \dots (1) \\ S l o p e o f (1) i s m_{1} = \frac{- 2}{\sqrt{3}} \\ S l o p e o f a l i n e p e r p e n d i c u l a r t o (1) i s m_{2} = \frac{\sqrt{3}}{2} \end{array}$