find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle tan^-1 (5/12) with the positive direction of x axis

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Solution

Here use the equation in normal form
$x \cos α + y \sin α = p H e r e p i s t h e p e r p e n d i c u l a r d i s \tan c e f r o m t h e o r i g i n \tan^{- 1} (\frac{5}{12}) = x S o \tan α = \frac{5}{12}, s o \sin α = \frac{5}{13}, \cos α = \frac{12}{13} h e n c e t h e e q u a t i o n i s x \times \frac{12}{13} + y \times \frac{5}{13} = 3 O r 12 x + 5 y - 39 = 0$

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find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle tan^-1 (5/12) with the positive direction of x axis

Here use the equation in normal form xcosα + ysinα = pHere p is the perpendicular distance from the origintan-1(512) =xSo tanα =512, so sinα =513, cosα =1213hence the equation is x×1213 + y×513 = 3Or 12x + 5y -39 = 0