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The distance from the origin to the line x cos theta + y sin theta - tan theta = 0 is

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Answer 1

Answer:

xcosθ+ysinθ=a

ysinθ=a−xcosθ

y=asinθ−xcosθsinθ

Thus the slope of the line , m1=−cosθsinθ

Let us take a line, l , perpendicularto the given line and passing through origin. Let it's slope be m2.

As the product of slopes of two perpendicular lines is −1. Thus m1m2=−1

−cosθsinθm2=−1

m2=sinθcosθ

Let the the equation of line l be y=m2x+c

As it passes through origin, (0,0) must be solution.

Thus, 0=m2×0+c

c=0

So the equation of the line is y=

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Nachiket Agrawal

Updated 3 years ago

Perpendicular Distance of a line L(x,y)=Ax+By+C=0 from point (h,k) is given by

d=Ah+Bk+CA2+B2−−−−−−−√=L(h,k)A2+B2−−−−−−−√

Let l(x,y)=xcosθ+ysinθ−a

Then distance of l(x,y)=0 from (0,0) is

|l(0,0)|cos2θ+sin2θ−−−−−−−−−−−√

=|−a|cos2θ+sin2θ−−−−−−−−−−−√

=|a|

Answer 2

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(cosθP,0),B=(cosθP)

Lines through A, B parallel to axes are 

x=cosθp,y=cosθp

These meet the line through origin and .l. to given line i.e., x sin θ - y cos θ = 0 in P and Q.

∴P(cosθP,cos2θPsinθ),Q(sin2θPcosθ,sinθP)

∴PQ2 = by distance formula

=P2cos22θ.[sin4θcos2θ1+cos4θsin2θ1]

=16p2cos22θ.(2sinθcosθ)41

∴P