The equationsAx+By+C=0andxcos⁡α+ysin⁡α−p=0will represent one and the same straight line if their corresponding coefficients are proportional:cos⁡αA=sin⁡αB=−pC=k(say)So thatcos⁡α=Aksin⁡α=Bk−p=CkNow,A2k2+B2k2=1k2=d1A2+B2k=±d1A2+B2If C&gt;0, thenk=−d1A2+B2Here I don't understand the portion after "If C&gt;0, then……"

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The equationsAx+By+C=0andxcos⁡α+ysin⁡α−p=0will represent one and the same straight line if their corresponding coefficients are proportional:cos⁡αA=sin⁡αB=−pC=k(say)So thatcos⁡α=Aksin⁡α=Bk−p=CkNow,A2k2+B2k2=1k2=d1A2+B2k=±d1A2+B2If C&amp;gt;0, thenk=−d1A2+B2Here I don&#039;t understand the portion after &quot;If C&amp;gt;0, then……&quot;

e4mot1ic5bf · Accepted Answer

By convention,p≥0, since it is the distance from the origin to the line. (The normal vector to the line is (cos⁡α,sin⁡α), and (x,y)⋅(cos⁡α,sin⁡α)=xcos⁡α+ysin⁡α gives this distance.) So if k=−pC and p and C are positive, then k must be negative.

mtakadamu9i5 · Accepted Answer

In your derivation you get to the point  k=±1A2+B2  where two solutions are possible, distinguished by their sign, and you have to decide which sign is the valid one. The relation for C was  −pC=k  So if p&amp;gt;0, then C and k have opposite signs, if p&amp;lt;0 then C and k have the same sign.

The equationsAx+By+C=0andx\cos \alpha + y\sin \alpha - p=0wi

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