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  • The equations of the diagonals of the parallelogram formed by the lines <img alt="\mathrm{L^{2}-aL=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL%5E%7B2%7D-aL%3D0%7D" /

The equations of the diagonals of the parallelogram formed by the lines \mathrm{L^{2}-aL=0} and \mathrm{L^{2}-aL^{\prime}=0}, where \mathrm{L \equiv x \cos \theta+y \sin \theta-p=0, L^{\prime} \equiv x \cos \theta^{\prime}+y \sin \theta^{\prime}-p^{\prime}=0 \text { are } x(\cos \theta+\cos \theta ^{\prime})}\mathrm{+y\left(\sin \theta+\sin \theta^{\prime}\right)-c=0 \text { and } x\left(\cos \theta-\cos \theta^{\prime}\right)+y\left(\sin \theta-\sin \theta^{\prime}\right)-k=0 . \text { Then } c+k=}

Option: 1

\mathrm{2p-a}


Option: 2

\mathrm{2p^{\prime}+a}


Option: 3

\mathrm{2p+a}


Option: 4

\mathrm{2p^{\prime}-a}


Answers (1)

best_answer

Equation of line AC is \mathrm{L}+\lambda \mathrm{L}^{\prime}=0 ........(1)

It also passes through L = a and L′ = a

\mathrm{\therefore a+\lambda a=0 \Rightarrow \lambda=-1}

Hence equation of diagonal AC is L – L′ = 0

\mathrm{\text { i.e. } x\left(\cos \theta-\cos \theta^{\prime}\right)+y\left(\sin \theta-\sin \theta^{\prime}\right)-p+p^{\prime}=0}

Equation of line BD through point of intersection of L = 0 and L′ = a is                

\mathrm{L+\lambda\left(L^{\prime}-a\right)=0}

It passes through D

\mathrm{\therefore a+\lambda(0-a)=0 \Rightarrow \lambda=1}

\therefore  Equation of diagonal BD is

\mathrm{x\left(\cos \theta+\cos \theta^{\prime}\right)+y\left(\sin \theta+\sin \theta^{\prime}\right)-p-p^{\prime}-a=0}

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Rishabh

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