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If one of the diagonals of the square is along the line x=2y and one of its

vertices

is (3,0), then its sides through this vertex are given by the equations.

Option: 1

\mathrm{y-3 x+9=0,3 y+x-3=0}


Option: 2

\mathrm{y+3 x+9=0,3 y+x-3=0}


Option: 3

\mathrm{y-3 x+9=0,3 y-x+3=0}


Option: 4

\mathrm{y-3 x+3=0,3 y+x+9=0}


Answers (1)

best_answer

a

Diagonal of the square is along 

      \mathrm{x=2y}                                    (1)

The point (3,0) does not lie on (1).

Let the side through this vertex be  \mathrm{\mathrm{y}-0=m(x-3)}

Angle between side (2) and diagonal (1) is \mathrm{45\degree}.

\mathrm{\therefore \quad \frac{m-1 / 2}{\tan ^{-1}} \frac{\Rightarrow m \cdot(1 / 2)}{1+m}= \pm 45^{\circ} \quad \Rightarrow m,-1 / 3}

\mathrm{\therefore } from (2) , the required sides are 

\mathrm{y-3 x+9=0 \text { and } 3 y+x-3=0}, which are given in (a). 

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Pankaj

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