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Three lines \mathrm{x+2 y+3=0 ; x+2 y-7=0} and \mathrm{2 x-y-4=0} form the three sides of two squares. The equation to the fourth side of each square is

Option: 1

2 \mathrm{x}-\mathrm{y}+14=0 \quad \& \quad 2 \mathrm{x}-\mathrm{y}+6=0


Option: 2

2 \mathrm{x}-\mathrm{y}+14=0 \quad \& \quad 2 \mathrm{x}-\mathrm{y}-6=0


Option: 3

\mathrm{2 x-y-14=0\: \&\: 2 x-y-6=0}


Option: 4

\mathrm{2 x-y-14=0 \: \& \: 2 x-y+6=0}


Answers (1)

best_answer

\mathrm{d}=\left|\frac{10}{\sqrt{5}}\right|

\mathrm{3^{\text {rd }}} side is parallel to the line \mathrm{2 x-y-4=0}

Hence line is \mathrm{2 x-y+\lambda=0}

Now \mathrm{\left|\frac{\lambda+4}{\sqrt{5}}\right|=\frac{10}{\sqrt{5}}}

\mathrm{\lambda+4= \pm 10}
\mathrm{\lambda=6 \text { or } \lambda=-14\Rightarrow (B) }

 

Posted by

Deependra Verma

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