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  • Three lines <img alt="\mathrm{x}+2 \mathrm{y}+3=0 ; \mathrm{x}+2 \mathrm{y}-7=0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%7D&plus;2%20%5Cmathrm%7By%7D&plus;3%3D0%2

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

Option: 1

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


Option: 2

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


Option: 3

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

\mathrm{\text{now}\begin{aligned} &\left|\frac{\lambda+4}{\sqrt{5}}\right|=\frac{10}{\sqrt{5}} \\ & \lambda+4= \pm 10 \\ & \lambda=6 \text { or } \lambda=-14 \Rightarrow \text { (B) }\end{aligned}}

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vinayak

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