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A line, with the slope greater tha one, passes through the point \mathrm{\mathrm{A}(4,3)} and inersects the line \mathrm{x-y-2=0} at the point \mathrm{B}. If the length of the line segment \mathrm{AB} is  \mathrm{\frac{\sqrt{29}}{3}}, then \mathrm{B} also lies on the line :

Option: 1

\mathrm{2 x+y=9} \\


Option: 2

\mathrm{3 x-2 y=7} \\


Option: 3

\mathrm{x+2 y=6} \\


Option: 4

\mathrm{2 x-3 y=3}


Answers (1)

best_answer

\mathrm{\text { Let } B\left(x_{1}, x_{1}-2\right)}

\mathrm{\sqrt{\left(x_{1}-4\right)^{2}+\left(x_{1}-2-3\right)^{2}}=\frac{\sqrt{29}}{3}}

Squaring on both sides

\mathrm{18 x_{1}^{2} -162 x_{1}+340=0} \\

\mathrm{x_{1} =\frac{51}{9}\quad \text { or }\quad x_{1}=\frac{10}{3}} \\

\mathrm{y_{1} =\frac{33}{9} \quad \text { or } \quad y_{1}=\frac{4}{3}}

option 3 will satisfy \mathrm{\left ( \frac{10}{3},\frac{4}{3} \right )}

Hence correct option is 3

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vinayak

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