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  • Let be the vertices of a triangle and <img alt="\mathrm{ ' B ', ' C ' }" src="htt

Let \mathrm{A, B, C} be the vertices of a triangle and \mathrm{ ' B ', ' C ' }lies on the line \mathrm{ 3 x-5 y=1 }. It is given that \mathrm{ \angle A=45^{\circ}}. The slope of the tangent at \mathrm{ ' \mathrm{B} ' }to the circumcircle of the \mathrm{ \triangle \mathrm{ABC} }, can be
 

Option: 1

\frac{1}{4}


Option: 2

\frac{1}{-4}


Option: 3

-4


Option: 4

\text{None of these}


Answers (1)

best_answer

Let the slope of the tangent at \mathrm{B} be \mathrm{m} and it makes \mathrm{45^{\circ} \: \: with \: \: 3 \mathrm{x}-5 \mathrm{y}=1}

\mathrm{ \therefore \tan 45^{\circ}=\left|\frac{m-\frac{3}{5}}{1+\frac{3 m}{5}}\right| }

\mathrm{ \Rightarrow \mathrm{m}=4, \quad-\frac{1}{4} . }

Hence option 2 is correct.



 

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manish

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