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Let ABC be a triangle with A(-3,1) and \angle ACB= \theta ,0< \theta < \frac{\pi }{2}. If the equation of the median through B is 2x+y-3= 0 and the equation of angle bisector of C is 7x-4y-1= 0, then \tan \theta is equal to :
Option: 1 \frac{3}{4}
Option: 2 \frac{4}{3}
Option: 3 2
Option: 4 \frac{1}{2}

Answers (1)

best_answer


Let C be (p,q)

\therefore E\: is\left ( \frac{p-3}{2},\frac{q+1}{2} \right )

It lies on 2x+y-3= 0
\Rightarrow p-3+\frac{q+1}{2}-3= 0
\Rightarrow 2p+q= 11----\left ( i \right )

Also (p,q) lies on 7x-4y-1= 0

7p-4q-1= 0------\left ( ii \right )

From (i) & (ii) \Rightarrow C (3,5)
\therefore Slope\, of\, AC= \frac{5-1}{3+3}= \frac{2}{3} &
 Slope\, of\, CF= \frac{7}{4}
Angle\, between\: AC and\, CF= \frac{\theta }{2}
\Rightarrow \tan \frac{\theta }{2}= \left |\frac{\frac{2}{3}-\frac{7}{4}}{1+\frac{2}{3},\frac{7}{4}} \right |\Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}
\Rightarrow \tan \theta =\frac{2\tan \frac{\theta }{2}}{1-\tan ^{2}\frac{\theta }{2}}=\frac{4}{3}

Posted by

Kuldeep Maurya

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