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A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \text { and } C\left(x_{3}, y_{3}\right)  are the vertices of a triangle, then the excentre with respect to B is ?

Option: 1

\left(\frac{a x_{1}-b x_{2}+c x_{3}}{a-b+c}, \frac{a y_{1}-b y_{2}+c y_{3}}{a-b+c}\right)


Option: 2

\left(\frac{a x_{1}+b x_{2}-c x_{3}}{a+b-c}, \frac{a y_{1}+b y_{2}-c y_{3}}{a+b-c}\right)


Option: 3

\left(\frac{a x_{1}-b x_{2}-c x_{3}}{a-b-c}, \frac{a y_{1}-b y_{2}-c y_{3}}{a-b-c}\right)


Option: 4

None of these 


Answers (1)

best_answer

Excenters of Triangle

Coordinates of I1 , I2 and I3 is given by

\\\mathrm{I_1\equiv \left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)}\\\\\mathrm{I_2\equiv\left(\frac{a x_{1}-b x_{2}+c x_{3}}{a-b+c}, \frac{a y_{1}-b y_{2}+c y_{3}}{a-b+c}\right)}\\\\\mathrm{I_3\equiv\left(\frac{a x_{1}+b x_{2}-c x_{3}}{a+b-c}, \frac{a y_{1}+b y_{2}-c y_{3}}{a+b-c}\right)}

-

 

From the above concept 

Correct answer is option 1

Posted by

jitender.kumar

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