question_answer

The vertices of a \[\Delta ABC\] are\[A(2,1)\], \[B\,(6,-2),\]\[C\,(8,\,9)\]. If AD is angle bisector, where D meets on BC, then coordinates of D are

A)  \[\left( \frac{20}{3},\frac{5}{3} \right)\]       

B)  \[(5,2)\]          

C)         \[(4,3)\]          

D)         \[\left( \frac{14}{3},\frac{7}{3} \right)\]                   

Correct Answer: A

Solution :

AD is the angle bisector of\[\angle BAC\]. So, by the angle bisector theorem in \[\Delta ABC,\]we have   \[\frac{AB}{AC}=\frac{BD}{DC}\]             ?..(i) Now, \[AB=5\]and \[AC=10\] \[\therefore \]   \[\frac{1}{2}=\frac{BD}{DC}\]    [using (i)] Thus, D divides BC in the ratio\[1:2\]. \[\therefore \]  \[D=\left( \frac{2\times 6+1\times 8}{2+1},\,\frac{-2\times 2+9\times 1}{2+1} \right)=\left( \frac{20}{3},\frac{5}{3} \right)\]