Given: △ABC, where A (2, 3), B (- 2, – 3), C (4, – 3).
Let AD be the bisector to ∠A meeting BC at D.
Then BD : DC = AB : AC
[∵ The bisector of vertical angle of triangle divides the base in the ratio of other two sides.]
Now D is a point which divides BC in the ratio \(\sqrt{13}:\sqrt{10}\) internally section formula (x, y) =
(By rationalising the denominator of the xcoordinate).