slope of AB =
slope of AC =
let the bisector of angle BAC intersects with BC at D.
hence slope of AD =
Equation of AD:
you get 8y = 7x + 4
multiply by 3 you get 24y = 21x + 12 ...(1)
Let E be the point lying on AD that is equidistant to both B and C.
Since BE = CE,
In the end you get y = -3 ...(2)
Substitute (2) into (1) you get x = -4
so the point in question is (-4, -3)