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)