The three bisector lines of the inner angles of a triangle (the bisectrices) intersect in a unique point, the incentre. Every point of the bisectrix of the angle with vertex at P is equidistant from the sides PQ and PR. Also every point of the bisectrix of Q is equidistant from the sides QR and QP. Hence its intersection I is simultaneously equidistant from the three sides, that is, I is unique and it is the centre of the circle inscribed into the triangle. In order to calculate the equation of a bisectrix, we take the sum of both unitary vectors of adjacent sides:
 
The incentre I is the intersection of the bisectrix passing through P, whose directing vector is u, and that passing through Q, with directing vector v:
I = P + k u = Q + m v k, m real
Arranging terms we find PQ as a linear combination of u and v:
k u - m v = Q - P = PQ
The coefficient k is:
Since all outer products are equal because they are twice the triangle area, this expression is simplified:
The centre of the inscribed circle is:
By taking common denominator and simplifying, we arrive at:
For example, let us calculate the centre of the circle inscribed in the triangle with vertices:
P = ( 0, 0 ) Q = ( 0, 3 ) R = ( 4, 0 )
| PQ | = 3 | QR | = 5 | RP | = 4