Now ∠P in △PQR and ∠P in △PQT have the same bisector. Since △PQT is isosceles, with PQ=PT, this bisector intersects QT at its midpoint, (−4,−17). Thus the slope of the bisector is −211 and its equation can be written in the form 11x+2y+78=0. Hence a+c=11+78=89.
OR
Consider the vectors PQ=−7−24 and PR=9−12. Let v=a+b be a vector parallel to the bisector of ∠P. Then the angle between vectors PQ and v is equal to the angle between vectors v and PR. Let ϕ be the measure of each of these angles. Then
∥v∥∥PQ∥v⋅PQ=cosϕ=∥v∥∥PR∥v⋅PR
giving
25−7a−24b=159a−12b
which simplifies to 11a+2b=0. Hence b=−211a and it follows that the bisector of ∠P has slope −211. The equation of the bisector can be written as 11x+2y+78=0, so a+c=11+78=89.
OR
Let α,β,γ, respectively, be the angles that PR,PQ and the bisector of ∠P make with the x-axis. (These angles are measured counterclockwise from the x-axis.) Let m be the slope of the bisector of ∠P. Then the slope of PR is tanα=−34, the slope of PQ is tanβ=724, and the slope of the bisector of ∠P is tanγ=m. Since α−γ=21∠P=γ−β, we have tan(α−γ)=tan(γ−β). Using the formula for the tangent of the difference of two angles gives
1+tanαtanγtanα−tanγ=1+tanγtanβtanγ−tanβ
which leads to
1−34m−34−m=1+724mm−724
The last equation has solutions m=−211 and m=112. The solution −211 is the slope of the internal bisector of ∠P. (Some other line has slope 112. Which one?) We then find that the equation of the bisector can be written in the form 11x+2y+78=0, and a+c=11+78=89.
