Problem:
A triangle has vertices P=(β8,5),Q=(β15,β19) and R=(1,β7). The equation of the bisector of β P can be written in the form ax+2y+c=0. Find a+c.
Solution:
Extend PR through R to T, where T is selected so that PQ=PT. Since PQ=25 and PR=15, the point T has coordinates
Now β P in β³PQR and β P in β³PQT have the same bisector. Since β³PQT is isosceles, with PQ=PT, this bisector intersects QTβ at its midpoint, (β4,β17). Thus the slope of the bisector is β211β and its equation can be written in the form 11x+2y+78=0. Hence a+c=11+78=89.
OR
Consider the vectors PQβ=β7ξ±β24ξ·β and PR=9ξ±β12ξ·β. Let v=aξ±+bξ·β be a vector parallel to the bisector of β P. Then the angle between vectors PQβ and v is equal to the angle between vectors v and PR. Let Ο be the measure of each of these angles. Then
which simplifies to 11a+2b=0. Hence b=β211βa and it follows that the bisector of β P has slope β211β. The equation of the bisector can be written as 11x+2y+78=0, so a+c=11+78=89β.
OR
Let Ξ±,Ξ²,Ξ³, respectively, be the angles that PR,PQβ and the bisector of β P make with the x-axis. (These angles are measured counterclockwise from the x-axis.) Let m be the slope of the bisector of β P. Then the slope of PR is tanΞ±=β34β, the slope of PQβ is tanΞ²=724β, and the slope of the bisector of β P is tanΞ³=m. Since Ξ±βΞ³=21ββ P=Ξ³βΞ², we have tan(Ξ±βΞ³)=tan(Ξ³βΞ²). Using the formula for the tangent of the difference of two angles gives
The last equation has solutions m=β211β and m=112β. The solution β211β is the slope of the internal bisector of β P. (Some other line has slope 112β. 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=89β.
