Problem:
Triangle ABC lies in the Cartesian plane and has area 70. The coordinates of B and C are (12,19) and (23,20), respectively, and the coordinates of A are (p,q). The line containing the median to side BC has slope −5 . Find the largest possible value of p+q.
Solution:
Let l be the line containing the median to side BC. Then l must contain the midpoint of BC, which is ((12+23)/2,(19+20)/2)=(35/2,39/2). Since l has the form y=−5x+b, substitute to find that b=107. Thus the coordinates of A are (p,−5p+107). Now compute p using the fact that the area of the triangle with coordinates (0,0),(x1,y1), and (x2,y2) is the absolute value of
(1/2)∣∣∣∣∣x1x2y1y2∣∣∣∣∣.
To use this formula, translate the point (12,19) to the origin, and, to preserve area, translate points A and C to A′=(p−12,−5p+107−19)=(p−12,−5p+88) and C′=(23−12,20−19)=(11,1), respectively. Apply the above formula to obtain (1/2)∣(p−12)⋅1−(−5p+88)⋅11∣=70, which yields ∣56p−980∣=140. Thus p=15 or p=20, and the corresponding values of q are 32 and 7, respectively. The largest possible value of p+q is 47 .
OR
Let M be the midpoint of BC. The coordinates of M are (35/2,39/2). An equation of line AM is y=−5x+107, so the coordinates of A can be represented as (p,−5p+107). Line BC has equation x=11y−197 or, equivalently, x−11y+197=0, so the distance from A to line BC is 12+112∣p−11(−5p+107)+197∣=122∣56p−980∣. The length of BC is 12+112, so 70=[△ABC]=21122. 122∣56p−980∣. Solve to obtain p=20 or p=15, so p+q=p−5p+107=−4p+107. Thus p+q=27 or 47 , and the maximum value of p+q is 47.