MathematicsMedium128×since 2002Q5541Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q(k1,k2)\left(k_{1}, k_{2}\right)(k1,k2), then k1+k2k_{1}+k_{2}k1+k2 is equal to :A2B47\frac{4}{7}74C27\frac{2}{7}72D4Check answerSkip